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this text. GUIDELINE When writing the interpretation of a percentile in the context of the given data, make sure the sentence contains the following information: • Information about the context of the situation being considered • The data value (value of the variable) that represents the percentile • The percentage of individuals or items with data values below the percentile • The percentage of individuals or items with data values above the percentile 100 Chapter 2 | Descriptive Statistics Example 2.20 On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation. Solution 2.20 • Twenty-five percent of students finished the exam in 35 minutes or less. • Seventy-five percent of students finished the exam in 35 minutes or more. • A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. If you take too long, you might not be able to finish. 2.20 For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation. Example 2.21 On a 20-question math test, the 70th percentile for number of correct answers was 16. Interpret the 70th percentile in the context of this situation. Solution 2.21 • Seventy percent of students answered 16 or fewer questions correctly. • Thirty percent of students answered 16 or more questions correctly. • A higher percentile could be considered good, as answering more questions correctly is desirable. 2.21 On a 60-point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation. Example 2.22 At a high school, it was found that the 30th percentile of number of hours that students spend studying per week is seven hours. Interpret the 30th percentile in the context of this situation. Solution 2.22 • Thirty percent of students study seven or fewer hours per week. • Seventy percent of students study seven or more hours per week. • In this example, there is not necessarily a good or bad value judgment associated with a higher or lower percentile, since the time a student studies per week is dependent on his/her needs. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 101 2.22 During a season, the 40 |
th percentile for points scored per player in a game is eight. Interpret the 40th percentile in the context of this situation. Example 2.23 A middle school is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown: 0 minutes, 40 minutes, 60 minutes, 30 minutes, 60 minutes, 10 minutes, 45 minutes, 30 minutes, 300 minutes, 90 minutes, 30 minutes, 120 minutes, 60 minutes, 0 minutes, 20 minutes Find the five values that make up the five number summary. Min = 0 Q1 = 20 Med = 40 Q3 = 60 Max = 300 Listing the data in ascending order gives the following: Figure 2.12 The minimum value is 0. The maximum value is 300. Since there are an odd number of data values, the median is the middle value of this data set as it is arranged in ascending order, or 40. The first quartile is the median of the lower half of the scores and does not include the median. The lower half has seven data values; the median of the lower half will equal the middle value of the lower half, or 20. The third quartile is the median of the upper half of the scores and does not include the median. The upper half also has seven data values; so the median of the upper half will equal the middle value of the upper half, or 60. If you were the principal, would you be justified in purchasing new fitness equipment? Since 75 percent of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment. However, the principal needs to be careful. The value 300 appears to be a potential outlier. Q3 + 1.5(IQR) = 60 + (1.5)(40) = 120. The value 300 is greater than 120, so it is a potential outlier. If we delete it and calculate the five values, we get the following values: Min = 0 Q1 = 20 102 Chapter 2 | Descriptive Statistics Q3 = 60 Max = 120 We still have 75 percent of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and |
60 minutes a day. However, 15 students is a small sample, and the principal should survey more students to be sure of his survey results. 2.4 | Box Plots Box plots, also called box-and-whisker plots or box-whisker plots, give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. As mentioned previously, a box plot is constructed from five values: the minimum value, the first quartile, the median, the third quartile, and the maximum value. We use these values to compare how close other data values are to them. To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box, and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The whiskers extend from the ends of the box to the smallest and largest data values. A box plot easily shows the range of a data set, which is the difference between the largest and smallest data values (or the difference between the maximum and minimum). Unless the median, first quartile, and third quartile are the same value, the median will lie inside the box or between the first and third quartiles. The box plot gives a good, quick picture of the data. NOTE You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values. Consider, again, this data set: 1, 1, 2, 2, 4, 6, 6.8, 7.2, 8, 8.3, 9, 10, 10, 11.5 The first quartile is two, the median is seven, and the third quartile is nine. The smallest value is one, and the largest value is 11.5. The following image shows the constructed box plot. NOTE See the calculator instructions on the TI website (https://education.ti.com/en/professional-development/ webinars-and-tutorials/technology-tutorials) or in the appendix. Figure 2.13 The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line. NOTE It is important |
to start a box plot with a scaled number line. Otherwise, the box plot may not be useful. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 103 Example 2.24 The following data are the heights of 40 students in a statistics class: 59, 60, 61, 62, 62, 63, 63, 64, 64, 64, 65, 65, 65, 65, 65, 65, 65, 65, 65, 66, 66, 67, 67, 68, 68, 69, 70, 70, 70, 70, 70, 71, 71, 72, 72, 73, 74, 74, 75, 77. Construct a box plot with the following properties. Calculator instructions for finding the five number summary follow this example: • Minimum value = 59 • Maximum value = 77 • Q1: First quartile = 64.5 • Q2: Second quartile or median = 66 • Q3: Third quartile = 70 Figure 2.14 a. Each quarter has approximately 25 percent of the data. b. The spreads of the four quarters are 64.5 – 59 = 5.5 (first quarter), 66 – 64.5 = 1.5 (second quarter), 70 – 66 = 4 (third quarter), and 77 – 70 = 7 (fourth quarter). So, the second quarter has the smallest spread, and the fourth quarter has the largest spread. c. Range = maximum value – minimum value = 77 – 59 = 18. d. Interquartile Range: IQR = Q3 – Q1 = 70 – 64.5 = 5.5. e. The interval 59–65 has more than 25 percent of the data, so it has more data in it than the interval 66–70, which has 25 percent of the data. f. The middle 50 percent (middle half) of the data has a range of 5.5 inches. To find the minimum, maximum, and quartiles: Enter data into the list editor (Pres STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, and then arrow down. Put the data values into the list L1. Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1. Press ENTER. Use the down and up arrow keys to scroll. Smallest value = 59. Largest value = 77. |
Q1: First quartile = 64.5. Q2: Second quartile or median = 66. Q3: Third quartile = 70. 104 Chapter 2 | Descriptive Statistics To construct the box plot: Press 4:Plotsoff. Press ENTER. Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER. Arrow down to Xlist: Press 2nd 1 for L1. Arrow down to Freq: Press ALPHA. Press 1. Press Zoom. Press 9: ZoomStat. Press TRACE and use the arrow keys to examine the box plot. 2.24 The following data are the number of pages in 40 books on a shelf. Construct a box plot using a graphing calculator and state the interquartile range. 136, 140, 178, 190, 205, 215, 217, 218, 232, 234, 240, 255, 270, 275, 290, 301, 303, 315, 317, 318, 326, 333, 343, 349, 360, 369, 377, 388, 391, 392, 398, 400, 402, 405, 408, 422, 429, 450, 475, 512 For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like the following: Figure 2.15 In this case, at least 25 percent of the values are equal to one. Twenty-five percent of the values are between one and five, inclusive. At least 25 percent of the values are equal to five. The top 25 percent of the values fall between five and seven, inclusive. Example 2.25 Test scores for Mr. Ramirez's class held during the day are as follows: 99, 56, 78, 55.5, 32, 90, 80, 81, 56, 59, 45, 77, 84.5, 84, 70, 72, 68, 32, 79, 90. Test scores for Ms. Park's |
class held during the evening are as follows: 98, 78, 68, 83, 81, 89, 88, 76, 65, 45, 98, 90, 80, 84.5, 85, 79, 78, 98, 90, 79, 81, 25.5. a. Find the smallest and largest values, the median, and the first and third quartile for Mr. Ramirez's class. b. Find the smallest and largest values, the median, and the first and third quartile for Ms. Park's class. c. For each data set, what percentage of the data is between the smallest value and the first quartile? the first quartile and the median? the median and the third quartile? the third quartile and the largest value? What percentage of the data is between the first quartile and the largest value? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 105 d. Create a box plot for each set of data. Use one number line for both box plots. e. Which box plot has the widest spread for the middle 50 percent of the data,the data between the first and third quartiles? What does this mean for that set of data in comparison to the other set of data? Solution 2.25 a. Min = 32 Q1 = 56 M = 74.5 Q3 = 82.5 Max = 99 b. Min = 25.5 Q1 = 78 M = 81 Q3 = 89 Max = 98 c. Mr. Ramirez's class: There are six data values ranging from 32 to 56: 30 percent. There are six data values ranging from 56 to 74.5: 30 percent. There are five data values ranging from 74.5 to 82.5: 25 percent. There are five data values ranging from 82.5 to 99: 25 percent. There are 16 data values between the first quartile, 56, and the largest value, 99: 75 percent. Ms. Park’s class: There are six data values ranging from 25.5 to 78: 27 percent. There are five data values ranging from 78 to the first instance of 81: 23 percent. There are six data values ranging from the second instance of 81 to 89: 27 percent. There are five data values ranging from 90 to 98: 23 percent. There are 17 values between the first quartile, 78, and the largest value, 98: 77 |
percent. d. Figure 2.16 e. The first data set has the wider spread for the middle 50 percent of the data. The IQR for the first data set is greater than the IQR for the second set. This means that there is more variability in the middle 50 percent of the first data set. 2.25 The following data set shows the heights in inches for the boys in a class of 40 students: 66, 66, 67, 67, 68, 68, 68, 68, 68, 69, 69, 69, 70, 71, 72, 72, 72, 73, 73, 74. The following data set shows the heights in inches for the girls in a class of 40 students: 61 61, 62, 62, 63, 63, 63, 65, 65, 65, 66, 66, 66, 67, 68, 68, 68, 69, 69, 69. Construct a box plot using a graphing calculator for each data set, and state which box plot has the wider spread for the middle 50 percent of the data. 106 Chapter 2 | Descriptive Statistics Example 2.26 Graph a box-and-whisker plot for the following data values shown: 10, 10, 10, 15, 35, 75, 90, 95, 100, 175, 420, 490, 515, 515, 790 The five numbers used to create a box-and-whisker plot are as follows: Min: 10 Q1: 15 Med: 95 Q3: 490 Max: 790 The following graph shows the box-and-whisker plot. Figure 2.17 2.26 Follow the steps you used to graph a box-and-whisker plot for the data values shown: 0, 5, 5, 15, 30, 30, 45, 50, 50, 60, 75, 110, 140, 240, 330 2.5 | Measures of the Center of the Data The center of a data set is also a way of describing location. The two most widely used measures of the center of the data are the mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. |
The mean is the most common measure of the center. NOTE The words mean and average are often used interchangeably. The substitution of one word for the other is common practice. The technical term is arithmetic mean and average is technically a center location. However, in practice among non statisticians, average is commonly accepted for arithmetic mean. When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an x with a bar over it (pronounced “x bar”): x ¯. The sample mean is a statistic. The Greek letter μ (pronounced "mew") represents the population mean. The population mean is a parameter. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random. To see that both ways of calculating the mean are the same, consider the following sample: 1, 1, 1, 2, 2, 3, 4, 4, 4, 4, 4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 107 ¯ 11 3(1) + 2(2) + 1(3) + 5(4) 11 = 2.7. x¯ = = 2.7 In the second example, the frequencies are 3(1) + 2(2) + 1(3) + 5(4). You can quickly find the location of the median by using the expression n + 1 2. The letter n is the total number of data values in the sample. As discussed earlier, if n is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the two middle values added together and divided by two after the data have been ordered. For example, if the total number of data values is 97, then n + 1 = 49. The median is the 49th value in the ordered data. If the total number of data values is 100, = 97 + 1 2 = 50.5. The median occurs midway between the 50th and 51st values. The location of the median and 2 then n + 1 2 = 100 + 1 2 the value of the median are not the same. The uppercase letter M is often |
used to represent the median. The next example illustrates the location of the median and the value of the median. Example 2.27 Data indicating the number of months a patient with a specific disease lives after taking a new antibody drug are as follows (smallest to largest): 3, 4, 8, 8, 10, 11, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 21, 22, 22, 24, 24, 25, 26, 26, 27, 27, 29, 29, 31, 32, 33, 33, 34, 34, 35, 37, 40, 44, 44, 47 Calculate the mean and the median. Solution 2.27 The calculation for the mean is x¯ = [3 + 4 + (8)(2) + 10 + 11 + 12 + 13 + 14 + (15)(2) + (16)(2) + (17)(2) + 18 + 21 + (22)(2) + (24)(2) + 25 + (26)(2) + (27)(2) + (29)(2) + 31 + 32 + (33)(2) + (34)(2) + 35 + 37 + 40 + (44)(2) + 47] / 40 = 23.6. To find the median, M, first use the formula for the location. The location is n + 1 2 = 40 + 1 2 = 20.5. Start from the smallest value and count up; the median is located between the 20th and 21st values (the two 24s): 3, 4, 8, 8, 10, 11, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 21, 22, 22, 24, 24, 25, 26, 26, 27, 27, 29, 29, 31, 32, 33, 33, 34, 34, 35, 37, 40, 44, 44, 47 M = 24 + 24 2 = 24 To find the mean and the median: Clear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER. Enter data into the list editor. Press STAT 1:EDIT. Put the data values into list L1. Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER. Press the down and up arrow keys to scroll. x¯ = 23.6, |
M = 24 108 Chapter 2 | Descriptive Statistics 2.27 The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median. 3, 4, 5, 7, 7, 7, 7, 8, 8, 9, 9, 10, 10, 10, 10, 10, 11, 12, 12, 13, 14, 14, 15, 15, 17, 17, 18, 19, 19, 19, 21, 21, 22, 22, 23, 24, 24, 24, 24 Example 2.28 Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn $30,000. Which is the better measure of the center: the mean or the median? Solution 2.28 x¯ = 5, 000, 000 + 49(30, 000) 50 = 129,400 M = 30,000 There are 49 people who earn $30,000 and one person who earns $5,000,000. The median is a better measure of the center than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data. 2.28 In a sample of 60 households, one house is worth $2,500,000. Half of the rest are worth $280,000, and all the others are worth $315,000. Which is the better measure of the center: the mean or the median? Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal. Example 2.29 Statistics exam scores for 20 students are as follows: 50, 53, 59, 59, 63, 63, 72, 72, 72, 72, 72, 76, 78, 81, 83, 84, 84, 84, 90, 93 Find the mode. Solution 2.29 The most frequent score is 72, which occurs five times. Mode = 72. 2.29 The number of books checked out from the library by 25 students are as follows: 0, 0, 0, 1, 2, 3, 3, 4 |
, 4, 5, 5, 7, 7, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12 Find the mode. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 109 Example 2.30 Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice. When is the mode the best measure of the center? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing. NOTE The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is red, red, red, green, green, yellow, purple, black, blue, the mode is red. Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software. 2.30 Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is $50,000, and the mean is $47,500. What would be the best measure of the center? The Law of Large Numbers and the Mean ¯ The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean x of the sample is very likely to get closer and closer to µ. This law is discussed in more detail later in the text. Sampling Distributions and Statistic of a Sampling Distribution You can think of a sampling distribution as a relative frequency distribution with a great many samples. See Chapter 1: Sampling and Data for a review of relative frequency. Suppose 30 randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below. Number of Movies Relative Frequency 0 1 2 3 4 Table 2.26 5 30 15 30 6 30 3 30 1 30 A relative frequency distribution includes the relative frequencies of a number of samples. Recall that a statistic |
is a number calculated from a sample. Statistic examples include the mean, the median, and the mode 110 Chapter 2 | Descriptive Statistics ¯ as well as others. The sample mean x is an example of a statistic that estimates the population mean μ. Calculating the Mean of Grouped Frequency Tables When only grouped data is available, you do not know the individual data values (we know only intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table, we can apply the basic. We simply need to modify the definition to fit within the restrictions definition of mean: mean = data sum number o f data values of a frequency table. Since we do not know the individual data values, we can instead find the midpoint of each interval. The midpoint is lower boundary + upper boundary 2. We can now modify the mean definition to be Mean o f Frequency Table = ∑ f m ∑ f, where f = the frequency of the interval, m = the midpoint of the interval, and sigma (∑) is read as "sigma" and means to sum up. So this formula says that we will sum the products of each midpoint and the corresponding frequency and divide by the sum of all of the frequencies. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 111 Example 2.31 A frequency table displaying Professor Blount’s last statistic test is shown. Find the best estimate of the class mean. Grade Interval Number of Students 50–56.5 56.5–62.5 62.5–68.5 68.5–74.5 74.5–80.5 80.5–86.5 86.5–92.5 92.5–98.5 Table 2.27 1 0 4 4 2 3 4 1 Solution 2.31 • Find the midpoints for all intervals. Grade Interval Midpoint 50–56.5 56.5–62.5 62.5–68.5 68.5–74.5 74.5–80.5 80.5–86.5 86.5–92.5 92.5–98.5 Table |
2.28 53.25 59.5 65.5 71.5 77.5 83.5 89.5 95.5 • Calculate the sum of the product of each interval frequency and midpoint. ∑ f m 53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.25 • μ = ∑ f m ∑ f = 1460.25 19 = 76.86 112 Chapter 2 | Descriptive Statistics 2.31 Maris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data: Hours Teenagers Spend on Video Games Number of Teenagers 0–3.5 3.5–7.5 7.5–11.5 11.5–15.5 15.5–19.5 Table 2.29 3 7 12 7 9 What is the best estimate for the mean number of hours spent playing video games? 2.6 | Skewness and the Mean, Median, and Mode Consider the following data set: 4, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 9, 10 This data set can be represented by the following histogram. Each interval has width 1, and each value is located in the middle of an interval. Figure 2.18 The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median. The histogram for the data: 4, 5, 6, 6, 6, 7, 7, 7, 7, 8 is not symmetrical. The right-hand side seems chopped off compared to the left-hand side. A distribution of this type is called skewed to |
the left because it is pulled out to the left. A skewed left distribution has more high values. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 113 Figure 2.19 The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so. The mean is pulled toward the tail in a skewed distribution. The histogram for the data: 6, 7, 7, 7, 7, 8, 8, 8, 9, 10 is also not symmetrical. It is skewed to the right. A skewed right distribution has more low values. Figure 2.20 The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, the mean is the largest, while the mode is the smallest. Again, the mean reflects the skewing the most. To summarize, generally if the distribution of data is skewed to the left, the mean is less than the median, which is often less than the mode. If the distribution of data is skewed to the right, the mode is often less than the median, which is less than the mean. Skewness and symmetry become important when we discuss probability distributions in later chapters. 114 Chapter 2 | Descriptive Statistics Example 2.32 Statistics are used to compare and sometimes identify authors. The following lists show a simple random sample that compares the letter counts for three authors. Terry: 7, 9, 3, 3, 3, 4, 1, 3, 2, 2 Davis: 3, 3, 3, 4, 1, 4, 3, 2, 3, 1 Maris: 2, 3, 4, 4, 4, 6, 6, 6, 8, 3 a. Make a dot plot for the three authors and compare the shapes. b. Calculate the mean for each. c. Calculate the median for each. d. Describe any pattern you notice between the shape and the measures of center. Solution 2.32 a. Figure 2.21 Terry’s distribution has a right (positive) skew. Figure 2.22 Davis’s distribution has a left (negative) skew. Figure 2.23 Maris’s distribution is symmetrically shaped |
. b. Terry’s mean is 3.7, Davis’s mean is 2.7, and Maris’s mean is 4.6. c. Terry’s median is 3, Davis’s median is 3, and Maris’s median is four. It would be helpful to manually calculate these descriptive statistics, using the given data sets and then compare to the graphs. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 115 d. It appears that the median is always closest to the high point (the mode), while the mean tends to be farther out on the tail. In a symmetrical distribution, the mean and the median are both centrally located close to the high point of the distribution. 2.32 Discuss the mean, median, and mode for each of the following problems. Is there a pattern between the shape and measure of the center? a. Figure 2.24 b. c. The Ages at Which Former U.S. Presidents Died Key: 8|0 means 80. Table 2.30 116 Chapter 2 | Descriptive Statistics Figure 2.25 2.7 | Measures of the Spread of the Data An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean. The standard deviation • provides a numerical measure of the overall amount of variation in a data set and • can be used to determine whether a particular data value is close to or far from the mean. The standard deviation provides a measure of the overall variation in a data set. The standard deviation is always positive or zero. The standard deviation is small when all the data are concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation. Suppose that we are studying the amount of time customers wait in line at the checkout at Supermarket A and Supermarket B. The average wait time at both supermarkets is five minutes. At Supermarket A, the standard deviation for the wait time is two minutes; at Supermarket B, the standard deviation for the wait time is |
four minutes. Because Supermarket B has a higher standard deviation, we know that there is more variation in the wait times at Supermarket B. Overall, wait times at Supermarket B are more spread out from the average whereas wait times at Supermarket A are more concentrated near the average. The standard deviation can be used to determine whether a data value is close to or far from the mean. Suppose that both Rosa and Binh shop at Supermarket A. Rosa waits at the checkout counter for seven minutes, and Binh waits for one minute. At Supermarket A, the mean waiting time is five minutes, and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean. A z-score is a standardized score that lets us compare data sets. It tells us how many standard deviations a data value is from the mean and is calculated as the ratio of the difference in a particular score and the population mean to the population standard deviation. We can use the given information to create the table below. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 117 Supermarket Population Standard Deviation, σ Individual Score, x Population Mean, μ Supermarket A Supermarket B Table 2.31 2 minutes 4 minutes 7, 1 5 5 Since Rosa and Binh only shop at Supermarket A, we can ignore the row for Supermarket B. We need the values from the first row to determine the number of standard deviations above or below the mean each individual wait time is; we can do so by calculating two different z-scores. Rosa waited for seven minutes, so the z-score representing this deviation from the population mean may be calculated as The z-score of one tells us that Rosa’s wait time is one standard deviation above the mean wait time of five minutes. Binh waited for one minute, so the z-score representing this deviation from the population mean may be calculated as. The z-score of −2 tells us that Binh’s wait time is two standard deviations below the mean wait time of five minutes. A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if they are more than two standard deviations away is more of an approximate rule of thumb than a rigid rule |
. In general, the shape of the distribution of the data affects how much of the data is farther away than two standard deviations. You will learn more about this in later chapters. The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because 5 + (1)(2) = 7. If one were also part of the data set, then one is two standard deviations to the left of five because 5 + (–2)(2) = 1. Figure 2.26 • In general, a value = mean + (#ofSTDEV)(standard deviation) • where #ofSTDEVs = the number of standard deviations • #ofSTDEV does not need to be an integer • One is two standard deviations less than the mean of five because 1 = 5 + (–2)(2). The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population as follows: • Sample: x = x¯ + ( # o f STDEV)(s) • Population: x = μ + ( # o f STDEV)(σ). The lowercase letter s represents the sample standard deviation and the Greek letter σ (lower case) represents the population standard deviation. ¯ The symbol x is the sample mean, and the Greek symbol μ is the population mean. Calculating the Standard Deviation If x is a number, then the difference x – mean is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols, a deviation is x – μ. For sample data, in symbols, a deviation is x – x¯. The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data 118 Chapter 2 | Descriptive Statistics from a sample. The calculations are similar but not identical. Therefore, the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lowercase letter s represents the sample standard deviation and the Greek letter σ (lowercase sigma) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ. To calculate the standard deviation, we need to calculate the variance |
first. The variance is the average of the squares of the deviations (the x – x¯ values for a sample or the x – μ values for a population). The symbol σ2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s2 represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations. If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample. Formulas for the Sample Standard Deviation • s = 2 Σ(x − x¯ ) n − 1 or s = 2 Σ f (x − x¯ ) n − 1 • For the sample standard deviation, the denominator is n−; that is, the sample size minus 1. Formulas for the Population Standard Deviation • σ = Σ(x − μ)2 N or σ = Σ f (x – μ)2 N • For the population standard deviation, the denominator is N, the number of items in the population. In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is one. If a value appears three times in the data set or population, f is three. Types of Variability in Samples When researchers study a population, they often use a sample, either for convenience or because it is not possible to access the entire population. Variability is the term used to describe the differences that may occur in these outcomes. Common types of variability include the following: • Observational or measurement variability • Natural variability • Induced variability • Sample variability Here are some examples to describe each type of variability: Example 1: Measurement variability Measurement variability occurs when there are differences in the instruments used to measure or in the people using those instruments. If we are gathering data on how long it takes for a ball to drop from a height by having students measure the time of the drop with a stopwatch, we may experience measurement variability if the two stopwatches used were made by different manufacturers. For example, one stopwatch |
measures to the nearest second, whereas the other one measures to the nearest tenth of a second. We also may experience measurement variability because two different people are gathering the data. Their reaction times in pressing the button on the stopwatch may differ; thus, the outcomes will vary accordingly. The differences in outcomes may be affected by measurement variability. Example 2: Natural variability Natural variability arises from the differences that naturally occur because members of a population differ from each other. For example, if we have two identical corn plants and we expose both plants to the same amount of water and sunlight, they may still grow at different rates simply because they are two different corn plants. The difference in outcomes may be explained by natural variability. Example 3: Induced variability Induced variability is the counterpart to natural variability. This occurs because we have artificially induced an element of variation that, by definition, was not present naturally. For example, we assign people to two different groups to study This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 119 memory, and we induce a variable in one group by limiting the amount of sleep they get. The difference in outcomes may be affected by induced variability. Example 4: Sample variability Sample variability occurs when multiple random samples are taken from the same population. For example, if I conduct four surveys of 50 people randomly selected from a given population, the differences in outcomes may be affected by sample variability. Sampling Variability of a Statistic The statistic of a sampling distribution was discussed in Descriptive Statistics: Measures of the Center of the Data. How much the statistic varies from one sample to another is known as the sampling variability of a statistic. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. The standard error is the standard deviation of the sampling distribution. In other words, it is the average standard deviation that results from repeated sampling. You will cover the standard error of the mean in the chapter The Central Limit Theorem (not now). The notation for the standard error of the mean is σ, where σ is the standard n deviation of the population and n is the size of the sample. NOTE In practice, use a calculator or computer software to calculate the standard deviation. If you are using a TI-83, 83+, or 84+ calculator, you need to select the appropriate standard deviation σx or sx from the |
summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However, you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. The calculator instructions appear at the end of this example. 120 Chapter 2 | Descriptive Statistics Example 2.33 In a fifth-grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of n = 20 fifth-grade students. The ages are rounded to the nearest half year. 9, 9.5, 9.5, 10, 10, 10, 10, 10.5, 10.5, 10.5, 10.5, 11, 11, 11, 11, 11, 11, 11.5, 11.5, 11.5 x¯ = 9 + 9.5(2) + 10(4) + 10.5(4) + 11(6) + 11.5(3) 20 = 10.525 The average age is 10.53 years, rounded to two places. The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s. Data Frequency Deviations Deviations2 (Frequency)(Deviations2) x 9 9.5 10 10.5 11 11.5 f 1 2 4 4 6 3 Table 2.32 (x – x¯ ) 9 – 10.525 = –1.525 (x – x¯ )2 (–1.525)2 = 2.325625 1 × 2.325625 = 2.325625 (f)(x – x¯ )2 9.5 – 10.525 = –1.025 (–1.025)2 = 1.050625 2 × 1.050625 = 2.101250 10 – 10.525 = –.525 (–.525)2 =.275625 4 ×.275625 = 1.1025 10.5 – 10.525 = –.025 (–.025)2 =.000625 4 ×.000625 =.0025 11 – 10.525 =.475 (.475)2 =.225625 6 ×.225625 = 1.35375 11.5 – 10.525 =.975 (.975)2 =. |
950625 3 ×.950625 = 2.851875 The total is 9.7375. The last column simply multiplies each squared deviation by the frequency for the corresponding data value. The sample variance, s2, is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s2 = 9.7375 20 − 1 =.5125 The sample standard deviation s is equal to the square root of the sample variance: s =.5125 =.715891, which is rounded to two decimal places, s =.72. Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy. • For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation). Verify the mean and standard deviation on a calculator or computer. Note that these formulas are derived by algebraically manipulating the z-score formulas, given either parameters or statistics. • For a sample: x = x¯ + (#ofSTDEVs)(s) • For a population: x = μ + (#ofSTDEVs)(σ) • For this example, use x = x¯ + (#ofSTDEVs)(s) because the data is from a sample a. Verify the mean and standard deviation on your calculator or computer. b. Find the value that is one standard deviation above the mean. Find ( x¯ + 1s). c. Find the value that is two standard deviations below the mean. Find ( x¯ – 2s). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 121 d. Find the values that are 1.5 standard deviations from (below and above) the mean. Solution 2.33 a. ◦ Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2. ◦ Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down. ◦ Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, |
6, 3) into list L2. Use the arrow keys to move around. ◦ Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER. ◦ x¯ = 10.525. ◦ Use Sx because this is sample data (not a population): Sx=.715891. b. c. d. ( x¯ + 1s) = 10.53 + (1)(.72) = 11.25 ( x¯ – 2s) = 10.53 – (2)(.72) = 9.09 ◦ ◦ ( x¯ – 1.5s) = 10.53 – (1.5)(.72) = 9.45 ( x¯ + 1.5s) = 10.53 + (1.5)(.72) = 11.61 2.33 On a baseball team, the ages of each of the players are as follows: 21, 21, 22, 23, 24, 24, 25, 25, 28, 29, 29, 31, 32, 33, 33, 34, 35, 36, 36, 36, 36, 38, 38, 38, 40 Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean. Explanation of the standard deviation calculation shown in the table The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11, which is indicated by the deviations.97 and.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. We can sum the always products of zero. to show that the deviations 1(−1.525) + 2(−1.025) + 4(−.525) + 4(−.025) + 6(.475) + 3(.975) = 0 For Example 2.33, there are n = 20 deviations. So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is |
the average squared deviation. and deviations frequencies sum of the the is The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data. Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n – 1). Why not divide by n? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (n – 1) gives a better estimate of the population variance. 122 NOTE Chapter 2 | Descriptive Statistics Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number that measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic. The standard deviation, s or σ, is either zero or larger than zero. Describing the data with reference to the spread is called variability. The variability in data depends on the method by which the outcomes are obtained, for example, by measuring or by random sampling. When the standard deviation is zero, there is no spread; that is, all the data values are equal to each other. The standard deviation is small when all the data are concentrated close to the mean and larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large. The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better feel for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful, but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot. Example 2.34 Use the following data (first exam scores) from Susan Dean's spring precalculus class: 33, 42, 49, 49, |
53, 55, 55, 61, 63, 67, 68, 68, 69, 69, 72, 73, 74, 78, 80, 83, 88, 88, 88, 90, 92, 94, 94, 94, 94, 96, 100 a. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. b. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator: i. The sample mean ii. The sample standard deviation iii. The median iv. The first quartile v. The third quartile vi. IQR c. Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart. Solution 2.34 a. See Table 2.33. b. Entering the data values into a list in your graphing calculator and then selecting Stat, Calc, and 1-Var Stats will produce the one-variable statistics you need. c. The x-axis goes from 32.5 to 100.5; the y-axis goes from –2.4 to 15 for the histogram. The number of intervals is 5, so the width of an interval is (100.5 – 32.5) divided by 5, equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, 32.5 + 13.6 = 46.1, 46.1 + 13.6 = 59.7, 59.7 + 13.6 = 73.3, 73.3 + 13.6 = 86.9, 86.9 + 13.6 = 100.5 = the ending value; no data values fall on an interval boundary. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 123 Figure 2.27 The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50 percent is greater (73 – 33 = 40) than the spread in the upper 50 percent (100 – 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50 percent of |
the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25 percent of the exam scores are Ds and Fs. Data Frequency Relative Frequency Cumulative Relative Frequency 33 42 49 53 55 61 63 67 68 69 72 73 74 78 80 83 88 90 Table 2.33.032.032.065.032.065.032.032.032.065.065.032.032.032.032.032.032.097.032.032.064.129.161.226.258.290.322.387.452.484.516.548.580.612.644.741.773 124 Chapter 2 | Descriptive Statistics Data Frequency Relative Frequency Cumulative Relative Frequency 92 94 96 100 1 4 1 1 Table 2.33.032.129.032.032.805.934.966.998 (Why isn't this value 1?) 2.34 The following data show the different types of pet food that stores in the area carry: 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12 Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator. Standard deviation of Grouped Frequency Tables Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula Mean o f Frequency Table = ∑ f m ∑ f, where f = interval frequencies and m = interval midpoints. Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how unusual individual data are when compared to the mean. Example 2.35 Find the standard deviation for the data in Table 2.34. Class Frequency, f Midpoint, m m2 0–2 3–5 6–8 9–11 12–14 15– |
17 1 6 10 7 0 2 Table 2.34 1 4 7 10 13 16 1 16 49 100 169 256 ¯ 2 x 7.58 7.58 7.58 7.58 7.58 7.58 fm2 Standard Deviation 1 96 490 700 0 512 3.5 3.5 3.5 3.5 3.5 3.5 For this data set, we have the mean, x¯ = 7.58, and the standard deviation, sx = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 125 =.58. While the formula for calculating the standard deviation is not complicated, s x = 2 f (m − x¯ ) n − 1, where sx = sample standard deviation, x¯ = sample mean; the calculations are tedious. It is usually best to use technology when performing the calculations. 2.35 Find the standard deviation for the data from the previous example: Class Frequency, f 0–2 3–5 6–8 9–11 12–14 15–17 1 6 10 7 0 2 Table 2.35 First, press the STAT key and select 1:Edit. Figure 2.28 Input the midpoint values into L1 and the frequencies into L2. 126 Chapter 2 | Descriptive Statistics Figure 2.29 Select STAT, CALC, and 1: 1-Var Stats. Figure 2.30 Select 2nd, then 1, then, 2nd, then 2 Enter. Figure 2.31 You will see displayed both a population standard deviation, σx, and the sample standard deviation, sx. Comparing Values from Different Data Sets As explained before, a z-score allows us to compare statistics from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading. • For each data value, calculate how many standard deviations away from its mean the value is. • In symbols, the formulas for calculating z-scores become the following. This OpenStax book is available for free at http://cnx.org/ |
content/col30309/1.8 Chapter 2 | Descriptive Statistics 127 z = x − x¯ s z = x − μ σ Sample Population Table 2.36 As shown in the table, when only a sample mean and sample standard deviation are given, the top formula is used. When the population mean and population standard deviation are given, the bottom formula is used. Example 2.36 Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school? Student GPA School Mean GPA School Standard Deviation John Ali 2.85 77 3.0 80 Table 2.37.7 10 Solution 2.36 For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer. z = # of STDEVs = value – mean standard deviation = x + μ σ For John, z = # o f STDEVs = 2.85 – 3.0.7 = − 0.21 For Ali, z = # o f STDEVs = 77 − 80 10 = − 0.3 John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean, while Ali's GPA is.3 standard deviations below his school's mean. John's z-score of –.21 is higher than Ali's z-score of –.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. The z-score representing John's score does not fall as far below the mean as the z-score representing Ali's score. 128 Chapter 2 | Descriptive Statistics 2.36 Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50-meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team? Swimmer Time (seconds) Team Mean Time Team Standard Deviation Angie Beth 26.2 27.3 Table 2.38 27.2 30.1.8 1.4 The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data. For any data set, no matter what the distribution of the data is, the following are true: • |
At least 75 percent of the data is within two standard deviations of the mean. • At least 89 percent of the data is within three standard deviations of the mean. • At least 95 percent of the data is within 4.5 standard deviations of the mean. • This is known as Chebyshev's Rule. A bell-shaped distribution is one that is normal and symmetric, meaning the curve can be folded along a line of symmetry drawn through the median, and the left and right sides of the curve would fold on each other symmetrically.. With a bellshaped distribution, the mean, median, and mode are all located at the same place. For data having a distribution that is bell-shaped and symmetric, the following are true: • Approximately 68 percent of the data is within one standard deviation of the mean. • Approximately 95 percent of the data is within two standard deviations of the mean. • More than 99 percent of the data is within three standard deviations of the mean. • This is known as the Empirical Rule. • It is important to note that this rule applies only when the shape of the distribution of the data is bell-shaped and symmetric; we will learn more about this when studying the Normal or Gaussian probability distribution in later chapters. 2.8 | Descriptive Statistics This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 129 2.1 Descriptive Statistics Student Learning Outcomes • The student will construct a histogram and a box plot. • The student will calculate univariate statistics. • The student will examine the graphs to interpret what the data imply. Collect the Data Record the number of pairs of shoes you own. 1. Randomly survey 30 classmates about the number of pairs of shoes they own. Record their values. _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ Table 2.39 Survey Results 2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil and scale the axes. Figure 2.32 3. Calculate the following values: a. b. x¯ = _____ s = _____ 4. Are the data discrete or continuous? How do you |
know? 5. In complete sentences, describe the shape of the histogram. 6. Are there any potential outliers? List the value(s) that could be outliers. Use a formula to check the end values to determine if they are potential outliers. 130 Chapter 2 | Descriptive Statistics Analyze the Data 1. Determine the following values: a. Min = _____ b. M = _____ c. Max = _____ d. Q1 = _____ e. Q3 = _____ f. IQR = _____ 2. Construct a box plot of data. 3. What does the shape of the box plot imply about the concentration of data? Use complete sentences. 4. Using the box plot, how can you determine if there are potential outliers? 5. How does the standard deviation help you to determine concentration of the data and whether there are potential outliers? 6. What does the IQR represent in this problem? 7. Show your work to find the value that is 1.5 standard deviations a. above the mean. b. below the mean. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 131 KEY TERMS box plot a graph that gives a quick picture of the middle 50 percent of the data first quartile the value that is the median of the lower half of the ordered data set frequency the number of times a value of the data occurs frequency polygon a data display that looks like a line graph but uses intervals to display ranges of large amounts of data frequency table a data representation in which grouped data are displayed along with the corresponding frequencies histogram a graphical representation in x-y form of the distribution of data in a data set; x represents the data and y represents the frequency, or relative frequency; the graph consists of contiguous rectangles interquartile range or IQR, is the range of the middle 50 percent of the data values; the IQR is found by subtracting the first quartile from the third quartile interval also called a class interval; an interval represents a range of data and is used when displaying large data sets mean a number that measures the central tendency of the data; a common name for mean is average. The term mean is a shortened form of arithmetic mean. By definition, the mean for a sample (denoted by x x¯ = the mean population (denoted Sum of all values in the sample Number of values in |
the sample and for by μ) a, ¯ ) is is μ = Sum of all values in the population Number of values in the population median a number that separates ordered data into halves; half the values are the same number or smaller than the median, and half the values are the same number or larger than the median The median may or may not be part of the data. midpoint the mean of an interval in a frequency table mode the value that appears most frequently in a set of data outlier an observation that does not fit the rest of the data paired data set two data sets that have a one-to-one relationship so that • both data sets are the same size, and • each data point in one data set is matched with exactly one point from the other set percentile a number that divides ordered data into hundredths; percentiles may or may not be part of the data. The median of the data is the second quartile and the 50th percentile The first and third quartiles are the 25th and the 75th percentiles, respectively. quartiles the numbers that separate the data into quarters; quartiles may or may not be part of the data; the second quartile is the median of the data relative frequency of all outcomes the ratio of the number of times a value of the data occurs in the set of all outcomes to the number skewed used to describe data that is not symmetrical; when the right side of a graph looks chopped off compared to the left side, we say it is skewed to the left. When the left side of the graph looks chopped off compared to the right side, we say the data are skewed to the right. Alternatively, when the lower values of the data are more spread out, we say the data are skewed to the left. When the greater values are more spread out, the data are skewed to the right. standard deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation 132 Chapter 2 | Descriptive Statistics variance mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as x – x¯ where x is a value of the data and x¯ variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and 1 is the sample mean; the sample CHAPTER REVIEW 2.1 Stem-and- |
Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends, that is, finding a general pattern in data sets, including temperature, sales, employment, company profit, or cost, over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Bar graphs are especially useful when categorical data are being used. 2.2 Histograms, Frequency Polygons, and Time Series Graphs A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values, and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually go on the y-axis with the frequency being graphed on the x-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time. 2.3 Measures of the Location of the Data The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50th percentile would be greater than 50 percent of the other observations in the set. Quartiles divide data into quarters. The first quartile (Q1) is the 25th percentile, the second quartile (Q2 or median) is the 50th percentile, and the third quartile (Q3) is the 75th percentile. The interquartile range, or IQR, is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q1 from Q3 and can help determine outliers by using the following two expressions. • Q3 + IQ |
R(1.5) • Q1 – IQR(1.5) 2.4 Box Plots Box plots are a type of graph that can help visually organize data. Before a box plot can be graphed, the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data. 2.5 Measures of the Center of the Data The mean and the median can be calculated to help you find the center of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges that lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set. 2.6 Skewness and the Mean, Median, and Mode Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A right (or positive) skewed distribution has a shape like Figure 2.19. A left (or negative) skewed distribution has a shape like Figure 2.20. A symmetrical distribution looks like Figure 2.18. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 133 2.7 Measures of the Spread of the Data The standard deviation can help you calculate the spread of data. There are different equations to use if you are calculating the standard deviation of a sample or of a population. • The standard deviation allows us to compare individual data or classes to the data set mean numerically. • s = ∑ (x − x¯ ) n − 1 2 or s = ∑ f (x − x¯ ) n − 1 2 is the formula for calculating the standard deviation of a sample. To calculate the standard |
deviation of a population, we would use the population mean, μ, and the formula σ = ∑ (x − μ)2 N or σ = ∑ f (x − μ)2 N. FORMULA REVIEW 2.3 Measures of the Location of the Data 2.5 Measures of the Center of the Data i = ⎛ ⎝ k 100 ⎞ ⎠(n + 1) where i = the ranking or position of a data value, k = the kth percentile, n = total number of data. Expression for finding the percentile of a data value ⎛ ⎝ x + 0.5y n (100) ⎞ ⎠ where x = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile, y = the number of data values equal to the data value for which you want to find the percentile, n = total number of data. PRACTICE where f = interval frequencies and m = μ = ∑ f m ∑ f interval midpoints. 2.7 Measures of the Spread of the Data s x = ∑ f m2 n − x¯ 2 s x = sample standard deviation x¯ = sample mean ⎛ ⎝x - x¯ ) s z = where and z = (x - μ) θ. 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs For each of the following data sets, create a stemplot and identify any outliers. 1. The miles-per-gallon ratings for 30 cars are shown below (lowest to highest): 19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43. 2. The height in feet of 25 trees is shown below (lowest to highest): 25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54. 3. The data are the prices of different laptops at an electronics store. Round each value to the nearest 10. 249, 249, |
260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610 4. The following data are daily high temperatures in a town for one month: 61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95. For the next three exercises, use the data to construct a line graph. 134 Chapter 2 | Descriptive Statistics 5. In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in Table 2.40. Number of Times in Store Frequency 1 2 3 4 5 Table 2.40 4 10 16 6 4 6. In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table 2.41. Years Since Last Purchase Frequency 0 1 2 3 4 5 Table 2.41 2 8 13 22 16 9 7. Several children were asked how many TV shows they watch each day. The results of the survey are shown in Table 2.42. Number of TV Shows Frequency 0 1 2 3 4 Table 2.42 12 18 36 7 2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 135 8. The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. Table 2.43 shows the four seasons, the number of students who have birthdays in each season, and the percentage of students in each group. Construct a bar graph showing the number of students. Seasons Number of Students Proportion of Population Spring Summer Autumn Winter 8 9 11 6 Table 2.43 24% 26% 32% 18% 9. Using the data from Mrs. Ramirez’s math class supplied in Exercise 2.8, construct a bar graph showing the percentages. 10. David County has six high schools. Each school sent students to participate in a county-wide science competition. Table 2.44 shows the percentage breakdown of competitors from each school and the percentage of the entire student population of the county that goes to each school. Construct a bar |
graph that shows the population percentage of competitors from each school. High School Science Competition Population Overall Student Population Alabaster 28.9% Concordia 7.6% Genoa 12.1% Mocksville 18.5% Tynneson 24.2% West End 8.7% Table 2.44 8.6% 23.2% 15.0% 14.3% 10.1% 28.8% 11. Use the data from the David County science competition supplied in Exercise 2.10. Construct a bar graph that shows the county-wide population percentage of students at each school. 2.2 Histograms, Frequency Polygons, and Time Series Graphs 12. 65 randomly selected car salespersons were asked the number of cars they generally sell in one week. 14 people answered that they generally sell three cars, 19 generally sell four cars, 12 generally sell five cars, nine generally sell six cars, and 11 generally sell seven cars. Complete the table. Data Value (Number of Cars) Frequency Relative Frequency Cumulative Relative Frequency Table 2.45 13. What does the frequency column in Table 2.45 sum to? Why? 136 Chapter 2 | Descriptive Statistics 14. What does the relative frequency column in Table 2.45 sum to? Why? 15. What is the difference between relative frequency and frequency for each data value in Table 2.45? 16. What is the difference between cumulative relative frequency and relative frequency for each data value? 17. To construct the histogram for the data in Table 2.45, determine appropriate minimum and maximum x- and y-values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling. Figure 2.33 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 137 18. Construct a frequency polygon for the following. a. b. c. Pulse Rates for Women Frequency 60–69 70–79 80–89 90–99 100–109 110–119 120–129 Table 2.46 12 14 11 1 1 0 1 Actual Speed in a 30-MPH Zone Frequency 42–45 46–49 50–53 54–57 58–61 Table 2.47 25 14 7 3 1 Tar (mg) in Nonfiltered Cigarettes Frequency 10–13 14–17 18–21 22–25 26–29 Table 2.48 1 0 15 7 2 |
138 Chapter 2 | Descriptive Statistics 19. Construct a frequency polygon from the frequency distribution for the 50 highest-ranked countries for depth of hunger. Depth of Hunger Frequency 230–259 260–289 290–319 320–349 350–379 380–409 410–439 Table 2.49 21 13 5 7 1 1 1 20. Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlaid frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men? Life Expectancy at Birth – Women Frequency 49–55 56–62 63–69 70–76 77–83 84–90 3 3 1 3 8 2 Table 2.50 Life Expectancy at Birth – Men Frequency 49–55 56–62 63–69 70–76 77–83 84–90 Table 2.51 3 3 1 1 7 5 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 139 21. Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births. Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 47,804 52,239 53,158 53,694 54,628 54,409 54,606 93,349 101,821 103,415 104,018 106,543 105,629 107,009 Male Total Table 2.52 Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354 Male Total Table 2.53 Sex/Year 1871 1870 1872 1871 1872 1827 1874 1875 Female 56,099 56,431 57,472 56,099 57,472 58,233 60,109 60,146 60,029 58,959 61, |
293 60,029 61,293 61,467 63,602 63,432 116,128 115,390 118,765 116,128 118,765 119,700 123,711 123,578 Male Total Table 2.54 22. The following data sets list full-time police per 100,000 citizens along with incidents of a certain crime per 100,000 citizens for the city of Detroit, Michigan, during the period from 1961 to 1973. Year Police 1961 1962 1963 1964 1965 1966 1967 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Incidents 8.6 8.9 8.52 8.89 13.07 14.57 21.36 Table 2.55 Year Police 1968 1969 1970 1971 1972 1973 295.99 319.87 341.43 356.59 376.69 390.19 Incidents 28.03 31.49 37.39 46.26 47.24 52.33 Table 2.56 a. Construct a double time series graph using a common x-axis for both sets of data. b. Which variable increased the fastest? Explain. c. Did Detroit’s increase in police officers have an impact on the incident rate? Explain. 140 Chapter 2 | Descriptive Statistics 2.3 Measures of the Location of the Data 23. Listed are 29 ages for Academy Award-winning best actors in order from smallest to largest: 18, 21, 22, 25, 26, 27, 29, 30, 31, 33, 36, 37, 41, 42, 47, 52, 55, 57, 58, 62, 64, 67, 69, 71, 72, 73, 74, 76, 77 a. Find the 40th percentile. b. Find the 78th percentile. 24. Listed are 32 ages for Academy Award-winning best actors in order from smallest to largest: 18, 18, 21, 22, 25, 26, 27, 29, 30, 31, 31, 33, 36, 37, 37, 41, 42, 47, 52, 55, 57, 58, 62, 64, 67, 69, 71, 72, 73, 74, 76, 77 a. Find the percentile of 37. b. Find the percentile of 72. 25. Jesse was ranked 37th in his graduating class of 180 students. At what percentile is Jesse’s ranking? 26. 27. a. For runners in a race, a low time means a faster run. The winners in a race |
have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race? b. The 20th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20th percentile in the context of the situation. c. A bicyclist in the 90th percentile of a bicycle race completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90th percentile in the context of the situation. a. For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race? b. The 40th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40th percentile in the context of the situation. 28. On an exam, would it be more desirable to earn a grade with a high or a low percentile? Explain. 29. Mina is waiting in line at the Department of Motor Vehicles. Her wait time of 32 minutes is the 85th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85th percentile in the context of this situation. 30. In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78th percentile. Should Li be pleased or upset by this result? Explain. 31. In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had $1,700 in damage and was in the 90th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain and write a sentence that interprets the 90th percentile in the context of this problem. 32. The University of California has two criteria used to set admission standards for freshman to be admitted to a college in the UC system: a. Students' GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an admissions index score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12 percent of high school students in the state. In this context, what percentile does the top 12 percent represent? b. Students whose GPAs are at or above the 96th percentile of all students at their high school are eligible, called eligible in the local context, even if they are |
not in the top 12 percent of all students in the state. What percentage of students from each high school are eligible in the local context? 33. Suppose that you are buying a house. You and your real estate agent have determined that the most expensive house you can afford is the 34th percentile. The 34th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34 percent of the houses or 66 percent of the houses? Use Exercise 2.25 to calculate the following values. 34. First quartile = ________ 35. Second quartile = median = 50th percentile = ________ 36. Third quartile = ________ 37. Interquartile range (IQR) = ________ – ________ = ________ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 141 38. 10th percentile = ________ 39. 70th percentile = ________ 2.4 Box Plots Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars, 19 generally sell four cars, 12 generally sell five cars, nine generally sell six cars, and 11 generally sell seven cars. 40. Construct a box plot below. Use a ruler to measure and scale accurately. 41. Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or concentrated in some areas but not in others? How can you tell? 2.5 Measures of the Center of the Data 42. Find the mean for the following frequency tables: a. b. c. Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 Table 2.57 Daily Low Temperature Frequency 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 89.5–99.5 Table 2.58 53 32 15 1 0 Points per Game Frequency 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 89.5–99.5 Table 2.59 14 32 15 23 2 Use the following information to answer the next three exercises: The following |
data show the lengths of boats moored in a marina. The data are ordered from smallest to largest: 16, 17, 19, 20, 20, 21, 23, 24, 25, 25, 25, 26, 26, 27, 27, 27, 28, 29, 30, 32, 33, 33, 34, 35, 37, 39, 40 43. Calculate the mean. 142 Chapter 2 | Descriptive Statistics 44. Identify the median. 45. Identify the mode. Use the following information to answer the next three exercises: Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars, 19 generally sell four cars, 12 generally sell five cars, nine generally sell six cars, and 11 generally sell seven cars. Calculate the following. 46. sample mean = x ¯ = ________ 47. median = ________ 48. mode = ________ 2.6 Skewness and the Mean, Median, and Mode Use the following information to answer the next three exercises. State whether the data are symmetrical, skewed to the left, or skewed to the right. 49. 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5 50. 16, 17, 19, 22, 22, 22, 22, 22, 23 51. 87, 87, 87, 87, 87, 88, 89, 89, 90, 91 52. When the data are skewed left, what is the typical relationship between the mean and median? 53. When the data are symmetrical, what is the typical relationship between the mean and median? 54. What word describes a distribution that has two modes? 55. Describe the shape of this distribution. Figure 2.34 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 143 56. Describe the relationship between the mode and the median of this distribution. Figure 2.35 57. Describe the relationship between the mean and the median of this distribution. Figure 2.36 58. Describe the shape of this distribution. Figure 2.37 144 Chapter 2 | Descriptive Statistics 59. Describe the relationship between the mode and the median of this distribution. Figure 2.38 60. Are |
the mean and the median the exact same in this distribution? Why or why not? Figure 2.39 61. Describe the shape of this distribution. Figure 2.40 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 145 62. Describe the relationship between the mode and the median of this distribution. Figure 2.41 63. Describe the relationship between the mean and the median of this distribution. Figure 2.42 64. The mean and median for the data are the same. 3, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7 Is the data perfectly symmetrical? Why or why not? 65. Which is the greatest, the mean, the mode, or the median of the data set? 11, 11, 12, 12, 12, 12, 13, 15, 17, 22, 22, 22 66. Which is the least, the mean, the mode, and the median of the data set? 56, 56, 56, 58, 59, 60, 62, 64, 64, 65, 67 67. Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median? Why? 68. In a perfectly symmetrical distribution, when would the mode be different from the mean and median? 2.7 Measures of the Spread of the Data For each of the examples given below, tell whether the differences in outcomes may be explained by measurement variability, natural variability, induced variability, or sampling variability. 146 Chapter 2 | Descriptive Statistics 69. Scientists randomly select five groups of 10 women from a population of 1,000 women to record their body fat percentage. The scientists compute the mean body fat percentage from each group. The differences in outcomes may be attributed to which type of variability? 70. A pharmaceutical company randomly assigns participants to one of two groups: one is a control group receiving a placebo, and another is a treatment group receiving a new drug to lower blood pressure. The differences in outcomes may be attributed to which type of variability? 71. Jaiqua and Harold are trying to determine how ramp steepness affects the speed of a ball rolling down the ramp. They measure the time it takes for the ball to roll down ramps of differing slopes. When Jaiqua rolls the ball and Harold works the stopwatch, they get different results than when |
Harold rolls the ball and Jaiqua works the stopwatch. The differences in outcomes may be attributed to which type of variability? 72. Twenty people begin the same workout program on the same day and continue for three months. During that time, all participants worked out for the same amount of time and did the same number of exercises and repetitions. Each person was weighed at both the beginning and the end of the program. The differences in outcomes regarding the amount of weight lost may be attributed to which type of variability? Use the following information to answer the next two exercises. The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles. 29, 37, 38, 40, 58, 67, 68, 69, 76, 86, 87, 95, 96, 96, 99, 106, 112, 127, 145, 150 73. Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth. 74. Find the value that is one standard deviation below the mean. 75. Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team? Baseball Player Batting Average Team Batting Average Team Standard Deviation Fredo Karl Table 2.60.158.177.166.189.012.015 76. Use Table 2.60 to find the value that is three standard deviations a. above the mean, and b. below the mean This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 147 77. Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/ 84. a. b. c. Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 Table 2.61 Daily Low Temperature Frequency 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 89.5–99.5 Table 2.62 53 32 15 1 0 Points per Game Frequency 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 89.5–99 |
.5 Table 2.63 14 32 15 23 2 HOMEWORK 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 78. Student grades on a chemistry exam were 77, 78, 76, 81, 86, 51, 79, 82, 84, and 99. a. Construct a stem-and-leaf plot of the data. b. Are there any potential outliers? If so, which scores are they? Why do you consider them outliers? 148 Chapter 2 | Descriptive Statistics 79. Table 2.64 contains the 2010 rates for a specific disease in U.S. states and Washington, DC. Percent (%) State Percent (%) State Percent (%) State Alabama Alaska Arizona Arkansas California Colorado Connecticut Delaware 32.2 24.5 24.3 30.1 24.0 21.0 22.5 28.0 Washington, DC 22.2 Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Table 2.64 26.6 29.6 22.7 26.5 28.2 29.6 28.4 29.4 Kentucky Louisiana Maine Maryland 31.3 31.0 26.8 27.1 North Dakota 27.2 Ohio Oklahoma Oregon 29.2 30.4 26.8 Massachusetts 23.0 Pennsylvania 28.6 Michigan Minnesota Mississippi Missouri Montana Nebraska Nevada 30.9 24.8 34.0 30.5 23.0 26.9 22.4 New Hampshire 25.0 New Jersey New Mexico New York 23.8 25.1 23.9 North Carolina 27.8 Rhode Island 25.5 South Carolina 31.5 South Dakota 27.3 Tennessee Texas Utah Vermont Virginia Washington 30.8 31.0 22.5 23.2 26.0 25.5 West Virginia 32.5 Wisconsin Wyoming 26.3 25.1 a. Use a random number generator to randomly pick eight states. Construct a bar graph of the rates of a specific disease of those eight states. b. Construct a bar graph for all the states beginning with the letter A. c. Construct a bar graph for all the states beginning with the letter M. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 149 2.2 Histograms, Frequency Polygons, and Time Series Graphs 150 Chapter 2 | Descriptive Statistics 80. Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month |
. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows: Number of Books Frequency Relative Frequency 0 1 2 3 4 5 6 8 10 12 16 12 8 6 2 2 Table 2.65 Publisher A Number of Books Frequency Relative Frequency 0 1 2 3 4 5 7 9 18 24 24 22 15 10 5 1 Table 2.66 Publisher B Number of Books Frequency Relative Frequency 0–1 2–3 4–5 6–7 8–9 20 35 12 2 1 Table 2.67 Publisher C a. Find the relative frequencies for each survey. Write them in the charts. b. Using either a graphing calculator or computer or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of 1. For Publisher C, make bar widths of 2. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical. c. d. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not? e. Make new histograms for Publisher A and Publisher B. This time, make bar widths of 2. f. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 151 81. Often, cruise ships conduct all onboard transactions, with the exception of souvenirs, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their onboard bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group: Amount ($) Frequency Relative Frequency 51–100 101–150 151–200 201–250 251–300 301–350 5 10 15 15 10 5 Table 2.68 Singles Amount ($) Frequency Relative Frequency 100–150 201–250 251–300 301–350 351–400 401–450 451–500 501–550 551–600 601–650 5 5 5 5 10 10 10 10 5 5 Table 2.69 Couples a. Fill in the relative frequency for each group. b. Construct a hist |
ogram for the singles group. Scale the x-axis by $50 widths. Use relative frequency on the y-axis. c. Construct a histogram for the couples group. Scale the x-axis by $50 widths. Use relative frequency on the y-axis. d. Compare the two graphs: i. List two similarities between the graphs. ii. List two differences between the graphs. iii. Overall, are the graphs more similar or different? e. Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x-axis by $50, scale it by $100. Use relative frequency on the y-axis. f. Compare the graph for the singles with the new graph for the couples: i. List two similarities between the graphs. ii. Overall, are the graphs more similar or different? g. How did scaling the couples graph differently change the way you compared it to the singles graph? h. Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences. 152 Chapter 2 | Descriptive Statistics 82. 25 randomly selected students were asked the number of movies they watched the previous week. The results are as follows: Number of Movies Frequency Relative Frequency Cumulative Relative Frequency 0 1 2 3 4 Table 2.70 5 9 6 4 1 a. Construct a histogram of the data. b. Complete the columns of the chart. Use the following information to answer the next two exercises: Suppose 111 people who shopped in a special T-shirt store were asked the number of T-shirts they own costing more than $19 each. 83. The percentage of people who own at most three T-shirts costing more than $19 each is approximately ________. a. 21 b. 59 c. 41 d. cannot be determined 84. If the data were collected by asking the first 111 people who entered the store, then the type of sampling is ________. a. cluster simple random b. c. stratified d. convenience This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 153 85. Following are the 2010 obesity rates by U.S. states and Washington, DC. Percent (%) State Percent (%) State Percent (%) State Alabama Alaska Arizona Arkansas California Colorado Connecticut Delaware 32 |
.2 24.5 24.3 30.1 24.0 21.0 22.5 28.0 Washington, DC 22.2 Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Table 2.71 26.6 29.6 22.7 26.5 28.2 29.6 28.4 29.4 Kentucky Louisiana Maine Maryland 31.3 31.0 26.8 27.1 North Dakota 27.2 Ohio Oklahoma Oregon 29.2 30.4 26.8 Massachusetts 23.0 Pennsylvania 28.6 Michigan Minnesota Mississippi Missouri Montana Nebraska Nevada 30.9 24.8 34.0 30.5 23.0 26.9 22.4 New Hampshire 25.0 New Jersey New Mexico New York 23.8 25.1 23.9 North Carolina 27.8 Rhode Island 25.5 South Carolina 31.5 South Dakota 27.3 Tennessee Texas Utah Vermont Virginia Washington 30.8 31.0 22.5 23.2 26.0 25.5 West Virginia 32.5 Wisconsin Wyoming 26.3 25.1 Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint—Label the x-axis with the states. 2.3 Measures of the Location of the Data 86. The median age for U.S. ethnicity A currently is 30.9 years; for U.S. ethnicity B, it is 42.3 years. a. Based on this information, give two reasons why ethnicity A median age could be lower than the ethnicity B median age. b. Does the lower median age for ethnicity A necessarily mean that ethnicity A die younger than ethnicity B? Why or why not? c. How might it be possible for ethnicity A and ethnicity B to die at approximately the same age but for the median age for ethnicity B to be higher? 154 Chapter 2 | Descriptive Statistics 87. Six hundred adult Americans were asked by telephone poll, "What do you think constitutes a middle-class income?" The results are in Table 2.72. Also, include the left endpoint but not the right endpoint. Salary ($) Relative Frequency < 20,000.02 20,000–25,000.09 25,000–30,000.19 30,000–40,000.26 40,000–50,000.18 50,000–75,000.17 75,000–99,999.02 100,000+.01 Table 2.72 a. What percentage of the survey answered "not sure"? b. What |
percentage think that middle class is from $25,000 to $50,000? c. Construct a histogram of the data. i. Should all bars have the same width, based on the data? Why or why not? ii. How should the < 20,000 and the 100,000+ intervals be handled? Why? d. Find the 40th and 80th percentiles. e. Construct a bar graph of the data. 88. Given the following box plot, answer the questions. Figure 2.43 a. Which quarter has the smallest spread of data? What is that spread? b. Which quarter has the largest spread of data? What is that spread? c. Find the interquartile range (IQR). d. Are there more data in the interval 5–10 or in the interval 10–13? How do you know this? e. Which interval has the fewest data in it? How do you know this? i. 0–2 ii. 2–4 iii. 10–12 iv. 12–13 v. need more information This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 155 89. The following box plot shows the ages of the U.S. population for 1990, the latest available year: Figure 2.44 a. Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)? How do you know? b. 12.6 percent are age 65 and over. Approximately what percentage of the population are working-age adults (above age 17 to age 65)? 2.4 Box Plots 90. In a survey of 20-year-olds in China, Germany, and the United States, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results: Figure 2.45 a. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected. b. Have more Americans or more Germans surveyed been to more than eight foreign countries? c. Compare the three box plots. What do they imply about the foreign travel of 20-year-old residents of the three countries when compared to each other? 91. Given the following box plot, answer the questions. Figure 2.46 a. Think of an example (in words) where the data might fit into the above box plot. In two to five sentences, write |
down the example. b. What does it mean to have the first and second quartiles so close together, while the second to third quartiles are far apart? 156 Chapter 2 | Descriptive Statistics 92. Given the following box plots, answer the questions. Figure 2.47 a. In complete sentences, explain why each statement is false. i. Data 1 has more data values above two than Data 2 has above two. ii. The data sets cannot have the same mode. iii. For Data 1, there are more data values below four than there are above four. b. For which group, Data 1 or Data 2, is the value of 7 more likely to be an outlier? Explain why in complete sentences. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 157 93. A survey was conducted of 130 purchasers of new black sports cars, 130 purchasers of new red sports cars, and 130 purchasers of new white sports cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results: Figure 2.48 a. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected for that car series. b. Which group is most likely to have an outlier? Explain how you determined that. c. Compare the three box plots. What do they imply about the age of purchasing a sports car from the series when compared to each other? d. Look at the red sports cars. Which quarter has the smallest spread of data? What is the spread? e. Look at the red sports cars. Which quarter has the largest spread of data? What is the spread? f. Look at the red sports cars. Estimate the interquartile range (IQR). g. Look at the red sports cars. Are there more data in the interval 31–38 or in the interval 45–55? How do you know this? h. Look at the red sports cars. Which interval has the fewest data in it? How do you know this? i. 31–35 ii. 38–41 iii. 41–64 94. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: Number of Movies Frequency 0 1 2 3 4 Table 2.73 5 9 6 4 1 Construct a box plot of the data. |
158 Chapter 2 | Descriptive Statistics 2.5 Measures of the Center of the Data 95. Scientists are studying a particular disease. They found that countries that have the highest rates of people who have ever been diagnosed with this disease range from 11.4 percent to 74.6 percent. Percentage of Population Diagnosed Number of Countries 11.4–20.45 20.45–29.45 29.45–38.45 38.45–47.45 47.45–56.45 56.45–65.45 65.45–74.45 74.45–83.45 Table 2.74 29 13 4 0 2 1 0 1 a. What is the best estimate of the average percentage affected by the disease for these countries? b. The United States has an average disease rate of 33.9 percent. Is this rate above average or below? c. How does the United States compare to other countries? 96. Table 2.75 gives the percentage of children under age five have been diagnosed with a medical condition. What is the best estimate for the mean percentage of children with the condition? Percentage of Children with the Condition Number of Countries 16–21.45 21.45–26.9 26.9–32.35 32.35–37.8 37.8–43.25 43.25–48.7 Table 2.75 23 4 9 7 6 1 2.6 Skewness and the Mean, Median, and Mode 97. The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years. a. What does it mean for the median age to rise? b. Give two reasons why the median age could rise. c. For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not? 2.7 Measures of the Spread of the Data Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005. • μ = 1,000 FTES • median = 1,014 FTES • σ = 474 FTES This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 159 • • first quartile = 528.5 FTES third |
quartile = 1,447.5 FTES • n = 29 years 98. A sample of 11 years is taken. About how many are expected to have an FTES of 1,014 or above? Explain how you determined your answer. 99. Seventy-five percent of all years have an FTES a. at or below ______. b. at or above ______. 100. The population standard deviation = ______. 101. What percentage of the FTES were from 528.5 to 1,447.5? How do you know? 102. What is the IQR? What does the IQR represent? 103. How many standard deviations away from the mean is the median? Additional Information: The population FTES for 2005–2006 through 2010–2011 was given in an updated report. The data are reported here. Year 2005–2006 2006–2007 2007–2008 2008–2009 2009–2010 2010–2011 Total FTES 1,585 1,690 1,735 1,935 2,021 1,890 Table 2.76 104. Calculate the mean, median, standard deviation, the first quartile, the third quartile, and the IQR. Round to one decimal place. 105. Construct a box plot for the FTES for 2005–2006 through 2010–2011 and a box plot for the FTES for 1976–1977 through 2004–2005. 106. Compare the IQR for the FTES for 1976–1977 through 2004–2005 with the IQR for the FTES for 2005-2006 through 2010–2011. Why do you suppose the IQRs are so different? 107. Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer. Student GPA School Average GPA School Standard Deviation Thuy Vichet Kamala 2.7 87 8.6 3.2 75 8 Table 2.77.8 20.4 108. A music school has budgeted to purchase three musical instruments. The school plans to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest when compared |
to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type? Justify your answer. 109. An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran one mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes. a. Why is Kenji considered a better runner than Nedda even though Nedda ran faster than he? b. Who is the fastest runner with respect to his or her class? Explain why. 160 Chapter 2 | Descriptive Statistics 110. Scientists are studying a particular disease. They found that countries that have the highest rates of people who have ever been diagnosed with this disease range from 11.4 percent to 74.6 percent. Percentage of Population with Disease Number of Countries 11.4–20.45 20.45–29.45 29.45–38.45 38.45–47.45 47.45–56.45 56.45–65.45 65.45–74.45 74.45–83.45 Table 2.78 29 13 4 0 2 1 0 1 What is the best estimate of the average percentage of people with the disease for these countries? What is the standard deviation for the listed rates? The United States has an average disease rate of 33.9 percent. Is this rate above average or below? How unusual is the U.S. obesity rate compared to the average rate? Explain. 111. Table 2.79 gives the percentage of children under age five diagnosed with a specific medical condition. Percentage of Children with the Condition Number of Countries 16–21.45 21.45–26.9 26.9–32.35 32.35–37.8 37.8–43.25 43.25–48.7 Table 2.79 23 4 9 7 6 1 What is the best estimate for the mean percentage of children with the condition? What is the standard deviation? Which interval(s) could be considered unusual? Explain. BRINGING IT TOGETHER: HOMEWORK This OpenStax book is available for |
free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 161 112. Santa Clara County, California, has approximately 27,873 Japanese Americans. Table 2.80 shows their ages by group and each age-group's percentage of the Japanese American community. Age-Group Percentage of Community 0–17 18–24 25–34 35–44 45–54 55–64 65+ Table 2.80 18.9 8.0 22.8 15.0 13.1 11.9 10.3 a. Construct a histogram of the Japanese American community in Santa Clara County. The bars will not be the same width for this example. Why not? What impact does this have on the reliability of the graph? b. What percentage of the community is under age 35? c. Which box plot most resembles the information above? Figure 2.49 162 Chapter 2 | Descriptive Statistics 113. Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information. Javier Ercilia ¯ 6.0 miles 6.0 miles x s 4.0 miles 7.0 miles Table 2.81 a. How can you determine which survey was correct? b. Explain what the difference in the results of the surveys implies about the data. c. If the two histograms depict the distribution of values for each supervisor, which one depicts Ercilia’s sample? How do you know? d. Figure 2.50 If the two box plots depict the distribution of values for each supervisor, which one depicts Ercilia’s sample? How do you know? Figure 2.51 Use the following information to answer the next three exercises: We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section. Number of Years Frequency Number of Years Frequency 1 3 1 1 4 22 23 26 40 42 7 14 15 18 19 Table 2.82 1 1 1 2 2 Total = 20 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 163 Number of Years Frequency Number of Years Frequency 20 3 Table 2.82 Total = 20 114. What is the IQR? a. 8 b. 11 c. |
15 d. 35 115. What is the mode? a. 19 b. 19.5 c. 14 and 20 d. 22.65 116. Is this a sample or the entire population? sample a. b. entire population c. neither 117. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: Number of Movies Frequency 0 1 2 3 4 Table 2.83 5 9 6 4 1 ¯. a. Find the sample mean x b. Find the approximate sample standard deviation, s. 164 Chapter 2 | Descriptive Statistics 118. Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows: X Frequency 12 12 0 1 Table 2.84 ¯ a. Find the sample mean, x b. Find the sample standard deviation, s. c. Construct a histogram of the data. d. Complete the columns of the chart. e. Find the first quartile. f. Find the median. g. Find the third quartile. h. Construct a box plot of the data. i. What percentage of the students owned at least five pairs? j. Find the 40th percentile. k. Find the 90th percentile. l. Construct a line graph of the data. m. Construct a stemplot of the data. 119. Following are the published weights (in pounds) of all of the football team members of the San Francisco 49ers from a previous year: 177, 205, 210, 210, 232, 205, 185, 185, 178, 210, 206, 212, 184, 174, 185, 242, 188, 212, 215, 247, 241, 223, 220, 260, 245, 259, 278, 270, 280, 295, 275, 285, 290, 272, 273, 280, 285, 286, 200, 215, 185, 230, 250, 241, 190, 260, 250, 302, 265, 290, 276, 228, 265 a. Organize the data from smallest to largest value. b. Find the median. c. Find the first quartile. d. Find the third quartile. e. Construct a box plot of the data. f. The middle 50 percent of the weights are from ________ to ________. g. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why? If our population |
included every team member who ever played for a California-based football team, would the above data be a sample of weights or the population of weights? Why? h. i. Assume the population was a California-based football team. Find i. ii. iii. iv. the population mean, μ, the population standard deviation, σ, and the weight that is two standard deviations below the mean. In addition, when the team's most famous quarterback, played football, he weighed 205 pounds. Also find how many standard deviations above or below the mean was he? j. That same year, the mean weight for a player from a Texas football team was 240.08 pounds with a standard deviation of 44.38 pounds. One player weighed in at 209 pounds. With respect to his team, who was lighter, the California quarterback or the Texas player? How did you determine your answer? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 165 120. One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The 12 change scores are as follows: 3, 8, –1, 2, 0, 5, –3, 1, –1, 6, 5, –2 a. What is the mean change score? b. What is the standard deviation for this population? c. What is the median change score? d. Find the change score that is 2.2 standard deviations below the mean. 121. Refer to Figure 2.52 to determine which of the following are true and which are false. Explain your solution to each part in complete sentences. Figure 2.52 a. The medians for all three graphs are the same. b. We cannot determine if any of the means for the three graphs are different. c. The standard deviation for Graph b is larger than the standard deviation for Graph a. d. We cannot determine if any of the third quartiles for the three graphs are different. 122. In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. |
Let X = the length (in days) of an engineering conference. a. Organize the data in a chart. b. Find the median, the first quartile, and the third quartile. c. Find the 65th percentile. d. Find the 10th percentile. e. Construct a box plot of the data. f. The middle 50 percent of the conferences last from ________ days to ________ days. g. Calculate the sample mean of days of engineering conferences. h. Calculate the sample standard deviation of days of engineering conferences. i. Find the mode. j. If you were planning an engineering conference, which would you choose as the length of the conference, mean, median, or mode? Explain why you made that choice. k. Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences. 123. A survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622 a. Organize the data into a chart with five intervals of equal width. Label the two columns Enrollment and Frequency. b. Construct a histogram of the data. c. If you were to build a new community college, which piece of information would be more valuable: the mode or the mean? d. Calculate the sample mean. e. Calculate the sample standard deviation. f. A school with an enrollment of 8,000 would be how many standard deviations away from the mean? 166 Chapter 2 | Descriptive Statistics Use the following information to answer the next two exercises. X = the number of days per week that 100 clients use a particular exercise facility. X Frequency 0 1 2 3 4 5 6 3 12 33 28 11 9 4 Table 2.85 124. The 80th percentile is ________. a. |
5 b. 80 c. 3 d. 4 125. The number that is 1.5 standard deviations below the mean is approximately ________. a. 0.7 b. 4.8 c. –2.8 d. cannot be determined This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 167 126. Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in Table 2.86. Number of Books Frequency Relative Frequency 0 1 2 3 4 5 7 9 Table 2.86 18 24 24 22 15 10 5 1 a. Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion. If a data value is identified as an outlier, what should be done about it? b. c. Are any data values farther than two standard deviations away from the mean? In some situations, statisticians may use this criterion to identify data values that are unusual, compared to the other data values. Note that this criterion is most appropriate to use for data that is mound shaped and symmetric rather than for skewed data. d. Do Parts a and c of this problem give the same answer? e. Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data? f. Based on the shape of the data, which is the most appropriate measure of center for this data, mean, median, or mode? REFERENCES 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs Burbary, K. (2011, March 7). Facebook demographics revisited – 2001 statistics. Social Media Today. Retrieved from http://www.kenburbary.com/2011/03/facebook-demographics-revisited-2011-statistics-2/ Centers for Disease Control and Prevention. (n.d.). Overweight and obesity: Adult obesity facts. Available online http://www.cdc.gov/obesity/data/adult.html CollegeBoard. (2013). The 9th annual AP report to the nation. Retrieved from http://apreport.collegeboard.org/goalsandfindings/promoting-equity 2.2 Histograms, Frequency Poly |
gons, and Time Series Graphs Bureau of Labor Statistics, U.S. Department of Labor. (n.d.). Consumer price index. Retrieved from https://www.bls.gov/ cpi/ CIA World Factbook. http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (n.d.). Demographics: Children under the age of 5 years underweight. Available at Centers for Disease Control and Prevention. (n.d.). Overweight and obesity: Adult obesity facts. Available online http://www.cdc.gov/obesity/data/adult.html Food and Agriculture Organization of http://www.fao.org/economic/ess/ess-fs/en/ the United Nations. (n.d.). Food security statistics. Retrieved from General Register Office for Scotland. Births time series data. (2013). Retrieved from http://www.gro-scotland.gov.uk/ statistics/theme/vital-events/births/time-series.html 168 Chapter 2 | Descriptive Statistics Gunst, R., and Mason, R. (1980). Regression analysis and its application: A data-oriented approach. Boca Raton, FL: CRC Press. Sandbox Networks. (2007). Presidents. Available online at http://www.factmonster.com/ipka/A0194030.html Scholastic. (2013). Timeline: Guide to the U.S. presidents. Retrieved from http://www.scholastic.com/teachers/article/ timeline-guide-us-presidents World Bank Group. (2013). DataBank: CO2 emissions (kt). Retrieved from http://databank.worldbank.org/data/home.aspx 2.3 Measures of the Location of the Data Cauchon, D., and Overberg, P. (2012). Census data shows minorities now a majority of U.S. births. USA Today. Retrieved from http://usatoday30.usatoday.com/news/nation/story/2012-05-17/minority-birthscensus/55029100/1 The Mercury News. (n.d.). Retrieved from http://www.mercurynews.com/ Time. (n.d.). Survey by Yankelovich Partners, Inc. U.S. Census Bureau. (1990). 1990 census. Retrieved from http://www |
.census.gov/main/www/cen1990.html U.S. Census Bureau. (n.d.). Data. Retrieved from http://www.census.gov/ 2.4 Box Plots West Magazine. (n.d.). Retrieved from https://westmagazine.net/ 2.5 Measures of the Center of the Data CIA World Factbook. r.aspx?t=50&v=2228&l=en (n.d.). Obesity – adult prevalence rate. Available at http://www.indexmundi.com/g/ World Bank Group. (n.d.). Retrieved from http://www.worldbank.org 2.7 Measures of the Spread of the Data King, B. (2005, Dec.). Graphically Speaking. Retrieved from http://www.ltcc.edu/web/about/institutional-research Microsoft Bookshelf. (n.d.). SOLUTIONS 1 Stem Leaf Table 2.87 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 169 Stem Leaf Table 2.88 3 5 Figure 2.53 7 Figure 2.54 Chapter 2 | Descriptive Statistics 170 9 Figure 2.55 11 Figure 2.56 13 65 15 The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value. 17 Answers will vary. One possible histogram is shown below. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 171 Figure 2.57 19 Find the midpoint for each class. These will be graphed on the x-axis. The frequency values will be graphed on the y-axis values. Figure 2.58 172 21 Chapter 2 | Descriptive Statistics Figure 2.59 23 a. The 40th percentile is 37 years. b. The 78th percentile is 70 years. 25 Jesse graduated 37th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37. x = 143 and y = 1. x +.5y (100) = 79.72. Jesse’s rank of 37 puts him at the 80th percentile. (100) = 143 +.5(1) n 180 27 a. For runners |
in a race, it is more desirable to have a high percentile for speed. A high percentile means a higher speed, which is faster. b. 40 percent of runners ran at speeds of 7.5 miles per hour or less (slower), and 60 percent of runners ran at speeds of 7.5 miles per hour or more (faster). 29 When waiting in line at the DMV, the 85th percentile would be a long wait time compared to the other people waiting. 85 percent of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85 percent of people at the DMV waited 32 minutes or less. 15 percent of people at the DMV waited 32 minutes or longer. 31 The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90 percent of the crash-tested cars had damage repair costs of $1,700 or less; only 10 percent had damage repair costs of $1,700 or more. 33 You can afford 34 percent of houses. 66 percent of the houses are too expensive for your budget. INTERPRETATION: 34 percent of houses cost $240,000 or less; 66 percent of houses cost $240,000 or more. 35 4 37 6 – 4 = 2 39 6 41 More than 25 percent of salespersons sell four cars in a typical week. You can see this concentration in the box plot because the first quartile is equal to the median. The top 25 percent and the bottom 25 percent are spread out evenly; the whiskers have the same length. 43 Mean: 16 + 17 + 19 + 20 + 20 + 21 + 23 + 24 + 25 + 25 + 25 + 26 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 32 + 33 + 33 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 173 + 34 + 35 + 37 + 39 + 40 = 738; 738 27 = 27.33 45 The most frequent lengths are 25 and 27, which occur three times. Mode = 25, 27 47 4 49 The data are symmetrical. The median is 3, and the mean is 2.85. They are close, and the mode lies close to the middle of the data, so the data are symmetrical. 51 The data are skewed |
right. The median is 87.5, and the mean is 88.2. Even though they are close, the mode lies to the left of the middle of the data, and there are many more instances of 87 than any other number, so the data are skewed right. 53 When the data are symmetrical, the mean and median are close or the same. 55 The distribution is skewed right because it looks pulled out to the right. 57 The mean is 4.1 and is slightly greater than the median, which is 4. 59 The mode and the median are the same. In this case, both 5. 61 The distribution is skewed left because it looks pulled out to the left. 63 Both the mean and the median are 6. 65 The mode is 12, the median is 13.5, and the mean is 15.1. The mean is the largest. 67 The mean tends to reflect skewing the most because it is affected the most by outliers. 69 sampling variability 70 induced variability 71 measurement variability 72 natural variability 73 s = 34.5 75 For Fredo: z =.158 –.166.012 = –0.67. For Karl: z =.177 –.189.015 = –.8. Fredo’s z score of –.67 is higher than Karl’s z score of –.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team. 77 a. s x = b. s x = c. s x = ∑ f m2 n ∑ f m2 n ∑ f m2 n 79 − x¯ 2 = 193,157.45 30 − 79.52 = 10.88 − x¯ 2 = 380,945.3 101 − 60.942 = 7.62 − x¯ 2 = 440,051.5 86 − 70.662 = 11.14 a. Example solution for using the random number generator for the TI-84+ to generate a simple random sample of eight states. Instructions are as follows. Number the entries in the table 1–51 (includes Washington, DC; numbered vertically) Press MATH Arrow over to PRB Press 5:randInt( Enter 51,1,8) Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different |
number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}. 174 Chapter 2 | Descriptive Statistics Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}. Figure 2.60 b. Figure 2.61 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 175 Figure 2.62 c. 81 Amount($) Frequency Relative Frequency 51–100 101–150 151–200 201–250 251–300 301–350 5 10 15 15 10 5 Table 2.89 Singles.08.17.25.25.17.08 Amount ($) Frequency Relative Frequency 100–150 201–250 251–300 301–350 351–400 401–450 451–500 501–550 551–600 601–650 5 5 5 5 10 10 10 10 5 5 Table 2.90 Couples.07.07.07.07.14.14.14.14.07.07 176 Chapter 2 | Descriptive Statistics a. See Table 2.69 and Table 2.69. b. In the following histogram, data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted, with the exception of the first interval, where both boundary values are included. Figure 2.63 c. In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted, with the exception of the first interval, where values on both boundaries are included. Figure 2.64 d. Compare the two graphs. i. Answers may vary. Possible answers include the following: ▪ Both graphs have a single peak. ▪ Both graphs use class intervals with width equal to $50 ii. Answers may vary. Possible answers include the following: ▪ The couples graph has a class interval with no values ▪ It takes almost twice as many class intervals to display the data for couples iii. Answers may vary. Possible answers include the following. The graphs are more similar than different because This OpenStax book is available for free at http://cnx.org/content/col |
30309/1.8 Chapter 2 | Descriptive Statistics 177 the overall patterns for the graphs are the same. e. Check student's solution. f. Compare the graph for the singles with the new graph for the couples: i. ▪ Both graphs have a single peak ▪ Both graphs display six class intervals ▪ Both graphs show the same general pattern ii. Answers may vary. Possible answers include the following. Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different. g. Answers may vary. Possible answers include the following. You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison. h. Answers may vary. Possible answers include the following. Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals. 83 c 85 Answers will vary. 87 a. 1 – (.02+.09+.19+.26+.18+.17+.02+.01) =.06 b..19+.26+.18 =.63 c. Check student’s solution. d. 40th percentile will fall between 30,000 and 40,000 80th percentile will fall between 50,000 and 75,000 e. Check student’s solution. 89 a. more children; the left whisker shows that 25 percent of the population are children 17 and younger; the right whisker shows that 25 percent of the population are adults 50 and older, so adults 65 and over represent less than 25 percent b. 62.4 percent 91 a. Answers will vary. Possible answer: State University conducted a survey to see how involved its students are in community service. The box plot shows the number of community service hours logged by participants over the past year. b. Because the first and second quartiles are close, the data in this quarter is very similar. There is not much variation in the values. The data in the third quarter is much more variable, or spread out. This is clear because the second quartile is so far away from the third quartile. 93 a. Each box plot is spread out more in the greater values. Each plot is skewed to the right, so the ages of the top |
50 percent of buyers are more variable than the ages of the lower 50 percent. b. The black sports car is most likely to have an outlier. It has the longest whisker. c. Comparing the median ages, younger people tend to buy the black sports car, while older people tend to buy the white sports car. However, this is not a rule, because there is so much variability in each data set. d. The second quarter has the smallest spread. There seems to be only a three-year difference between the first quartile and the median. e. The third quarter has the largest spread. There seems to be approximately a 14-year difference between the median and the third quartile. 178 Chapter 2 | Descriptive Statistics f. IQR ~ 17 years g. There is not enough information to tell. Each interval lies within a quarter, so we cannot tell exactly where the data in that quarter is are concentrated. h. The interval from 31 to 35 years has the fewest data values. Twenty-five percent of the values fall in the interval 38 to 41, and 25 percent fall between 41 and 64. Since 25 percent of values fall between 31 and 38, we know that fewer than 25 percent fall between 31 and 35. 96 the mean percentage, x¯ = 1,328.65 50 = 26.75 98 The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the sixth number in order. Six years will have totals at or below the median. 100 474 FTES 102 919 104 • mean = 1,809.3 • median = 1,812.5 • • • • standard deviation = 151.2 first quartile = 1,690 third quartile = 1,935 IQR = 245 106 Hint: think about the number of years covered by each time period and what happened to higher education during those periods. 108 For pianos, the cost of the piano is.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type. 110 • x¯ = 23.32 • Using the TI 83/84, we |
obtain a standard deviation of: s x = 12.95. • The obesity rate of the United States is 10.58 percent higher than the average obesity rate. • Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the disease percentage that is one standard deviation from the mean. The U.S. disease rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, although 34 percent have the disease, does not have an unusually high percentage of people with the disease. 112 a. For graph, check student's solution. b. 49.7 percent of the community is under the age of 35 c. Based on the information in the table, graph (a) most closely represents the data. 114 a 116 b 117 a. 1.48 b. 1.12 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 2 | Descriptive Statistics 179 119 a. 174, 177, 178, 184, 185, 185, 185, 185, 188, 190, 200, 205, 205, 206, 210, 210, 210, 212, 212, 215, 215, 220, 223, 228, 230, 232, 241, 241, 242, 245, 247, 250, 250, 259, 260, 260, 265, 265, 270, 272, 273, 275, 276, 278, 280, 280, 285, 285, 286, 290, 290, 295, 302 b. 241 c. 205.5 d. 272.5 e. f. 205.5, 272.5 g. sample h. population i. i. 236.34 ii. 37.50 iii. 161.34 iv..84 standard deviations below the mean j. young true true true false 121 a. b. c. d. 123 a. b. Check student’s solution. c. mode d. 8,628.74 e. 6,943.88 f. –0.09 Enrollment Frequency 1,000–5,000 5,000–10,000 10 16 10,000–15,000 3 15,000–20,000 3 20,000–25,000 1 25,000–30,000 2 Table 2.91 180 125 a Chapter 2 | Descriptive Statistics This OpenStax book is available for free at http://cnx.org/ |
content/col30309/1.8 Chapter 3 | Probability Topics 181 3 | PROBABILITY TOPICS Figure 3.1 Meteor showers are rare, but the probability of them occurring can be calculated. (credit: Navicore/flickr) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Understand and use the terminology of probability • Determine whether two events are mutually exclusive and whether two events are independent • Calculate probabilities using the addition rules and multiplication rules • Construct and interpret contingency tables • Construct and interpret Venn diagrams • Construct and interpret tree diagrams It is often necessary to guess about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs. You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach. 182 Chapter 3 | Probability Topics How likely is it that a randomly chosen person in your class has change in his or her pocket? Would you say that it is very likely? Somewhat likely? Not likely? How likely is it that a randomly chosen person in your class has ridden a bus in the past month? If a person is chosen at random from your classroom and you know that he or she has ridden a bus in the past month, do you think that person is more likely or less likely to have change? Probability theory allows us to measure how likely—or unlikely—a given result is. Your instructor will survey your class. Count the number of students in the class today. • Raise your hand if you have any change in your pocket or purse. Record the number of raised hands. • Raise your hand if you rode a bus within the past month. Record the number of raised hands. • Raise your hand if you answered yes to BOTH of the first two questions. Record the number of raised |
hands. Use the class data as estimates of the following probabilities. P(change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. P(bus) means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers. • Find P(change). • Find P(bus). • Find P(change AND bus). Find the probability that a randomly chosen student in your class has change in his/her pocket or purse and rode a bus within the last month. • Find P(change|bus). Find the probability that a randomly chosen student has change given that he or she rode a bus within the last month. Count all the students who rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus. 3.1 | Terminology Probability is a measure that is associated with how certain we are of results, or outcomes, of a particular activity. When the activity is a planned operation carried out under controlled conditions, it is called an experiment. If the result is not predetermined, then the experiment is said to be a chance experiment. Each time the experiment is attempted is called a trial. Examples of chance experiments include the following: • • flipping a fair coin, spinning a spinner, • drawing a marble at random from a bag, and • rolling a pair of dice. A result of an experiment is called an outcome. The sample space of an experiment is the set, or collection, of all possible outcomes. There are four main ways to represent a sample space: Systematic List of Outcomes heads (H) HH Flipping a Fair Coin Flipping Two Fair Coins Table 3.1 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 183 Flipping a Fair Coin Flipping Two Fair Coins tails (T) HT TH TT Figure 3.2 Figure 3.3 Tree Diagram* Venn Diagram* Figure 3.4 Figure 3.5 Set Notation S = {H, T} S = {HH, HT, TH, TT} Table 3.1 *We will investigate tree diagrams and Venn diagrams in Section 3.5. Note—when represented as a set, the sample space is denoted with an uppercase S. |
An event is any combination of outcomes. It is a subset of the sample space, so uppercase letters like A and B are commonly used to represent events. For example, if the experiment is to flip three fair coins, event A might be getting at most one head. The probability of an event A is written P(A), and 0 ≤ P(A) ≤ 1.P(A) = 0 means the event A can never happen. P(A) = 1 means the event A always happens. P(A) = 0.5 means the event A is equally likely to occur or not to occur. Figure 3.6 If two outcomes or events are equally likely, then they have equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer. To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. This is known as the theoretical probability of A. 184 Chapter 3 | Probability Topics Theoretical Probability of Event A P(A) = Number of outcomes in event A Total number of possible outcomes. For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. Let A represent the outcome getting one head. There are two outcomes that meet this condition {HT, TH}, so P(A) = 2 4 = 1 2 =.5. Theoretical probability is not sufficient in all situations, however. Suppose we want to calculate the probability that a randomly selected car will run a red light at a given intersection. In this case, we need to look at events that have occurred, not theoretical possibilities. We could install a traffic camera and count the number of times that cars failed to stop when the light was red and the total number of cars that passed through the intersection for a period of time. These data will allow us to calculate the experimental, or empirical, probability that a car |
runs the red light. Experimental Probability of Event A P(A) = Number of times event A occurs. Total number of trials While theoretical and experimental methods provide two different ways to calculate probability, these methods are closely related. If you flip one fair coin, there is one way to obtain heads and two possible outcomes. So, the theoretical probability of heads is 1 2. Probability does not predict short-term results, however. If an experiment involves flipping a coin 10 times, you should not expect exactly five heads and five tails. The probability of any outcome measures the long-term relative frequency of that outcome. If you continue to flip the coin (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches.5 (the probability of heads).This important characteristic of probability experiments is known as the law of large numbers, which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed, or empirical, relative frequency will approach the theoretical probability. Suppose you roll one fair, six-sided die with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P(E) = 2. If you were to roll the die only a few times, you would not be 6 surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, 2 6 of the rolls would result in an outcome of at least five. You would not expect exactly 2 6, but the long-term relative frequency of obtaining this result would approach the theoretical probability of 2 6 as the number of repetitions grows larger and larger. It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one-euro coin and discovered that in 250 trials, a head was obtained 56 percent of the time and a tail was obtained 44 percent of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at |
the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely. OR Event An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are not listed twice. AND Event An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 185 The complement of event A is denoted A′ (read "A prime"). A′ consists of all outcomes that are not in A. Notice that P(A) + P(A′) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then, A′ = {5, 6}. P(A) = 4 6, P(A′) = 2 6, and P(A) + P(A′) = 4 6 + 2 6 = 1. The conditional probability of A given B is written P(A|B), read "the probability of A, given B." P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional probability |
reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A|B) is P(A|B) = P(A AND B) P(B) where P(B) is greater than zero. For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = {2, 3} and B = {2, 4, 6}. P(A|B) represents the probability that a randomly selected outcome is in A given that it is in B. We know that the outcome must lie in B, so there are three possible outcomes. There is only one outcome in B that also lies in A, so P(A|B) = 1 3. We get the same result by using the formula. Remember that S has six outcomes. P(A|B) = P(A AND B) P(B) = (the number of outcomes that are 2 or 3 and even in S) 6 (the number of outcomes that are even in S Understanding Terminology and Symbols It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any. Example 3.1 The sample space S is the whole numbers starting at one and less than 20. a. S = ________ Let event A = the even numbers and event B = numbers greater than 13. b. A = ________, B = ________ c. P(A) = ________, P(B) = ________ d. A AND B = ________, A OR B = ________ e. P(A AND B) = ________, P(A OR B) = ________ f. A′ = ________, P(A′) = ________ g. P(A) + P(A′) = ________ h. P(A|B) = ________, P(B|A) = ________; are the probabilities equal? Solution 3.1 a. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 |
, 13, 14, 15, 16, 17, 18, 19} b. A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19} c. P(A) = number of outcomes in A number of outcomes in S = 9 19, P(B) = P(A) = number of outcomes in B number of outcomes in S = 6 19 d. The set A AND B contains all outcomes that lie in both sets A and B, so A AND B = {14,16,18}, The set A OR B contains all outcomes that lie either of the sets A or B, so A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}. e. P(A AND B) = 3 19, P(A OR B) = 12 19 186 Chapter 3 | Probability Topics f. A' consists of all outcomes in the sample space, S, that DO NOT lie in A, so A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P(A′) = 10 19. g. P(A) + P(A′) = 9 19 + 10 19 = 1 h. P(A|B) = P(AANDB) P(B) = 3 19 6 19 = 3 6, P(B|A) = P(AANDB) P(A) = 3 19 9 19 = 3 9, No, the probabilities are not equal. 3.1 The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)). a. S = ________ Let event A = the sum is even and event B = the first number is prime. b. A = ________, B = ________ c. P(A) = ________, P(B) = ________ d. A AND B = ________, A OR B = ________ e. P(A AND B) = ________, P(A OR B) = ________ f. B′ = ________, P(B′) = ________ g. P(A) + P(A′) = ________ h. P(A|B) = ________, P |
(B|A) = ________; are the probabilities equal? Example 3.2 A fair, six-sided die is rolled. The sample space, S, is {1, 2, 3, 4, 5, 6}. Describe each event and calculate its probability. a. Event T = the outcome is two. b. Event A = the outcome is an even number. c. Event B = the outcome is less than four. d. The complement of A e. A GIVEN B f. B GIVEN A g. A AND B h. A OR B i. A OR B′ j. Event N = the outcome is a prime number. k. Event I = the outcome is seven. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 187 Solution 3.2 a. T = {2}, P(T) = number of outcomes in T number of outcomes in S = 1 6 b. A = {2, 4, 6}, P(A) = 3 6 = 1 2 c. B = {1, 2, 3}, P(B) = 3 6 = 1 2 d. A′ = {1, 3, 5}, P(A′) = 3 6 = 1 2 e. A|B = {2}, There are three outcomes in B, and only 1 of these lies in A, so P(A|B) = 1 3 f. B|A = {2}, There are three outcomes in A, and only 1 of these lies in B, so P(B|A) = 1 3 g. A AND B = {2}, P(A AND B) = 1 6 h. A OR B = {1, 2, 3, 4, 6}, P(A OR B) = 5 6 i. A OR B′ = {2, 4, 5, 6}, P(A OR B′) = 4 6 = 2 3 j. N = {2, 3, 5}, P(N) = 1 2 k. It is impossible to roll a die and get an outcome of 7, so P(7) = 0. Example 3.3 Table 3.2 describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right or left-handed. Right-Handed Left-Handed Males 43 Females 44 Table 3.2 9 |
4 Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities: a. P(M) b. P(F) c. P(R) d. P(L) e. P(M AND R) f. P(F AND L) g. P(M OR F) h. P(M OR R) 188 Chapter 3 | Probability Topics i. P(F OR L) j. P(M') k. P(R|M) l. P(F|L) m. P(L|F) Solution 3.3 a. P(M) = number of males total number of subjects = 43 + 9 43 + 9 + 44 + 4 = 52 100 =.52 b. P(F) = number of females total number of subjects = 44 + 4 43 + 9 + 44 + 4 = 48 100 =.48 c. P(R) = number of right-handed subjects total number of subjects = 43 + 44 43 + 9 + 44 + 4 = 87 100 =.87 d. P(L) = number of left-handed subjects total number of subjects = 9 + 4 43 + 9 + 44 + 4 = 13 100 =.13 e. P(MandR) = number of male, right-handed subjects total number of subjects = 43 100 =.43 f. P(FandL) = number of female, left-handed subjects total number of subjects = 4 100 =.04 g. P(MorF) = number of subjects that are male or female total number of subjects = 52 + 48 100 = 100 100 = 1 h. P(MorR) = number of subjects that are male or right-handed total number of subjects = 43 + 9 + 44 100 = 96 100 =.96 i. P(ForL) = number of subjects that are female or left-handed total number of subjects = 44 + 4 + 9 100 = 57 100 =.57 j. P(M') = number of subjects who are not male total number of subjects = 44 + 4 43 + 9 + 44 + 4 = 48 100 =.48 k. P(R|M) = l. P(F|L) = m. P(L|F) = P(RandM) P(M) = 0.43 0. |
52 =.8269 (rounded to four decimal places) P(FandL) P(L) = 0.04 0.13 P(LandF) P(F) = 0.04 0.48 =.3077 (rounded to four decimal places) =.0833 (rounded to four decimal places) 3.2 | Independent and Mutually Exclusive Events Independent and mutually exclusive do not mean the same thing. Independent Events Two events are independent if the following are true: • P(A|B) = P(A) • P(B|A) = P(B) • P(A AND B) = P(A)P(B) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 189 Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are not independent, then we say that they are dependent events. Sampling may be done with replacement or without replacement. • With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick. A bag contains four blue and three white marbles. James draws one marble from the bag at random, records the color, and replaces the marble. The probability of drawing blue is 4 7. When James draws a marble from the bag a second time, the probability of drawing blue is still 4 7 white marbles.. James replaced the marble after the first draw, so there are still four blue and three Figure 3.7 • Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent. The bag still contains four blue and three white marbles. Maria draws one marble from the bag at random, records the color, and sets the marble aside. The probability of |
drawing blue on the first draw is 4 7. Suppose Maria draws a blue marble and sets it aside. When she draws a marble from the bag a second time, there are now three blue and three white marbles. So, the probability of drawing blue is now 3 6. Removing the first marble without replacing it influences the probabilities = 1 2 on the second draw. 190 Chapter 3 | Probability Topics Figure 3.8 If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise. Example 3.4 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. Clubs and spades are black, while diamonds and hearts are red cards. There are 13 cards in each suit consisting of A (ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Figure 3.9 a. Sampling with replacement Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the Q of spades. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 191 You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the 10 of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are {Q of spades, 10 of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck. b. Sampling without replacement Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the K of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are {K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. 3.4 You have a fair, well- |
shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random. a. Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement? b. Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement? Example 3.5 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. a. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD. b. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH, 7D, 6D, KH. Which of a. or b. did you sample with replacement and which did you sample without replacement? Solution 3.5 a. Because you do not put any cards back, the deck changes after each draw. These events are dependent, and this is sampling without replacement; b. Because you put each card back before picking the next one, the deck never changes. These events are independent, so this is sampling with replacement. 3.5 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you |
sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement. a. QS, 1D, 1C, QD b. KH, 7D, 6D, KH c. QS, 7D, 6D, KS 192 Chapter 3 | Probability Topics Mutually Exclusive Events A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0. For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P(A AND B) = 2 10 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive. If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms. Example 3.6 Flip two fair coins. This is an experiment. The sample space is {HH, HT, TH, TT}, where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads. • Let A = the event of getting at most one tail. At most one tail means zero or one tail. Then A can be written as {HH, HT, TH}. The outcome HH shows zero tails. HT and TH each show one tail. • Let B = the event of getting all tails. B can be written as {TT}. B is the complement event of A, so B = A′. Also, P(A) + P(B) = P(A) + P(A′) = 1. • The probabilities for A and for B are P(A) = 3 4 and P(B) = 1 4. • Let C = the event of getting all heads. C = {HH}. Since B = {TT |
}, P(B AND C) = 0. B and C are mutually exclusive. (B and C have no members in common because you cannot have all tails and all heads at the same time.) • Let D = event of getting more than one tail. D = {TT}. P(D) = 1 4. • Let E = event of getting a head on the first roll. This implies you can get either a head or tail on the second roll. E = {HT, HH}. P(E) = 2 4. • Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = {HT, TH, TT}. P(F) = 3 4. 3.6 Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. Example 3.7 Flip two fair coins. Find the probabilities of the events. a. Let F = the event of getting at most one tail (zero or one tail). b. Let G = the event of getting two faces that are the same. c. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 193 d. Are F and G mutually exclusive? e. Let J = the event of getting all tails. Are J and H mutually exclusive? Solution 3.7 Look at the sample space in Example 3.6. a. Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = 3 4. b. Two faces are the same if HH or TT show up. P(G) = 2 4. c. A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P(H) = 2 4. d. F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive. e. Getting all tails occurs when tails shows up on both coins (TT). H’s outcomes are HH and HT. J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive. 3.7 A |
box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events: a. Let F = the event of getting the white ball twice. b. Let G = the event of getting two balls of different colors. c. Let H = the event of getting white on the first pick. d. Are F and G mutually exclusive? e. Are G and H mutually exclusive? Example 3.8 Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}. • Find the complement of A, A′. The complement of A, A′, is B because A and B together make up the sample space. P(A) + P(B) = P(A) + P(A′) = 1. Also, P(A) = 3 6 and P(B) = 3 6. • Let event C = odd faces larger than two. Then C = {3, 5}. Let event D = all even faces smaller than five. Then D = {2, 4}. P(C AND D) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events. • Let event E = all faces less than five. E = {1, 2, 3, 4}. Are C and E mutually exclusive events? Answer yes or no. Why or why not? Solution 3.8 No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = 1 6. To be mutually exclusive, P(C AND E) must be zero. • Find P(C|A). This is a conditional probability. Recall that event C is {3, 5} and event A is {1, 3, 5}. To find P(C|A), find the probability of C using the sample space A. You have reduced the sample space from the 194 Chapter 3 | Probability Topics original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P(C|A) = 2 3. 3. |
8 Let event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German. Suppose P(A) = 0.4 and P(B) =.2. P(A AND B) =.08. Are events A and B independent? Hint—You must show one of the following: • P(A|B) = P(A) • P(B|A) • P(A AND B) = P(A)P(B) Example 3.9 Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) =.6, P(H) =.5, and P(G AND H) =.3. Are G and H independent? If G and H are independent, then you must show ONE of the following: • P(G|H) = P(G) • P(H|G) = P(H) • P(G AND H) = P(G)P(H) NOTE The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information. a. Show that P(G|H) = P(G). Solution 3.9 P(G|H) = P(G AND H) P(H) =.3.5 =.6 = P(G) b. Show P(G AND H) = P(G)P(H). Solution 3.9 P(G)P(H) = (.6)(.5) =.3 = P(G AND H) Since G and H are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent, that is, they are dependent, then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that P(H|G) = P(H) to show that G and H are independent events. 3.9 In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4. This Open |
Stax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 195 • R = a red marble • G = a green marble • O = an odd-numbered marble • The sample space is S = {R1, R2, R3, R4, R5, R6, G1, G2, G3, G4}. S has 10 outcomes. What is P(G AND O)? Example 3.10 Let event C = taking an English class. Let event D = taking a speech class. Suppose P(C) =.75, P(D) =.3, P(C|D) =.75 and P(C AND D) =.225. Justify your answers to the following questions numerically. a. Are C and D independent? b. Are C and D mutually exclusive? c. What is P(D|C)? Solution 3.10 a. Yes, because P(C|D) =.75 = P(C). b. No, because P(C AND D) is not equal to zero. c. P(D|C) = P(C AND D) P(C) = 0.225.75 =.3 3.10 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) =.40, P(D) =.30 and P(B AND D) =.20. a. Find P(B|D). b. Find P(D|B). c. Are B and D independent? d. Are B and D mutually exclusive? Example 3.11 In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn. The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes. • P(R) = 3 8. P(B) = |
5 8. P(R AND B) = 0. You cannot draw one card that is both red and blue. • P(E) = 3 8. There are three even-numbered cards, R2, B2, and B4. 196 Chapter 3 | Probability Topics • P(E|B = 2 5. There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 and B4. • P(B|E) = 2 3 are blue; B2 and B4.. There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, two • The events R and B are mutually exclusive because P(R AND B) = 0. • Let G = card with a number greater than 3. G = {B4, B5}. P(G) = 2 8. Let H = blue card numbered between one and four, inclusive. H = {B1, B2, B3, B4}. P(G|H) = 1 4. The only card in H that has a number greater than three is B4. Since 2 8 = 1 4, P(G) = P(G|H), which means that G and H are independent. 3.11 In a basketball arena, • 70 percent of the fans are rooting for the home team, • 25 percent of the fans are wearing blue, • 20 percent of the fans are wearing blue and are rooting for the away team, and • Of the fans rooting for the away team, 67 percent are wearing blue. Let A be the event that a fan is rooting for the away team. Let B be the event that a fan is wearing blue. Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive? Example 3.12 In a particular class, 60 percent of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75 percent have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent? The following probabilities are given in this example: • P(F) = 0.60; P( |
L) = 0.50 • P(F AND L) = 0.45 • P(L|F) = 0.75 NOTE The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(F|L) yet, so you cannot use the second condition. Solution 1 Check whether P(F AND L) = P(F)P(L). We are given that P(F AND L) = 0.45, but P(F)P(L) = (.60)(.50) =.30. The events of being female and having long hair are not independent because P(F AND L) does not equal P(F)P(L). Solution 2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 197 Check whether P(L|F) equals P(L). We are given that P(L|F) =.75, but P(L) =.50; they are not equal. The events of being female and having long hair are not independent. Interpretation of Results The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair. 3.12 Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street. • P(I) =.44 and P(F) =.55 • P(I AND F) = 0 because Mark will take only one route to work. What is the probability of P(I OR F)? Example 3.13 a. Toss one fair coin (the coin has two sides, H and T). The outcomes are ________. Count the outcomes. There are ________ outcomes. b. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5, or 6 dots on a side). The outcomes are ________. Count the outcomes. There are ________ outcomes. c. Multiply the two numbers of outcomes. The answer is ________. d. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in Part c is the number of outcomes (size of the sample space). List the outcomes. Hint—Two of the outcomes are H1 and |
T6. e. Event A = heads (H) on the coin followed by an even number (2, 4, 6) on the die. A = {________}. Find P(A). f. Event B = heads on the coin followed by a three on the die. B = {________}. Find P(B). g. Are A and B mutually exclusive? Hint—What is P(A AND B)? If P(A AND B) = 0, then A and B are mutually exclusive. h. Are A and B independent? Hint—Is P(A AND B) = P(A)P(B)? If P(A AND B) = P(A)P(B), then A and B are independent. If not, then they are dependent. Solution 3.13 a. H and T; 2 b. 1, 2, 3, 4, 5, 6; 6 c. 2(6) = 12 d. Make a systematic list of possible outcomes. Start by listing all possible outcomes when the coin shows tails (T). Then list the outcomes that are possible when the coin shows heads (H): T1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6 e. A = {H2, H4, H6}; P(A) = number of outcomes inA number of possible outcomes = 3 12 f. B = {H3}; P(B) = 1 12 g. Yes, because P(A AND B) = 0 198 Chapter 3 | Probability Topics h. P(A AND B) = 0. P(A)P(B) = ⎛ ⎝ ⎞ ⎠ 3 12 ⎛ ⎝ 1 12 ⎞ ⎠. P(A AND B) does not equal P(A)P(B), so A and B are dependent. 3.13 A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, and S the event of picking the white ball in the second drawing. a. Compute P(T). b. Compute P(T|F). c. Are T and F independent? |
d. Are F and S mutually exclusive? e. Are F and S independent? 3.3 | Two Basic Rules of Probability In calculating probability, there are two rules to consider when you are determining if two events are independent or dependent and if they are mutually exclusive or not. The Multiplication Rule If A and B are two events defined on a sample space, then P(A AND B) = P(B)P(A|B). This equation can be rewritten as P(A AND B) = P(B)P(A|B), the multiplication rule. If A and B are independent, then P(A|B) = P(A). In this special case, P(A AND B) = P(A|B)P(B) becomes P(A AND B) = P(A)P(B). A bag contains four green marbles, three red marbles, and two yellow marbles. Mark draws two marbles from the bag without replacement. The probability that he draws a yellow marble and then a green marble is P⎛ ⎝yellow and green⎞ ⎠ ⋅ P⎛ ⎝green | yellow⎞ ⎠ ⎝yellow⎞ ⋅ 4 8 ⎠ = P⎛ = 2 9 = 1 9 Notice that P⎛ ⎝green | yellow⎞ ⎠ = 4 8. After the yellow marble is drawn, there are four green marbles in the bag and eight marbles in all. The Addition Rule If A and B are defined on a sample space, then P(A OR B) = P(A) + P(B) − P(A AND B). Draw one card from a standard deck of playing cards. Let H = the card is a heart, and let J = the card is a jack. These events are not mutually exclusive because a card can be both a heart and a jack. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 199 = 13 52 P(H or J) = P(H) + P(J) − P(H and J) + 4 52 = 16 52 = 4 13 − 1 52 ≈.3077 If A and B are mutually exclusive, then P(A AND B) = 0. Then P( |
A OR B) = P(A) + P(B) − P(A AND B) becomes P(A OR B) = P(A) + P(B). Draw one card from a standard deck of playing cards. Let H = the card is a heart and S = the card is a spade. These events are mutually exclusive because a card cannot be a heart and a spade at the same time. The probability that the card is a heart or a spade is P(H or S) = P(H) + P(S) + 13 52 = 13 52 = 26 52 = 1 2 =.5 Example 3.14 Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska. • Klaus can only afford one vacation. The probability that he chooses A is P(A) =.6 and the probability that he chooses B is P(B) =.35. • P(A AND B) = 0 because Klaus can only afford to take one vacation. • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) =.6 +.35 =.95. Note that the probability that he does not choose to go anywhere on vacation must be.05. Example 3.15 Carlos plays college soccer. He makes a goal 65 percent of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) =.65. B = the event Carlos is successful on his second attempt. P(B) =.65. Carlos tends to shoot in streaks. The probability that he makes the second goal given that he made the first goal is.90. a. What is the probability that he makes both goals? Solution 3.15 a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) =.90: P(B AND A) = P(B|A) P(A) = (.90)(.65) =.585. Carlos makes the first and second goals with probability.585. b. What is the probability that Carlos makes either the first goal or the second goal? Solution 3.15 b. The problem is asking you to find P(A OR B). 200 Chapter 3 | Probability |
Topics P(A OR B) = P(A) + P(B) − P(A AND B) =.65 +.65 −.585 =.715 Carlos makes either the first goal or the second goal with probability.715. c. Are A and B independent? Solution 3.15 c. No, they are not, because P(B AND A) =.585. P(B)P(A) = (.65)(.65) =.423.423 ≠.585 = P(B AND A) So, P(B AND A) is not equal to P(B)P(A). d. Are A and B mutually exclusive? Solution 3.15 d. No, they are not because P(A and B) =.585. To be mutually exclusive, P(A AND B) must equal zero. 3.15 Helen plays basketball. For free throws, she makes the shot 75 percent of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) =.75. D = the event Helen makes the second shot. P(D) =.75. The probability that Helen makes the second free throw given that she made the first is.85. What is the probability that Helen makes both free throws? Example 3.16 A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Fortyseven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. a. What is the probability that the member is a novice swimmer? Solution 3.16 a. There are 150 members; 75 of these are advanced, and 47 of these are intermediate swimmers. So there are 150 − 75 − 47 = 28 novice swimmers. The probability that a randomly selected swimmer is a novice is 28 150. b. What is the probability that the member practices four times a week? Solution 3.16 b. 40 + 30 + 10 150 = 80 150 c. What is the probability that the member is an advanced swimmer and practices four times a week? This OpenStax book is available for free at http://cnx.org/content/col30309/1 |
.8 Chapter 3 | Probability Topics 201 Solution 3.16 c. There are 40 advanced swimmers who practice four times per week, so the probability is 40 150. d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and being an intermediate swimmer mutually exclusive? Why or why not? Solution 3.16 d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. e. Are being a novice swimmer and practicing four times a week independent events? Why or why not? Solution 3.16 e. No, these are not independent events. P(novice AND practices four times per week) =.0667 P(novice)P(practices four times per week) =.0996.0667 ≠.0996 3.16 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college are on their school's sports teams. Thirty of the seniors going directly to work are on their school's sports teams. Five of the seniors taking a gap year are on their schools sports teams. What is the probability that a senior is taking a gap year? Example 3.17 Felicity attends a school in Modesto, CA. The probability that Felicity enrolls in a math class is.2 and the probability that she enrolls in a speech class is.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is.25. Let M = math class, S = speech class, and M|S = math given speech. a. What is the probability that Felicity enrolls in math and speech? Find P(M AND S) = P(M|S)P(S). b. What is the probability that Felicity enrolls in math or speech classes? Find P(M OR S) = P(M) + P(S) − P(M AND S). c. Are M and S independent? Is P(M|S) = P(M)? d. Are M and S mutually exclusive? Is P(M AND S) = 0? Solution 3.17 a. P(M AND S) = P(M|S)P( |
S) =.25(.65) =.1625 b. P(M OR S) = P(M) + P(S) − P(M AND S) =.2 +.65 −.1625 =.6875 c. No, P(M|S) =.25 and P(M) =.2. d. No, P(M AND S) =.1625. 202 Chapter 3 | Probability Topics 3.17 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) =.40, P(D) =.30, and P(D|B) =.5. a. Find P(B AND D). b. Find P(B OR D). Example 3.18 Researchers are studying one particular type of disease that affects women more often than men. Studies show that about one woman in seven (approximately 14.3 percent) who live to be 90 will develop the disease. Suppose that of those women who develop this disease, a test is negative 2 percent of the time. Also suppose that in the general population of women, the test for the disease is negative about 85 percent of the time. Let B = woman develops the disease and let N = tests negative. Suppose one woman is selected at random. a. What is the probability that the woman develops the disease? What is the probability that woman tests negative? Solution 3.18 a. P(B) =.143; P(N) =.85 b. Given that the woman develops the disease, what is the probability that she tests negative? Solution 3.18 b. Among women who develop the disease, the test is negative 2 percent of the time, so P(N|B) =.02 c. What is the probability that the woman has the disease AND tests negative? Solution 3.18 c. P(B AND N) = P(B)P(N|B) = (.143)(.02) =.0029 d. What is the probability that the woman has the disease OR tests negative? Solution 3.18 d. P(B OR N) = P(B) + P(N) − P(B AND N) =.143 +.85 −.0029 =.9901 e. Are having the disease and testing negative independent events? Solution 3.18 e. No. P(N |
) =.85; P(N|B) =.02. So, P(N|B) does not equal P(N). f. Are having the disease and testing negative mutually exclusive? Solution 3.18 f. No. P(B AND N) =.0029. For B and N to be mutually exclusive, P(B AND N) must be zero. 3.18 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 203 The remainder are taking a gap year. Fifty of the seniors going to college are on their school's sports teams. Thirty of the seniors going directly to work are on their school's sports teams. Five of the seniors taking a gap year are on their school's sports teams. What is the probability that a senior is going to college and plays sports? Example 3.19 Refer to the information in Example 3.18. P = tests positive. a. Given that a woman develops the disease, what is the probability that she tests positive? Find P(P|B) = 1 − P(N|B). b. What is the probability that a woman develops the disease and tests positive? Find P(B AND P) = P(P|B)P(B). c. What is the probability that a woman does not develop the disease? Find P(B′) = 1 − P(B). d. What is the probability that a woman tests positive for the disease? Find P(P) = 1 − P(N). Solution 3.19 a. P(P|B) = 1 − P(N|B) = 1 −.02 =.98 b. P(B AND P) = P(P|B)P(B) =.98(.143) =.1401 c. P(B') = 1 − P(B) = 1 −.143 =.857 d. P(P) = 1 − P(N) = 1 −.85 =.15 3.19 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) =.40, P(D) =.30, and P(D|B) = |
.5. a. Find P(B′). b. Find P(D AND B). c. Find P(B|D). d. Find P(D AND B′). e. Find P(D|B′). 3.4 | Contingency Tables A two-way table provides a way of portraying data that can facilitate calculating probabilities. When used to calculate probabilities, a two-way table is often called a contingency table. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. We used two-way tables in Chapters 1 and 2 to calculate marginal and conditional distributions. These tables organize data in a way that supports the calculation of relative frequency and, therefore, experimental (empirical) probability. Later on, we will use contingency tables again, but in another manner. Example 3.20 Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data: 204 Chapter 3 | Probability Topics Speeding Violation in the Last Year No Speeding Violation in the Last Year Total Uses a cell phone while driving Does not use a cell phone while driving Total Table 3.3 25 45 70 280 405 685 305 450 755 The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755. Using the table, calculate the following probabilities: a. Find P(Person uses a cell phone while driving). b. Find P(Person had no violation in the last year). c. Find P(Person had no violation in the last year and uses a cell phone while driving). d. Find P(Person uses a cell phone while driving or person had no violation in the last year). e. Find P(Person uses a cell phone while driving given person had a violation in the last year). f. Find P(Person had no violation last year given person does not use a cell phone while driving). Solution 3.20 a. This is the same as the marginal distribution (Section 1.2). P⎛ ⎝Person uses a cell phone while driving⎞ ⎠ = number who use cell phones while driving number in study = 305 755 ≈.4040 b. The marginal distribution is P⎛ ⎝Person had no violation in the last year� |
� ⎠ = number who had no violation number in study = 685 755 ≈.9073. c. Find the number of participants who satisfy both conditions. P(Person had no violation in the last year AND uses a cell phone while driving) = number who had no violation AND uses cell phone while driving number in study = 280 755 ≈.3709 d. To find this probability, you need to identify how many participants use a cell phone while driving OR have no violation in the past year OR both. P⎛ ⎝Person uses a cell phone while driving OR had no violation in the last year⎞ ⎠ = 25 + 405 + 280 755 = 710 755 ≈.9404 e. This is a conditional probability. You are given that the person had no violation in the last year, so you need only consider the values in that column of data. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 205 (Person uses a cell phone while driving GIVEN the person had a violation in the last year) = number who used cell phone AND had a violation number in study who had a violation in the last year = 25 70 ≈.3571 f. For this conditional probability, consider only values in the row labeled “Does not use a cell phone while driving.” P⎛ ⎝Person had no violation last year GIVEN person does not use cell phone while driving⎞ ⎠ = 405 450 =.9 3.20 Table 3.4 shows the number of athletes who stretch before exercising and how many had injuries within the past year. Injury in Past Year No Injury in Past Year Total Stretches 55 Does not stretch 231 Total 286 Table 3.4 295 219 514 350 450 800 a. What is P(Athlete stretches before exercising)? b. What is P(Athlete stretches before exercising|no injury in the last year)? Example 3.21 Table 3.5 shows a random sample of 100 hikers and the areas of hiking they prefer. Sex The Coastline Near Lakes and Streams On Mountain Peaks Total Female 18 Male Total ___ ___ 16 ___ 41 Table 3.5 Hiking Area Preference ___ 14 ___ 45 55 ___ a. Complete the table. Solution 3.21 a. There are 45 females in the sample; 18 prefer the |
coastline and 16 prefer hiking near lakes and streams. So, we know there are 45 − 18 − 16 = 11 female students who prefer hiking on mountain peaks. Continue reasoning in this way to complete the table. 206 Chapter 3 | Probability Topics Sex The Coastline Near Lakes and Streams On Mountain Peaks Total Female 18 Male Total 16 34 16 25 41 Table 3.6 Hiking Area Preference 11 14 25 45 55 100 b. Are the events being female and preferring the coastline independent events? Let F = being female and let C = preferring the coastline. 1. Find P(F AND C). 2. Find P(F)P(C). Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent. Solution 3.21 b. 1. P(F AND C) = 18 100 =.18 2. P(F)P(C) = ⎛ ⎝ 45 100 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 34 100 = (.45)(.34) =.153 P(F AND C) ≠ P(F)P(C), so the events F and C are not independent. c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams. 1. What word tells you this is a conditional? 2. Is the sample space for this problem all 100 hikers? If not, what is it? 3. Fill in the blanks and calculate the probability: P(_____|_____) = _____. Solution 3.21 c. 1. The word given tells you that this is a conditional. 2. No, the sample space for this problem is the 41 hikers who prefer lakes and streams. 3. Find the conditional probability P(M|L). Because it is given that the person prefers hiking near lakes and streams, you need only consider the values in the column labeled "Near Lakes and Streams." P(M|L) = 25 41 d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks. 1. Find P(F). 2. Find P(P). 3. Find P(F AND P). 4. Find P(F OR P). This OpenStax book is available |
for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 207 Solution 3.21 d. 1. P(F) = 45 100 2. P(P) = 25 100 3. P(F AND P) = number of hikers that are both female AND prefers mountain peaks number of hikers in study = 11 100 4. P(F OR P) = P(F) + P(P) − P(F AND P) = 45 100 + 25 100 - 11 100 = 59 100 3.21 Table 3.7 shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path. Gender Lake Path Hilly Path Wooded Path Total Female Male Total 45 26 71 Table 3.7 38 52 90 27 12 39 110 90 200 a. Out of the males, what is the probability that the cyclist prefers a hilly path? b. Are the events being male and preferring the hilly path independent events? 208 Chapter 3 | Probability Topics Example 3.22 Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1 5 and the probability he is not caught is 4 5. If he goes out the second door, the probability he gets caught by Alissa is 1 4 and the probability he is not caught is 3 4. The probability that Alissa catches Muddy coming out of the third door is 1 2 and the probability she does not catch Muddy is 1 2 Muddy will choose any of the three doors, so the probability of choosing each door is 1 3. It is equally likely that. Caught or Not Door One Door Two Door Three Total Caught Not Caught 1 15 4 15 1 12 3 12 1 6 1 6 Total ____ ____ ____ ____ ____ 1 Table 3.8 Door Choice • The first entry 1 15 = ⎛ ⎝ is P(Door One AND Caught). • The entry 4 15 = ⎛ ⎝ Verify the remaining entries. is P(Door One AND Not Caught). a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. Solution 3.22 a. Caught or Not Door One Door Two Door Three Total Caught Not Caught Total 1 15 4 15 5 15 Table 3.9 Door |
Choice 1 12 3 12 4 12 1 6 1 6 2 6 19 60 41 60 1 b. What is the probability that Alissa does not catch Muddy? Solution 3.22 b. 41 60 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 209 c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa? Solution 3.22 c. This is a conditional probability, so consider only probabilities in the row labeled "Caught." Choosing Door One and choosing Door Two are mutually exclusive, so P⎛ ⎝Choosing Door One OR Choosing Door Two AND Caught ⎞ ⎠ = 1 15 + 1 12 = 9 60. Use the formula for conditional probability P(A|B) = P(AANDB) P(B). P⎛ ⎝Door One OR Door Two|Caught ⎞ ⎠ = P⎛ ⎝Door One OR Door Two AND Caught ⎞ ⎝Caught⎞ P⎛ ⎠ ⎠ = 9 60 19 60 = 9 19.. Example 3.23 Table 3.10 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the United States. Year Crime A Crime B Crime C Crime D Total 2008 145.7 2009 133.1 2010 119.3 2011 113.7 Total 732.1 717.7 701 702.2 29.7 29.1 27.7 26.8 314.7 259.2 239.1 229.6 Table 3.10 U.S. Crime Index Rates Per 100,000 Inhabitants 2008–2011 TOTAL each column and each row. Total data = 4,520.7. a. Find P(2009 AND Crime A). b. Find P(2010 AND Crime B). c. Find P(2010 OR Crime B). d. Find P(2011|Crime A). e. Find P(Crime D|2008). Solution 3.23 a. 133.1 4,520.7 B) = 1,087.1 4,520.7 =.0294, b. 701 4,520.7 =.1551, c. P(2010 OR Crime B) = P(2010) + P(Crime B) – P |
(2010 AND Crime + 2,852.9 4,520.7 − 701 4,520.7 =.7165, d. 113.7 511.8 =.2222, e. 314.7 1,222.2 =.2575 3.23 Table 3.11 relates the weights and heights of a group of individuals participating in an observational study. 210 Chapter 3 | Probability Topics Ages Tall Medium Short Totals 28 51 25 14 28 9 Under 18 18 20 12 18–50 51+ Totals Table 3.11 a. Find the total for each row and column. b. Find the probability that a randomly chosen individual from this group is tall. c. Find the probability that a randomly chosen individual from this group is Under 18 and tall. d. Find the probability that a randomly chosen individual from this group is tall given that the individual is Under 18. e. Find the probability that a randomly chosen individual from this group is Under 18 given that the individual is tall. f. Find the probability a randomly chosen individual from this group is tall and age 51+. g. Are the events under 18 and tall independent? 3.5 | Tree and Venn Diagrams Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities. Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of branches that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram: Example 3.24 In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. With replacement means that you put the first ball back in the urn before you select the second ball. Therefore, you are selecting from exactly the same group each time, so each draw is independent. The tree diagram shows all the possible outcomes. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 211 Figure 3.10 Total = 64 + 24 + 24 + 9 = 121. The first set of branches represents the first draw. There are 8 ways to draw a blue marble and 3 ways to draw a red one. The |
second set of branches represents the second draw. Regardless of the choice on the first draw, there are again eight ways to draw a blue marble and 3 ways to draw a red one. Read down each branch to see the total number of possible outcomes. For example, there are 8 ways to get a blue marble on the first draw, and eight ways to get one on the second draw, so there are 8 × 8 = 64 different ways to draw two blue marbles in succession. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as follows: R1R1, R1R2, R1R3, R2R1, R2R2, R2R3, R3R1, R3R2, R3R3. The other outcomes are similar. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space. a. List the 24 BR outcomes: B1R1, B1R2, B1R3,... Solution 3.24 a. We know that there will be 24 different possible outcomes because there are eight ways to draw blue and three ways to draw red. Make a systematic list of possible outcomes that consist of a blue marble on the first draw and a red marble on the second draw. B1R1, B1R2, B1R3 B2R1, B2R2, B2R3 B3R1, B3R2, B3R3 B4R1, B4R2, B4R3 B5R1, B5R2, B5R3 B6R1, B6R2, B6R3 B7R1, B7R2, B7R3 B8R1, B8R2, B8R3 b. Calculate P(RR). Solution 3.24 b. You can use the tree diagram. There are nine ways to draw two reds and 121 possible outcomes. So, P(RR) = 9 121.. Each draw is independent, so you can also use the formula: P |
(RR) = P(R)P(R) = ⎛ ⎝ 3 11 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 3 11 = 9 121. 212 Chapter 3 | Probability Topics c. Calculate P(RB OR BR). Solution 3.24 c. The tree diagram shows that there are 24 ways to draw RB and 24 ways to draw BR. There are 121 possible outcomes, so P(RB or BR) = 24 + 24. = 48 121 121 The events RB and BR are mutually exclusive, so P(RB OR BR) = P(RB) + P(BR) = P(R)P(B) + P(B)P(R 11 8 11 8 11 3 11 = 48 121. d. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw). Solution 3.24 d. Follow the path on the tree. There are three ways to get a red marble on the first draw and eight ways to get a blue on the second draw. There are 3 × 8 = 24 ways to draw red then blue, so P(RB) = 24 121. Can you think of another way to find this probability? P(R on 1st draw AND B on 2nd draw) = P(RB) = ⎛ ⎝ 3 11 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ 8 11 = 24 121 e. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw). Solution 3.24 e. Given that a blue marble is selected first, we need only follow the left set of branches on the tree diagram. In this case, there are three ways to obtain red on the second draw and 11 possible outcomes. Figure 3.11 P(R on 2nd draw GIVEN B on 1st) = P(R on 2nd | B on 1st) = 3 11 You can also use the formula This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 213 P(R on 2nd | B on 1st) = P(R on 2nd AND B on 1st) P(B on 1st) = 24 121 64 + 24 121 = 24 88 = 3 11. f. Using the tree diagram, calculate P |
(BB). Solution 3.24 f. P(BB) = 64 121 g. Using the tree diagram, calculate P(B on the 2nd draw GIVEN R on the first draw). Solution 3.24 g. P(B on 2nd draw|R on 1st draw) = 8 11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. Twenty-four of the 33 outcomes have B on the second draw. The probability is then 24 33. 3.24 In a standard deck, there are 52 cards. Twelve cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF). Figure 3.12 Example 3.25 An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. Without replacement means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two ⎞ corresponding branches, for example, P(RR) = ⎛ ⎠ = 6 ⎝ 110 3 11 2 10 ⎛ ⎞ ⎠ ⎝. 214 Chapter 3 | Probability Topics Figure 3.13 Total = 56 + 24 + 24 + 6 110 = 110 110 = 1. NOTE If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are 10 marbles left in the urn. Calculate the following probabilities using the tree diagram: a. P(RR) = ________ Solution 3.25 a. P(RR) = ⎛ 3 ⎝ 11 ⎛ ⎞ ⎠ ⎝ 2 10 ⎞ ⎠ = 6 110 b. Fill in the blanks. P(RB OR BR) = ⎛ ⎝ 3 11 ⎛ ⎞ � |
�� ⎝ 8 10 ⎞ ⎠ + (________)(________) = 48 110 Solution 3.25 b. P(RB OR BR) = P(RB) + P(BR) = P(R on 1st) P(B on 2nd) + P(B on 1st) P(R on 2nd) = ⎛ ⎝ 3 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 8 10 + ⎛ ⎝ 8 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 3 10 = 48 110 c. Because this is a conditional probability, we restrict the sample space to consider only those outcomes that have a blue marble in the first draw. Look at the second level of the tree to see that P(R on 2nd|B on 1st) = 3 10 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 3 | Probability Topics 215 Solution 3.25 c. P(R on 2nd|B on 1st) = 3 10 d. Fill in the blanks. P(R on 1st AND B on 2nd) = P(RB) = (________)(________) = 24 100 Solution 3.25 d. P(R on 1st AND B on 2nd) = P(RB) = ⎛ ⎝ 3 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 8 10 = 24 100 e. Find P(BB). Solution 3.25 e. P(BB) = ⎛ 8 ⎝ 11 ⎛ ⎞ ⎠ ⎝ ⎞ ⎠ 7 10 f. Find P(B on 2nd|R on 1st). Solution 3.25 f. Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = 8 10. If we are using probabilities, we can label the tree in the following general way: • P(R|R) here means P(R on 2nd|R on 1st) • P(B|R) here means P(B on 2nd|R on 1st) • P(R|B) here means P(R on 2nd|B on 1st) • P( |