Geometric Mean The mean (Arithmetic), median and mode are all measures of the “center” of the data, the “average”. They are all in their own way trying to measure the “common” point within the data, that which is “normal”. In the case of the arithmetic mean this is solved by finding the value from which all points are equal linear distances. We can imagine that all the data values are combined through addition and then distributed back to each data point in equal amounts. The sum of all the values is what is redistributed in equal amounts such that the total sum remains the same. The geometric mean redistributes not the sum of the values but the product of multiplying all the individual values and then redistributing them in equal portions such that the total product remains the same. This can be seen from the formula for the geometric mean, x ~ : (Pronounced x-tilde) x ~ = ( ∏ i = 1 n x i ) 1 n = x 1 · x 2 ··· x n n = ( x 1 · x 2 ··· x n ) 1 n where π is another mathematical operator, that tells us to multiply all the x i numbers in the same way capital Greek sigma tells us to add all the x i numbers. Remember that a fractional exponent is calling for the nth root of the number thus an exponent of 1/3 is the cube root of the number. The geometric mean answers the question, "if all the quantities had the same value, what would that value have to be in order to achieve the same product?” The geometric mean gets its name from the fact that when redistributed in this way the sides form a geometric shape for which all sides have the same length. To see this, take the example of the numbers 10, 51.2 and 8. The geometric mean is the product of multiplying these three numbers together (4,096) and taking the cube root because there are three numbers among which this product is to be distributed. Thus the geometric mean of these three numbers is 16. This describes a cube 16x16x16 and has a volume of 4,096 units. The geometric mean is relevant in Economics and Finance for dealing with growth: growth of markets, in investment, population and other variables the growth in which there is an interest. Imagine that our box of 4,096 units (perhaps dollars) is the value of an investment after three years and that the investment returns in percents were the three numbers in our example. The geometric mean will provide us with the answer to the question, what is the average rate of return: 16 percent. The arithmetic mean of these three numbers is 23.6 percent. The reason for this difference, 16 versus 23.6, is that the arithmetic mean is additive and thus does not account for the interest on the interest, compound interest, embedded in the investment growth process. The same issue arises when asking for the average rate of growth of a population or sales or market penetration, etc., knowing the annual rates of growth. The formula for the geometric mean rate of return, or any other growth rate, is: r s = ( x 1 · x 2 ··· x n ) 1 n - 1 Manipulating the formula for the geometric mean can also provide a calculation of the average rate of growth between two periods knowing only the initial value a 0 and the ending value a n and the number of periods, n . The following formula provides this information: ( a n a 0 ) 1 n = x ~ Finally, we note that the formula for the geometric mean requires that all numbers be positive, greater than zero. The reason of course is that the root of a negative number is undefined for use outside of mathematical theory. There are ways to avoid this problem however. In the case of rates of return and other simple growth problems we can convert the negative values to meaningful positive equivalent values. Imagine that the annual returns for the past three years are +12%, -8%, and +2%. Using the decimal multiplier equivalents of 1.12, 0.92, and 1.02, allows us to compute a geometric mean of 1.0167. Subtracting 1 from this value gives the geometric mean of +1.67% as a net rate of population growth (or financial return). From this example we can see that the geometric mean provides us with this formula for calculating the geometric (mean) rate of return for a series of annual rates of return: r s = x ~ - 1 where r s is average rate of return and x ~ is the geometric mean of the returns during some number of time periods. Note that the length of each time period must be the same. As a general rule one should convert the percent values to its decimal equivalent multiplier. It is important to recognize that when dealing with percents, the geometric mean of percent values does not equal the geometric mean of the decimal multiplier equivalents and it is the decimal multiplier equivalent geometric mean that is relevant. Formula Review The Geometric Mean: x ~ = ( ∏ i = 1 n x i ) 1 n = x 1 · x 2 ··· x n n = ( x 1 · x 2 ··· x n ) 1 n What is the geometric mean of the data set given? 5, 10, 20 10 What is the geometric mean of the data set given? 9.000, 15.00, 21.00 14.15 What is the geometric mean of the data set given? 7.0, 10.0, 39.2 14 What is the geometric mean of the data set given? 17.00, 10.00, 19.00 14.78 What is the average rate of return for the values that follow? 1.0, 2.0, 1.5 44% What is the average rate of return for the values that follow? 0.80, 2.0, 5.0 100% What is the average rate of return for the values that follow? 0.90, 1.1, 1.2 6% What is the average rate of return for the values that follow? 4.2, 4.3, 4.5 33% Homework An investment grows from $10,000 to $22,000 in five years. What is the average rate of return? 17% An initial investment of $20,000 grows at a rate of 9% for five years. What is its final value? $30,772.48 A culture contains 1,300 bacteria. The bacteria grow to 2,000 in 10 hours. What is the rate at which the bacteria grow per hour to the nearest tenth of a percent? 4.4% An investment of $3,000 grows at a rate of 5% for one year, then at a rate of 8% for three years. What is the average rate of return to the nearest hundredth of a percent? 7.24% An investment of $10,000 goes down to $9,500 in four years. What is the average return per year to the nearest hundredth of a percent? -1.27%

Math


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 following histogram. Each interval has width one, and each value is located in the middle of an interval. The histogram in 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 in 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 side. A distribution of this type is called skewed to the left because it is pulled out to the left. We can formally measure the skewness of a distribution just as we can mathematically measure the center weight of the data or its general "speadness". The mathematical formula for skewness is: a 3 = ∑ ( x i − x – ) 3 n s 3 . The greater the deviation from zero indicates a greater degree of skewness. If the skewness is negative then the distribution is skewed left as in . A positive measure of skewness indicates right skewness such as . 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 histogram for the data: 6 7 7 7 7 8 8 8 9 10 shown in , is also not symmetrical. It is skewed to the right . 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. The mean is affected by outliers that do not influence the median. Therefore, when the distribution of data is skewed to the left, the mean is often less than the median. When the distribution is skewed to the right, the mean is often greater than the median. In symmetric distributions, we expect the mean and median to be approximately equal in value. This is an important connection between the shape of the distribution and the relationship of the mean and median. It is not, however, true for every data set. The most common exceptions occur in sets of discrete data. As with the mean, median and mode, and as we will see shortly, the variance, there are mathematical formulas that give us precise measures of these characteristics of the distribution of the data. Again looking at the formula for skewness we see that this is a relationship between the mean of the data and the individual observations cubed. a 3 = ∑ ( x i − x – ) 3 n s 3 where s is the sample standard deviation of the data, X i , and x ¯ is the arithmetic mean and n is the sample size. Formally the arithmetic mean is known as the first moment of the distribution. The second moment we will see is the variance, and skewness is the third moment. The variance measures the squared differences of the data from the mean and skewness measures the cubed differences of the data from the mean. While a variance can never be a negative number, the measure of skewness can and this is how we determine if the data are skewed right of left. The skewness for a normal distribution is zero, and any symmetric data should have skewness near zero. Negative values for the skewness indicate data that are skewed left and positive values for the skewness indicate data that are skewed right. By skewed left, we mean that the left tail is long relative to the right tail. Similarly, skewed right means that the right tail is long relative to the left tail. The skewness characterizes the degree of asymmetry of a distribution around its mean. While the mean and standard deviation are dimensional quantities (this is why we will take the square root of the variance ) that is, have the same units as the measured quantities X i , the skewness is conventionally defined in such a way as to make it nondimensional . It is a pure number that characterizes only the shape of the distribution. A positive value of skewness signifies a distribution with an asymmetric tail extending out towards more positive X and a negative value signifies a distribution whose tail extends out towards more negative X. A zero measure of skewness will indicate a symmetrical distribution. Skewness and symmetry become important when we discuss probability distributions in later chapters. Chapter Review 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 left (or negative) skewed distribution has a shape like . A right (or positive) skewed distribution has a shape like . A symmetrical distribution looks like . Formula Review Formula for skewness: a 3 = ∑ ( x i − x ¯ ) 3 n s 3 Formula for Coefficient of Variation: C V = s x ¯ · 100 conditioned upon x ¯ ≠ 0 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. 1 1 1 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 5 5 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. 16 17 19 22 22 22 22 22 23 87 87 87 87 87 88 89 89 90 91 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. When the data are skewed left, what is the typical relationship between the mean and median? When the data are symmetrical, what is the typical relationship between the mean and median? When the data are symmetrical, the mean and median are close or the same. What word describes a distribution that has two modes? Describe the shape of this distribution. The distribution is skewed right because it looks pulled out to the right. Describe the relationship between the mode and the median of this distribution. Describe the relationship between the mean and the median of this distribution. The mean is 4.1 and is slightly greater than the median, which is four. Describe the shape of this distribution. Describe the relationship between the mode and the median of this distribution. The mode and the median are the same. In this case, they are both five. Are the mean and the median the exact same in this distribution? Why or why not? Describe the shape of this distribution. The distribution is skewed left because it looks pulled out to the left. Describe the relationship between the mode and the median of this distribution. Describe the relationship between the mean and the median of this distribution. The mean and the median are both six. 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? 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 The mode is 12, the median is 12.5, and the mean is 15.1. The mean is the largest. 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 Of the three measures, which tends to reflect skewing the most, the mean, the mode, or the median? Why? The mean tends to reflect skewing the most because it is affected the most by outliers. In a perfectly symmetrical distribution, when would the mode be different from the mean and median? Homework The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years. What does it mean for the median age to rise? Give two reasons why the median age could rise. For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not?

Math


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 the data are all 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; wait times at supermarket A are more concentrated near the average. Calculating the Standard Deviation If x is a number, then the difference " x minus the 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 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 lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) 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 σ 2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s 2 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. Formally, the variance is the second moment of the distribution or the first moment around the mean. Remember that the mean is the first moment of the distribution. 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 = Σ ( x − x – ) 2 n − 1 or s = Σ f ( x − x – ) 2 n − 1 or s = ( ∑ i = 1 n x 2 ) - n x ¯ 2 n - 1 For the sample standard deviation, the denominator is n - 1 , 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 or σ = ∑ i = 1 N x i 2 N - μ 2 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. Two important observations concerning the variance and standard deviation: the deviations are measured from the mean and the deviations are squared. In principle, the deviations could be measured from any point, however, our interest is measurement from the center weight of the data, what is the "normal" or most usual value of the observation. Later we will be trying to measure the "unusualness" of an observation or a sample mean and thus we need a measure from the mean. The second observation is that the deviations are squared. This does two things, first it makes the deviations all positive and second it changes the units of measurement from that of the mean and the original observations. If the data are weights then the mean is measured in pounds, but the variance is measured in pounds-squared. One reason to use the standard deviation is to return to the original units of measurement by taking the square root of the variance. Further, when the deviations are squared it explodes their value. For example, a deviation of 10 from the mean when squared is 100, but a deviation of 100 from the mean is 10,000. What this does is place great weight on outliers when calculating the variance. Types of Variability in Samples When trying to study a population, a sample is often used, 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 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. In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of the 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 Freq. Deviations Deviations 2 (Freq.)(Deviations 2 ) x f ( x – x – ) ( x – x – ) 2 ( f )( x – x – ) 2 9 1 9 – 10.525 = –1.525 (–1.525) 2 = 2.325625 1 × 2.325625 = 2.325625 9.5 2 9.5 – 10.525 = –1.025 (–1.025) 2 = 1.050625 2 × 1.050625 = 2.101250 10 4 10 – 10.525 = –0.525 (–0.525) 2 = 0.275625 4 × 0.275625 = 1.1025 10.5 4 10.5 – 10.525 = –0.025 (–0.025) 2 = 0.000625 4 × 0.000625 = 0.0025 11 6 11 – 10.525 = 0.475 (0.475) 2 = 0.225625 6 × 0.225625 = 1.35375 11.5 3 11.5 – 10.525 = 0.975 (0.975) 2 = 0.950625 3 × 0.950625 = 2.851875 The total is 9.7375 The sample variance, s 2 , is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s 2 = 9.7375 20 − 1 = 0.5125 The sample standard deviation s is equal to the square root of the sample variance: s = 0.5125 = 0.715891 , which is rounded to two decimal places, s = 0.72. Try It 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 0.97 and 0.47. 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 . (For , 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. By squaring the deviations we are placing an extreme penalty on observations that are far from the mean; these observations get greater weight in the calculations of the variance. We will see later on that the variance (standard deviation) plays the critical role in determining our conclusions in inferential statistics. We can begin now by using the standard deviation as a measure of "unusualness." "How did you do on the test?" "Terrific! Two standard deviations above the mean." This, we will see, is an unusually good exam grade. 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. This estimate requires us to use an estimate of the population mean rather than the actual population mean. Based on the theoretical mathematics that lies behind these calculations, dividing by ( n – 1) gives a better estimate of the population variance. 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 upon 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, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is 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. Use the following data (first exam scores) from Professor Dean's spring pre-calculus 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 Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. Calculate the following to one decimal place: The sample mean The sample standard deviation The median The first quartile The third quartile IQR See The sample mean = 73.5 The sample standard deviation = 17.9 The median = 73 The first quartile = 61 The third quartile = 90 IQR = 90 – 61 = 29 Data Frequency Relative frequency Cumulative relative frequency 33 1 0.032 0.032 42 1 0.032 0.064 49 2 0.065 0.129 53 1 0.032 0.161 55 2 0.065 0.226 61 1 0.032 0.258 63 1 0.032 0.29 67 1 0.032 0.322 68 2 0.065 0.387 69 2 0.065 0.452 72 1 0.032 0.484 73 1 0.032 0.516 74 1 0.032 0.548 78 1 0.032 0.580 80 1 0.032 0.612 83 1 0.032 0.644 88 3 0.097 0.741 90 1 0.032 0.773 92 1 0.032 0.805 94 4 0.129 0.934 96 1 0.032 0.966 100 1 0.032 0.998 (Why isn't this value 1? ANSWER: Rounding) Try It The following data show the different types of pet food 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: M e a n o f F r e q u e n c y T a b l e = ∑ 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 is compared to the mean. Find the standard deviation for the data in . Class Frequency, f Midpoint, m f · m f ( m - x ¯ ) 2 0–2 1 1 1 · 1 = 1 1 ( 1 - 6.88 ) 2 = 34.57 3–5 6 4 6 · 4 = 24 6 ( 4 - 6.88 ) 2 = 49.77 6-8 10 7 10 · 7 = 70 10 ( 7 - 6.88 ) 2 = 0.14 9-11 7 10 7 · 10 = 70 7 ( 10 - 6.88 ) 2 = 68.14 12-14 0 13 0 · 13 = 0 0 ( 13 - 6.88 ) 2 = 0 n = 24 x ¯ = 165 24 = 6.88 s 2 = 152.62 24 - 1 = 6.64 For this data set, we have the mean, x – = 6.88 and the standard deviation, s x = 2.58. This means that a randomly selected data value would be expected to be 2.58 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost three standard deviations from the mean. While the formula for calculating the standard deviation is not complicated, s x = Σ ( m − x – ) 2 f n − 1 where s x = sample standard deviation, x – = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations. Try It Find the standard deviation for the data from the previous example Class Frequency, f 0–2 1 3–5 6 6–8 10 9–11 7 12–14 0 15–17 2 Comparing Values from Different Data Sets The standard deviation is useful when comparing data values that come 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 x, calculate how many standard deviations away from its mean the value is. Use the formula: x = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs. # o f S T D E V s = x – mean standard deviation Compare the results of this calculation. #ofSTDEVs is often called a " z -score"; we can use the symbol z . In symbols, the formulas become: Sample x = x ¯ + zs z = x − x – s Population x = μ + zσ z = x − μ σ 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 2.85 3.0 0.7 Ali 77 80 10 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 S T D E V s = 2.85 - 3.0 0.7 = - 0.21 For Ali, z = # o f S T D E V s = 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 0.3 standard deviations below his school's mean. John's z -score of –0.21 is higher than Ali's z -score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. Try It 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 26.2 27.2 0.8 Beth 27.3 30.1 1.4 For Angie: z = 26 .2 – 27 .2 0 .8 = –1.25 For Beth: z = 27 .3 – 30 .1 1. 4 = –2 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: At least 75% of the data is within two standard deviations of the mean. At least 89% of the data is within three standard deviations of the mean. At least 95% of the data is within 4.5 standard deviations of the mean. This is known as Chebyshev's Rule. For data having a Normal Distribution, which we will examine in great detail later: Approximately 68% of the data is within one standard deviation of the mean. Approximately 95% of the data is within two standard deviations of the mean. More than 99% 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 only applies 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. Coefficient of Variation Another useful way to compare distributions besides simple comparisons of means or standard deviations is to adjust for differences in the scale of the data being measured. Quite simply, a large variation in data with a large mean is different than the same variation in data with a small mean. To adjust for the scale of the underlying data the Coefficient of Variation (CV) has been developed. Mathematically: C V = s x – * 100 conditioned upon x – ≠ 0, where s is the standard deviation of the data and x – is the mean. We can see that this measures the variability of the underlying data as a percentage of the mean value; the center weight of the data set. This measure is useful in comparing risk where an adjustment is warranted because of differences in scale of two data sets. In effect, the scale is changed to common scale, percentage differences, and allows direct comparison of the two or more magnitudes of variation of different data sets. References Data from Microsoft Bookshelf. King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013). Chapter Review The standard deviation can help you calculate the spread of data. There are different equations to use if 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 – ) 2 n − 1 or s = ∑  f ( x − x – ) 2 n − 1 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 s x = ∑ f m 2 n − x – 2 where s x = sample standard deviation x – = sample mean Formulas for Sample Standard Deviation s = Σ ( x − x – ) 2 n − 1 or s = Σ f ( x − x – ) 2 n − 1 or s = ( ∑ i = 1 n x 2 ) - n x ¯ 2 n - 1 For the sample standard deviation, the denominator is n - 1 , that is the sample size - 1. Formulas for Population Standard Deviation σ = Σ ( x − μ ) 2 N or σ = Σ f ( x – μ ) 2 N or σ = ∑ i = 1 N x i 2 N - μ 2 For the population standard deviation, the denominator is N , the number of items in the population. 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 Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth. s = 34.5 Find the value that is one standard deviation below the mean. 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 0.158 0.166 0.012 Karl 0.177 0.189 0.015 For Fredo: z = 0.158 – 0.166 0.012 = –0.67 For Karl: z = 0.177 – 0.189 0.015 = –0.8 Fredo’s z -score of –0.67 is higher than Karl’s z -score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team. Use to find the value that is three standard deviations: a above the mean b below the mean Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84 . Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84. 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 Daily low temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 Points per game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2 s x = ∑ f m 2 n − x – 2 = 193157.45 30 − 79.5 2 = 10.88 s x = ∑ f m 2 n − x – 2 = 380945.3 101 − 60.94 2 = 7.62 s x = ∑ f m 2 n − x – 2 = 440051.5 86 − 70.66 2 = 11.14 Homework 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 over a 29-year period. μ = 1000 FTES median = 1,014 FTES σ = 474 FTES first quartile = 528.5 FTES third quartile = 1,447.5 FTES n = 29 years A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer. 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 6th number in order. Six years will have totals at or above the median. 75% of all years have an FTES: at or below: _____ at or above: _____ The population standard deviation = _____ 474 FTES What percent of the FTES were from 528.5 to 1447.5? How do you know? What is the IQR ? What does the IQR represent? 919 How many standard deviations away from the mean is the median? Additional Information: The population for a specific six-year period was given in an updated report. The data are reported here. Year 1 2 3 4 5 6 Total FTES 1,585 1,690 1,735 1,935 2,021 1,890 Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR . Round to one decimal place. mean = 1,809.3 median = 1,812.5 standard deviation = 151.2 first quartile = 1,690 third quartile = 1,935 IQR = 245 Compare the IQR for the FTES for the previous 29-year period with the IQR for the six-year period shown in . Why do you suppose the IQR s are so different? Hint: Think about the number of years covered by each time period and what happened to higher education during those periods. 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 2.7 3.2 0.8 Vichet 87 75 20 Kamala 8.6 8 0.4 A music school has budgeted to purchase three musical instruments. They plan 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. For pianos, the cost of the piano is 0.4 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. 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 1 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. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he? Who is the fastest runner with respect to their class? Explain why. A selection of countries has poverty rates that range from 11.4% to 74.6%. This data is summarized in Table 14 . Percent of population Number of countries 11.4–20.45 29 20.45–29.45 13 29.45–38.45 4 38.45–47.45 0 47.45–56.45 2 56.45–65.45 1 65.45–74.45 0 74.45–83.45 1 What is the best estimate of the average poverty percentage for these countries? What is the standard deviation for the listed poverty rates? The United States has an average poverty rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ poverty rate compared to the average rate? Explain. x – = 23.32 Using the TI 83/84, we obtain a standard deviation of: s x = 12.95. The poverty rate of the United States is 10.58% higher than the average poverty rate. Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the poverty percentage that is one standard deviation from the mean. The United States poverty rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States does not have an unusually high percentage of people experiencing poverty. gives the percent of children under five considered to be underweight. Percent of underweight children Number of countries 16–21.45 23 21.45–26.9 4 26.9–32.35 9 32.35–37.8 7 37.8–43.25 6 43.25–48.7 1 What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain. Bringing It Together Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: # of movies Frequency 0 5 1 9 2 6 3 4 4 1 Find the sample mean x ¯ . Find the approximate sample standard deviation, s . 1.48 1.12 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 1 2 2 5 3 8 4 12 5 12 6 0 7 1 Find the sample mean x ¯ Find the sample standard deviation, s Construct a histogram of the data. Complete the columns of the chart. Find the first quartile. Find the median. Find the third quartile. What percent of the students owned at least five pairs? Find the 40 th percentile. Find the 90 th percentile. Construct a line graph of the data Construct a stemplot of the data Following are the published weights (in pounds) of all of the 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 Organize the data from smallest to largest value. Find the median. Find the first quartile. Find the third quartile. The middle 50% of the weights are from _______ to _______. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why? Assume the population was the San Francisco 49ers. Find: the population mean, μ . the population standard deviation, σ . the weight that is two standard deviations below the mean. When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he? That same year, the mean weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer? 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; 302 241 205.5 272.5 205.5, 272.5 sample 236.34 37.50 161.34 0.84 std. dev. below the mean Young 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 What is the mean change score? What is the standard deviation for this population? What is the median change score? Find the change score that is 2.2 standard deviations below the mean. Refer to determine which of the following are true and which are false. Explain your solution to each part in complete sentences. The medians for both graphs are the same. We cannot determine if any of the means for both graphs is different. The standard deviation for graph b is larger than the standard deviation for graph a. We cannot determine if any of the third quartiles for both graphs is different. True True True False 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. Organize the data in a chart. Find the median, the first quartile, and the third quartile. Find the 65 th percentile. Find the 10 th percentile. The middle 50% of the conferences last from _______ days to _______ days. Calculate the sample mean of days of engineering conferences. Calculate the sample standard deviation of days of engineering conferences. Find the mode. 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. Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences. A survey of enrollment at 35 community colleges across the United States yielded the following figures: 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622 Organize the data into a chart with six intervals of equal width from 0 to 30,000. Label the two columns "Enrollment" and "Frequency." Construct a histogram of the data. If you were to build a new community college, which piece of information would be more valuable: the mode or the mean? Calculate the sample mean. Calculate the sample standard deviation. A school with an enrollment of 8000 would be how many standard deviations away from the mean? Enrollment Frequency 1000-5000 10 5000-10000 16 10000-15000 3 15000-20000 3 20000-25000 1 25000-30000 2 Answers may vary. mode 8628.74 6943.88 –0.09 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 3 1 12 2 33 3 28 4 11 5 9 6 4 The 80 th percentile is _____ 5 80 3 4 The number that is 1.5 standard deviations BELOW the mean is approximately _____ 0.7 4.8 –2.8 Cannot be determined a 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 the . # of books Freq. Rel. Freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 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? Are any data values further than two standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.) Do parts a and c of this problem give the same answer? Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data? Based on the shape of the data which is the most appropriate measure of center for this data: mean, median or mode? 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. 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 – is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.

Math


Introduction Meteor showers are rare, but the probability of them occurring can be calculated. (credit: modification of work “A meteor during the peak of the 2009 Leonid Meteor Shower” by Navicore/ Wikimedia Commons, CC BY 3.0) 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.

Math


Terminology Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment. A result of an experiment is called an outcome . The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = { H , T } where H = heads and T = tails are the outcomes. An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P ( A ). The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). 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. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads). Equally likely means that each outcome of an experiment occurs with 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. 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. A = getting one head. There are two outcomes that meet this condition { HT , TH }, so P ( A ) = 2 4 = 0.5. 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 6 . If you were to roll the die only a few times, you would not be 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 . 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. This important characteristic of probability experiments is known as the law of large numbers (discussed in detail later), 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 relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.) 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% of the time and a tail was obtained 44% 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 . The event A OR B can also be written as A Union B, with notation as follows: A ∪ B For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. Then A ∪ 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. The event A AND B can also be written as A Intersection B , with notation as follows: A ∩ B For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A ∩ B = {4, 5}. 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 ). P ( A | B ) is the probability that event A will occur given that the event B has already occurred. A conditional 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 ∩ 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 = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P ( A | B ), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S ). We get the same result by using the formula. Remember that S has six outcomes. P ( A | B ) = P ( A ∩ 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 ) 6 = 1 6 3 6 = 1 3 Odds The odds of an event presents the probability as a ratio of success to failure. This is common in various gambling formats. Mathematically, the odds of an event can be defined as: P ( A ) 1 − P ( A ) where P ( A ) is the probability of success and of course 1 − P ( A ) is the probability of failure. Odds are always quoted as "numerator to denominator," e.g. 2 to 1. Here the probability of winning is twice that of losing; thus, the probability of winning is 0.66. A probability of winning of 0.60 would generate odds in favor of winning of 3 to 2. While the calculation of odds can be useful in gambling venues in determining payoff amounts, it is not helpful for understanding probability or statistical theory. 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. The sample space S is the whole numbers starting at one and less than 20. S = _____________________________ Let event A = the even numbers and event B = numbers greater than 13. A = _____________________, B = _____________________ P ( A ) = _____________, P ( B ) = ________________ A ∩ B = ____________________, A OR B = ________________ P ( A ∩ B ) = _________, P ( A ∪ B ) = _____________ A′ = _____________, P ( A′ ) = _____________ P ( A ) + P ( A′ ) = ____________ P ( A | B ) = ___________, P ( B | A ) = _____________; are the probabilities equal? S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} A = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19} P ( A ) = 9 19 , P ( B ) = 6 19 A ∩ B = {14,16,18}, A OR B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19} P ( A ∩ B ) = 3 19 , P ( A ∪ B ) = 12 19 A′ = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; P ( A′ ) = 10 19 P ( A ) + P ( A′ ) = 1 ( 9 19 + 10 19 = 1) P ( A | B ) = P ( A ∩ B ) P ( B ) = 3 6 , P ( B | A ) = P ( A ∩ B ) P ( A ) = 3 9 , No Try It 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)). S = _____________________________ Let event A = the sum is even and event B = the first number is prime. A = _____________________, B = _____________________ P ( A ) = _____________, P ( B ) = ________________ A ∩ B = ____________________, A ∪ B = ________________ P ( A ∩ B ) = _________, P ( A ∪ B ) = _____________ B′ = _____________, P ( B′ ) = _____________ P ( A ) + P ( A′ ) = ____________ P ( A | B ) = ___________, P ( B | A ) = _____________; are the probabilities equal? S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)} B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} P ( A ) = 1 2 , P ( B ) = 2 3 A ∩ B = {(2,2), (2,4), (3,1), (3,3)} A ∪ B = {(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} P ( A ∩ B ) = 1 3 , P ( A ∪ B ) = 5 6 B′ = {(1,1), (1,2), (1,3), (1,4)}, P ( B′ ) = 1 3 P ( B ) + P ( B′ ) = 1 P ( A | B ) = P ( A ∩ B ) P ( B ) = 1 2 , P ( B | A ) = P ( A ∩ B ) P ( B ) = 2 3 , No. A fair, six-sided die is rolled. Describe the sample space S , identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up). Event T = the outcome is two. Event A = the outcome is an even number. Event B = the outcome is less than four. The complement of A . A | B B | A A ∩ B A ∪ B A ∪ B′ Event N = the outcome is a prime number. Event I = the outcome is seven. T = {2}, P ( T ) = 1 6 A = {2, 4, 6}, P ( A ) = 1 2 B = {1, 2, 3}, P ( B ) = 1 2 A′ = {1, 3, 5}, P ( A′ ) = 1 2 A | B = {2}, P ( A | B ) = 1 3 B | A = {2}, P ( B | A ) = 1 3 A ∩ B = {2}, P ( A ∩ B ) = 1 6 A ∪ B = {1, 2, 3, 4, 6}, P ( A ∪ B ) = 5 6 A ∪ B′ = {2, 4, 5, 6}, P ( A ∪ B′ ) = 2 3 N = {2, 3, 5}, P ( N ) = 1 2 A six-sided die does not have seven dots. P (7) = 0. Try It A number is randomly selected between 1 to 10. Describe the sample space S , identify each of the following events with a subset of S, and compute its probability (an outcome is the number selected). Event F = the outcome is 5. Event A = the number is more than 6. Event B = the number is odd. The complement of B . A GIVEN B . B GIVEN A . A OR B . A’ OR B . Event P = the outcome is a composite number. Event M = the number is a multiple of 3. F = 5 , P ( F ) = 1 10 A = 7 , 8 , 9 , P ( A ) = 3 10 B = 1 , 3 , 5 , 7 , 9 , P ( B ) = 5 10 B ' = 2 , 4 , 6 , 8 , 10 , P ( B ' ) = 5 10 A | B = 7 , 9 , P A | B = 2 5 B | A = 7 , 9 , P ( B | A ) = 1 2 A O R B = 1 , 3 , 5 , 7 , 8 , 9 , 10 , P ( A O R B ) = 7 10 A ' O R B = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 9 , P ( A ' O R B ) = 4 5 P = 4 , 6 , 8 , 9 , 10 , P ( P ) = 1 2 M = 3 , 6 , 9 , P ( M ) = 3 10 describes the distribution of a random sample S of 100 individuals, organized by sex assigned at birth and whether they are right- or left-handed. Right-handed Left-handed Males 43 9 Females 44 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: P ( M ) P ( F ) P ( R ) P ( L ) P ( M ∩ R ) P ( F ∩ L ) P ( M ∪ F ) P ( M ∪ R ) P ( F ∪ L ) P ( M' ) P ( R | M ) P ( F | L ) P ( L | F ) P ( M ) = 0.52 P ( F ) = 0.48 P ( R ) = 0.87 P ( L ) = 0.13 P ( M ∩ R ) = 0.43 P ( F ∩ L ) = 0.04 P ( M ∪ F ) = 1 P ( M ∪ R ) = 0.96 P ( F ∪ L ) = 0.57 P ( M' ) = 0.48 P ( R | M ) = 0.8269 (rounded to four decimal places) P ( F | L ) = 0.3077 (rounded to four decimal places) P ( L | F ) = 0.0833 Try It The table describes the distribution of a random sample S of 100 individuals, organized by gender and whether they like tea or coffee. Tea Coffee Men 22 26 Women 16 36 Let's denote the events M = the subject is a man, W = the subject is a woman, T = the subject likes tea, and C = the subject likes coffee. Compute the following probabilities: P ( M ) P ( W ) P ( T ) P ( C ) P ( M AND T ) P ( W and C ) P ( M OR W ) P ( M OR T ) P ( W OR C ) P ( M ') P ( T | M ) P ( W | C ) P ( C | W ) P ( M ) = 0 . 48 P ( W ) = 0 . 52 P ( T ) = 0 . 38 P ( C ) = 0 . 62 P ( M A N D T ) = 0 . 22 P ( W a n d C ) = 0 . 36 P ( M O R W ) = 1 P ( M O R T ) = 0 . 64 P ( W O R C ) = 0 . 78 P ( M ' ) = 0 . 52 P ( T | M ) = 0 . 4583 (rounded to four decimal places) P ( W | C ) = 0 . 5806 (rounded to four decimal places) P ( C | W ) = 0 . 6923 (rounded to four decimal places) References “Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013). Chapter Review In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive. Formula Review A and B are events P ( S ) = 1 where S is the sample space 0 ≤ P ( A ) ≤ 1 P ( A | B ) = P ( A ∩ B ) P ( B ) In a particular college class, there are both men and women students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) Let W be the event that a student is a woman. Let M be the event that a student is a man. Let S be the event that a student has short hair. Let L be the event that a student has long hair. The probability that a student does not have long hair. The probability that a student is a man or has short hair. The probability that a student is a woman and has long hair. The probability that a student is a man, given that the student has long hair. The probability that a student has long hair, given that the student is a man. Of all the women students, the probability that a student has short hair. Of all students with long hair, the probability that a student is a woman. The probability that a student is a woman or has long hair. The probability that a randomly selected student is a man with short hair. The probability that a student is a woman. P ( L′ ) = P ( S ) P ( M ∪ S ) P ( W ∩ L ) P ( M | L ) P ( L | M ) P ( S | W ) P ( W | L ) P ( W ∪ L ) P ( M ∩ S ) P ( W ) Use the following information to answer the next four exercises. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti. One party favor is chosen from the box at random. Let H = the event of getting a hat. Let N = the event of getting a noisemaker. Let F = the event of getting a finger trap. Let C = the event of getting a bag of confetti. Find P ( H ). Find P ( N ). P ( N ) = 15 42 = 5 14 = 0.36 Find P ( F ). Find P ( C ). P ( C ) = 5 42 = 0.12 Use the following information to answer the next six exercises. A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. One jelly bean is chosen from the box at random. Let B = the event of getting a blue jelly bean Let G = the event of getting a green jelly bean. Let O = the event of getting an orange jelly bean. Let P = the event of getting a purple jelly bean. Let R = the event of getting a red jelly bean. Let Y = the event of getting a yellow jelly bean. Find P ( B ). Find P ( G ). P ( G ) = 20 150 = 2 15 = 0.13 Find P ( P ). Find P ( R ). P ( R ) = 22 150 = 11 75 = 0.15 Find P ( Y ). Find P ( O ). P ( O ) = 150 - 22 - 38 - 20 - 28 - 26 150 = 16 150 = 8 75 = 0.11 Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region). Let A = the event that a country is in Asia. Let E = the event that a country is in Europe. Let F = the event that a country is in Africa. Let N = the event that a country is in North America. Let O = the event that a country is in Oceania. Let S = the event that a country is in South America. Find P ( A ). Find P ( E ). P ( E ) = 47 194 = 0.24 Find P ( F ). Find P ( N ). P ( N ) = 23 194 = 0.12 Find P ( O ). Find P ( S ). P ( S ) = 12 194 = 6 97 = 0.06 What is the probability of drawing a red card in a standard deck of 52 cards? What is the probability of drawing a club in a standard deck of 52 cards? 13 52 = 1 4 = 0.25 What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six? What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six? 3 6 = 1 2 = 0.5 Use the following information to answer the next two exercises. You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area. Let B = the event of landing on blue. Let R = the event of landing on red. Let G = the event of landing on green. Let Y = the event of landing on yellow. If you land on Y , you get the biggest prize. Find P ( Y ). If you land on red, you don’t get a prize. What is P ( R )? P ( R ) = 4 8 = 0.5 Use the following information to answer the next ten exercises. On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let I = the event that a player in an infielder. Let O = the event that a player is an outfielder. Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter. Write the symbols for the probability that a player is not an outfielder. Write the symbols for the probability that a player is an outfielder or is a great hitter. P ( O ∪ H ) Write the symbols for the probability that a player is an infielder and is not a great hitter. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder. P ( H | I ) Write the symbols for the probability that a player is an infielder, given that the player is a great hitter. Write the symbols for the probability that of all the outfielders, a player is not a great hitter. P ( N | O ) Write the symbols for the probability that of all the great hitters, a player is an outfielder. Write the symbols for the probability that a player is an infielder or is not a great hitter. P ( I ∪ N ) Write the symbols for the probability that a player is an outfielder and is a great hitter. Write the symbols for the probability that a player is an infielder. P ( I ) What is the word for the set of all possible outcomes? What is conditional probability? The likelihood that an event will occur given that another event has already occurred. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book Let F = event that book is fiction Let N = event that book is nonfiction What is the sample space? What is the sum of the probabilities of an event and its complement? 1 Use the following information to answer the next two exercises. You are rolling a fair, six-sided number cube. Let E = the event that it lands on an even number. Let M = the event that it lands on a multiple of three. What does P ( E | M ) mean in words? What does P ( E ∪ M ) mean in words? the probability of landing on an even number or a multiple of three Homework The graph in displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045. Define three events in the graph. Describe in words what the entry 40 means. Describe in words the complement of the entry in question 2. Describe in words what the entry 30 means. Out of the men and women, what percent are men? Out of the women, what percent disapprove of Mayor Ford? Out of all the age groups, what percent approve of Mayor Ford? Find P (Approve | Men). Out of the age groups, what percent are more than 44 years old? Find P (Approve | Age < 35). Explain what is wrong with the following statements. Use complete sentences. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100% A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs. Conditional Probability the likelihood that an event will occur given that another event has already occurred Equally Likely Each outcome of an experiment has the same probability. Event a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by S . An event is an arbitrary subset in S . It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as A , B , C , and so on. Experiment a planned activity carried out under controlled conditions Outcome a particular result of an experiment Probability a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let S denote the sample space and A and B are two events in S . Then: 0 ≤ P ( A ) ≤ 1 If A and B are any two mutually exclusive events, then P ( A ∪ B ) = P ( A ) + P (B). P ( S ) = 1 Sample Space the set of all possible outcomes of an experiment Intersection: the ∩ Event An outcome is in the event A ∩ B if the outcome is in both A ∩ B at the same time. Conditional Probability of A | B P ( A | B ) is the probability that event A will occur given that the event B has already occurred. Union: the ∪ Event An outcome is in the event A ∪ B if the outcome is in A or is in B or is in both A and B .

Math


Independent and Mutually Exclusive Events Independent and mutually exclusive do not mean the same thing. Independent Events Two events are independent if one of the following are true: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A ∩ B ) = P ( A ) P ( B ) Two events A and B are independent 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 . 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. 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. If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise . 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. 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. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten 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, ten 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. The probability of picking the three of diamonds is called a conditional probability because it is conditioned on what was picked first. This is true also of the probability of picking the J of spades. The probability of picking the J of spades is actually conditioned on both the previous picks. Try It 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. 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? 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? With replacement No 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 you pick four cards, but do not put any cards back into the deck. Your cards are QS , 1 D , 1 C , QD . Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH , 7 D , 6 D , KH . Which of a. or b. did you sample with replacement and which did you sample without replacement? a. Without replacement; b. With replacement Try It 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. QS , 1 D , 1 C , QD KH , 7 D , 6 D , KH QS , 7 D , 6 D , KS without replacement: 1. Possible; 2. Impossible, 3. Possible with replacement: 1. Possible; 2. Possible, 3. Possible Mutually Exclusive Events A and B are mutually exclusive events if they cannot occur at the same time. Said another way, If A occurred then B cannot occur and vise-a-versa. This means that A and B do not share any outcomes and P ( A ∩ 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 ∩ B = {4, 5}. P ( A ∩ 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 ∩ 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. 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 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 ∩ 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 Try It Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. Try It Solutions The sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is { BB , BR , RB , RR }. Event A = Getting at least one black card = { BB , BR , RB } P ( A ) = 3 4 = 0.75 Flip two fair coins. Find the probabilities of the events. Let F = the event of getting at most one tail (zero or one tail). Let G = the event of getting two faces that are the same. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. Are F and G mutually exclusive? Let J = the event of getting all tails. Are J and H mutually exclusive? Look at the sample space in . Zero (0) or one (1) tails occur when the outcomes HH , TH , HT show up. P ( F ) = 3 4 Two faces are the same if HH or TT show up. P ( G ) = 2 4 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 F and G share HH so P ( F ∩ G ) is not equal to zero (0). F and G are not mutually exclusive. 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 ∩ H ) = 0. J and H are mutually exclusive. Try It 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: Let F = the event of getting the white ball twice. Let G = the event of getting two balls of different colors. Let H = the event of getting white on the first pick. Are F and G mutually exclusive? Are G and H mutually exclusive? P ( F ) = 1 4 P ( G ) = 1 2 P ( H ) = 1 2 Yes No 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 ∩ 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? No. C = {3, 5} and E = {1, 2, 3, 4}. P ( C ∩ E ) = 1 6 . To be mutually exclusive, P ( C ∩ E ) must be zero. Find P ( C | A ) . This is a conditional probability. Recall that the 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 original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P ( C | A ) = 2 3 . Try It Let event A = learning Spanish. Let event B = learning German. Then A ∩ B = learning Spanish and German. Suppose P ( A ) = 0.4 and P ( B ) = 0.2 . P ( A ∩ B ) = 0.08 . Are events A and B independent? Hint: You must show ONE of the following: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) P ( A ∩ B ) = P ( A ) P ( B ) P ( A | B ) = P ( A ∩ B ) P ( B ) = 0. 08 0.2 = 0 .4 = P ( A ) The events are independent because P ( A | B ) = P ( A ). Let event G = taking a math class. Let event H = taking a science class. Then, G ∩ H = taking a math class and a science class. Suppose P ( G ) = 0.6 , P ( H ) = 0.5 , and P ( G ∩ H ) = 0.3 . 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 ∩ 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 ) . P ( G | H ) = P ( G ∩ H ) P ( H ) = 0 .3 0 .5 = 0.6 = P ( G ) b. Show P ( G ∩ H ) = P ( G ) P ( H ) . P ( G ) P ( H ) = ( 0.6 ) ( 0.5 ) = 0.3 = P ( G ∩ H ) Since G and H are independent, knowing that a person is taking a science class does not change the chance that they are 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 they are taking math. For practice, show that P ( H | G ) = P ( H ) to show that G and H are independent events. Try It 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. R = a red marble G = a green marble O = an odd-numbered marble The sample space is S = { R 1, R 2, R 3, R 4, R 5, R 6, G 1, G 2, G 3, G 4}. S has ten outcomes. What is P ( G ∩ O ) ? Event G and O = { G 1, G 3} P ( G ∩ O ) = 2 10 = 0.2 Let event C = taking an English class. Let event D = taking a speech class. Suppose P ( C ) = 0.75 , P ( D ) = 0.3 , P ( C | D ) = 0.75 and P ( C ∩ D ) = 0.225 . Justify your answers to the following questions numerically. Are C and D independent? Are C and D mutually exclusive? What is P ( D | C ) ? Yes, because P ( C | D ) = P ( C ) . No, because P ( C ∩ D ) is not equal to zero. P ( D | C ) = P ( C ∩ D ) P ( C ) = 0 .225 0.75 = 0.3 Try It 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 ) = 0.40 , P ( D ) = 0.30 and P ( B ∩ D ) = 0.20 . Find P ( B | D ) . Find P ( D | B ) . Are B and D independent? Are B and D mutually exclusive? P ( B | D ) = 0.6667 P ( D | B ) = 0.5 No No 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 = R 1, R 2, R 3, B 1, B 2, B 3, B 4, B 5. S has eight outcomes. P ( R ) = 3 8 . P ( B ) = 5 8 . P ( R ∩ B ) = 0 . (You cannot draw one card that is both red and blue.) P ( E ) = 3 8 . (There are three even-numbered cards, R 2, B 2, and B 4.) P ( E | B ) = 2 5 . (There are five blue cards: B 1, B 2, B 3, B 4, and B 5. Out of the blue cards, there are two even cards; B 2 and B 4.) P ( B | E ) = 2 3 . (There are three even-numbered cards: R 2, B 2, and B 4. Out of the even-numbered cards, to are blue; B 2 and B 4.) The events R and B are mutually exclusive because P ( R ∩ B ) = 0 . Let G = card with a number greater than 3. G = { B 4, B 5}. P ( G ) = 2 8 . Let H = blue card numbered between one and four, inclusive. H = { B 1, B 2, B 3, B 4}. P ( G | H ) = 1 4 . (The only card in H that has a number greater than three is B 4.) Since 2 8 = 1 4 , P ( G ) = P ( G | H ) , which means that G and H are independent. Try It In a basketball arena, 70% of the fans are rooting for the home team. 25% of the fans are wearing blue. 20% of the fans are wearing blue and are rooting for the away team. Of the fans rooting for the away team, 67% 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? P ( B | A ) = 0.67 P ( B ) = 0.25 So P ( B ) does not equal P ( B | A ) which means that B and A are not independent (wearing blue and rooting for the away team are not independent). They are also not mutually exclusive, because P ( B ∩ A ) = 0.20 , not 0. In a particular college class, 60% 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% 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 ∩ 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. Check whether P ( F ∩ L ) = P ( F ) P ( L ) . We are given that P ( F ∩ L ) = 0.45 , but P ( F ) P ( L ) = ( 0.60 ) ( 0.50 ) = 0.30 . The events of being female and having long hair are not independent because P ( F ∩ L ) does not equal P ( F ) P ( L ) . Check whether P ( L | F ) equals P ( L ) . We are given that P ( L | F ) = 0.75 , but P ( L ) = 0.50 ; 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. Try It Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street. P ( I ) = 0.44 and P ( F ) = 0.56 P ( I ∩ F ) = 0 because Mark will take only one route to work. What is the probability of P ( I ∪ F ) ? Because P ( I ∩ F ) = 0 , P ( I ∪ F ) = P ( I ) + P ( F ) − P ( I ∩ F ) = 0.44 + 0.56 − 0 = 1 Toss one fair coin (the coin has two sides, H and T ). The outcomes are ________. Count the outcomes. There are ____ outcomes. 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. Multiply the two numbers of outcomes. The answer is _______. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer to c is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H 1 and T 6.) Event A = heads ( H ) on the coin followed by an even number (2, 4, 6) on the die. A = {_________________}. Find P ( A ). Event B = heads on the coin followed by a three on the die. B = {________}. Find P ( B ). Are A and B mutually exclusive? (Hint: What is P ( A ∩ B ) ? If P ( A ∩ B ) = 0 , then A and B are mutually exclusive.) Are A and B independent? (Hint: Is P ( A ∩ B ) = P ( A ) P ( B ) ? If P ( A ∩ B ) = P ( A ) P ( B ) , then A and B are independent. If not, then they are dependent). H and T ; 2 1, 2, 3, 4, 5, 6; 6 2(6) = 12 T 1, T 2, T 3, T 4, T 5, T 6, H 1, H 2, H 3, H 4, H 5, H 6 A = { H 2, H 4, H 6}; P ( A ) = 3 12 B = { H 3}; P ( B ) = 1 12 Yes, because P ( A ∩ B ) = 0 P ( A ∩ B ) = 0 . P ( A ) P ( B ) = 3 12 · 1 12 = 1 48 . P ( A ∩ B ) does not equal P ( A ) P ( B ) , so A and B are dependent. Try It 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, S the event of picking the white ball in the second drawing. Compute P ( T ) . Compute P ( T | F ) . Are T and F independent?. Are F and S mutually exclusive? Are F and S independent? P ( T ) = 1 4 P ( T | F ) = 1 2 No No Yes References Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013). Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013). Chapter Review Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. Formula Review If A and B are independent, P ( A ∩ B ) = P ( A ) P ( B ) , P ( A | B ) = P ( A ) and P ( B | A ) = P ( B ) . If A and B are mutually exclusive, P ( A ∪ B ) = P ( A ) + P ( B ) and P ( A ∩ B ) = 0 . E and F are mutually exclusive events. P ( E ) = 0.4 ; P ( F ) = 0.5 . Find P ( E ∣ F ) . J and K are independent events. P ( J | K ) = 0.3 . Find P ( J ) . P ( J ) = 0.3 U and V are mutually exclusive events. P ( U ) = 0.26; P ( V ) = 0.37. Find: P ( U ∩ V ) = P ( U | V ) = P ( U ∪ V ) = Q and R are independent events. P ( Q ) = 0.4 and P ( Q ∩ R ) = 0.1 . Find P ( R ) . P ( Q ∩ R ) = P ( Q ) P ( R ) 0.1 = (0.4) P ( R ) P ( R ) = 0.25 Homework Use the following information to answer the next 12 exercises. In a certain year, Gallup interviewed more than 170,000 employed Americans 18 years of age or older. The graph shown is compiled from data collected by Gallup. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score. Find the probability that the selected occupation has an Emotional Health Index Score of 82.7. Find the probability that the selected occupation has an Emotional Health Index Score of 81.0. 0 Find the probability that the selected occupation has an Emotional Health Index Score more than 81. Find the probability that the selected occupation has an Emotional Health Index Score between 80.5 and 82. 0.3571 If we know an occupation has an Emotional Health Index Score of 81.5 or more, what is the probability that it is 82.7? What is the probability that an occupation has an Emotional Health Index Score of 80.7 or 82.7? 0.2142 What is the probability that an occupation has an Emotional Health Index Score less than 80.2 given that it is already less than 81? What occupation has the highest emotional index score? Physician (83.7) What occupation has the lowest emotional index score? What is the range of the data? 83.7 − 79.6 = 4.1 Compute the average EHIS. If all occupations are equally likely for a certain individual, what is the probability that they will have an occupation with lower than average EHIS? P (Occupation < 81.3) = 0.5 Bringing It Together A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data are compiled into . Shirt # ≤ 210 211–250 251–290 290≤ 1–33 21 5 0 0 34–66 6 18 7 4 66–99 6 12 22 5 For the following, suppose that you randomly select one player from the 49ers or Cowboys. If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P (Shirt# 1–33|≤ 210 pounds)? The probability that someone develops cancer during their lifetime is 0.4567. The probability that a person has at least one false positive test result (meaning the test comes back for cancer when the person does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a person develops cancer and P = person has at least one false positive. P ( C ) = ______ P ( P | C ) = ______ P ( P | C' ) = ______ If a test comes up positive, based upon numerical values, can you assume that person has cancer? Justify numerically and explain why or why not. P ( C ) = 0.4567 not enough information not enough information No, because over half (0.51) of men have at least one false positive text Given events G and H : P ( G ) = 0.43; P ( H ) = 0.26; P ( H ∩ G ) = 0.14 Find P ( H ∪ G ) . Find the probability of the complement of event ( H ∩ G ) . Find the probability of the complement of event ( H ∪ G ) . Given events J and K : P ( J ) = 0.18 ; P ( K ) = 0.37 ; P ( J ∪ K ) = 0.45 Find P ( J ∩ K ) . Find the probability of the complement of event ( J ∩ K ) . Find the probability of the complement of event ( J ∩ K ) . ( J ∪ K ) = P ( J ) + P ( K ) − P ( J ∩ K ) ; 0.45 = 0.18 + 0.37 − P ( J ∩ K ) ; solve to find P ( J ∩ K ) = 0.10 P ( NOT ( J ∩ K ) ) = 1 − P ( J ∩ K ) = 1 − 0.10 = 0.90 P ( NOT ( J ∪ K ) ) = 1 − P ( J ∪ K ) = 1 − 0.45 = 0.55

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