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: In this table, median household income (in $1000s) from a random sample of 100 counties that had population gains are shown on the left. Median incomes from a random sample of 50 counties that had no population gain are shown on the right. Population: Gain Population: No Gain | 2 |12 | 2 |79 4| 3 |1234 99999987555| 3... |
may vary a little. The counties with population gains tend to have higher income (median of about $45,000) versus counties without a gain (median of about $40,000). The variability is also slightly larger for the population gain group. This is evident in the IQR, which is about 50% bigger in the gain group. Both distr... |
that we could plot using dot plots, scatterplots, or box plots, but these miss the true nature of the data. Rather, when we encounter geographic data, we should create an intensity map, where colors are used to show higher and lower values of a variable. Figures 2.22 and 2.23 shows intensity maps for poverty rate in p... |
. SUMMARIZING DATA (a) (b) Figure 2.23: (a) Intensity map of homeownership rate (percent). (b) Intensity map of median household income ($1000s). Explore dozens of intensity maps using American Community Survey data on Tableau Public. <55%73%91%Homeownership Rate$19$47>$75Median Household Income 2.2. NUMERICAL SUMMARIE... |
rst quartile Q1. IQR = Q3 − Q1 • Range is also sometimes used as a measure of spread. The range of a data set is defined as the difference between the maximum value and the minimum value, i.e. max − min. • Outliers are observations that are extreme relative to the rest of the data. Two rules of thumb for identifying obse... |
78, 81, 88, 69, 77, 79 Draw a histogram of these data and describe the distribution. 2.9 Smoking habits of UK residents, Part II. A random sample of 5 smokers from the data set discussed in Exercise 2.7 is provided below. gender Female Male Female Female Female age maritalStatus 51 24 33 17 76 Married Single Married S... |
off, or those who have about the average number of days off? 2.12 Medians and IQRs. For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics; simply state how the medians and IQRs compare. Make sure to explain your reasoning. (a) (1) 3, 5, 6, 7, 9 (2... |
��ects might be a concern. The index is calculated for five major air pollutants regulated by the Clean Air Act and takes values from 0 to 300, where a higher value indicates lower air quality. AQI was reported for a sample of 91 days in 2011 in Durham, NC. The relative frequency histogram below shows the distribution o... |
males. 2.20 Distributions and appropriate statistics, Part II. For each of the following, state whether you expect the distribution to be symmetric, right skewed, or left skewed. Also specify whether the mean or median would best represent a typical observation in the data, and whether the variability of observations ... |
the two measures? 2.22 Midrange. The midrange of a distribution is defined as the average of the maximum and the minimum of that distribution. Is this statistic robust to outliers and extreme skew? Explain your reasoning (1)$60k$62.5k$65k$67.5k$70k04812(2)$60k$110k$160k$210k$260k04812 100 CHAPTER 2. SUMMARIZING DATA 2.... |
distribution where appropriate. 2. Find probabilities and percentiles using the normal approximation. 3. Find the value that corresponds to a given percentile when the distribution is approximately normal. 2.3.1 Normal distribution model Among all the distributions we see in practice, one is overwhelmingly the most co... |
shows the mean and standard deviation for total scores on the SAT and ACT. The distribution of SAT and ACT scores are both nearly normal. Suppose Ann scored 1300 on her SAT and Tom scored 24 on his ACT. Who performed better? As we saw in section 2.2.3, we can use Z-scores to compare observations from different distribu... |
curve It’s very useful in statistics to be able to identify areas of distributions, especially tail areas. For instance, what percent of people have an SAT score below Ann’s score of 1300? This is the same as Ann’s percentile. We previously determined that a score of 1300 corresponds to a Z-score of 1 and that SAT sco... |
to familiarize themselves with one of the options above before continuing on to the applications that follow. −3−2−10123 2.3. NORMAL DISTRIBUTION 105 2.3.4 Normal probability examples Combined SAT scores are approximated well by a normal model with mean 1100 and standard deviation 200. EXAMPLE 2.59 What is the probabi... |
0.6736” in Example 2.59. 70011001500 106 CHAPTER 2. SUMMARIZING DATA EXAMPLE 2.61 Edward earned a 1030 on his SAT. What is his percentile? First, a picture is needed. Edward’s percentile is the proportion of people who do not get as high as a 1030. These are the scores to the left of 1030. Identifying the mean µ = 1100... |
scored at the 72nd percentile on the SAT. What was her SAT score?46 IF THE DATA ARE NOT NEARLY NORMAL, DON’T USE THE NORMAL APPROXIMATION Before using the normal approximation method, verify that the data or distribution is approximately normal. If it is not, the normal approximation will give incorrect results. Also ... |
. Similarly there should be an area of about 0.95 between Z = −2 and Z = 2.47 It is possible for a normal random variable to fall 4, 5, or even more standard deviations from the mean. However, these occurrences are very rare if the data are nearly normal. The probability of being further than 4 standard deviations from... |
, as shown in the left panel of Figure 2.31. The sample mean ¯x and standard deviation s are used as the parameters of the best fitting normal curve. The closer this curve fits the histogram, the more reasonable the normal model assumption. Another more common method is examining a normal probability plot,49 shown in the... |
Practice 2.68. 2.3.7 Technology: finding normal probabilities Get started quickly with a Desmos Normal Calculator that we’ve put together (visit openintro.org/ahss/desmos). Height (inches)70758085llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll... |
set lower to -5. • If finding just an upper tail area, set upper to 5. 4. Leave µ as 0 and σ as 1. 5. Down arrow, choose Paste, and hit ENTER. TI-83: Do steps 1-2, then enter the lower bound and upper bound separated by a comma, e.g. normalcdf(2, 5), and hit ENTER. CASIO FX-9750GII: FINDING AREA UNDER THE NORMAL CURVE ... |
1 < Z < 1) = 0.6827. Similarly, P (−2 < Z < 2) = 0.9545 and P (−3 < Z < 3) = 0.9973. 52Lower bound is -1.5 and upper bound is 1.5. The area under the normal curve between -1.5 and 1.5 = P (−1.5 < Z < 1.5) = 0.866. Note that is not simply the average of 0.6827 and 0.9545, as the normal curve is not a rectangle. 53normal... |
253. This means that Z = −0.253 corresponds to the 40th percentile, that is, P (Z < −0.253) = 0.40. GUIDED PRACTICE 2.74 Find the Z-score such that 20 percent of the area is to the right of that Z-score.55 54Now we want to shade to the right. Therefore our lower bound will be 2 and the upper bound will be +5 (or a numb... |
DATA Exercises 2.25 Area under the curve, Part I. What percent of a standard normal distribution N (µ = 0, σ = 1) is found in each region? Be sure to draw a graph. (a) Z < −1.35 (b) Z > 1.48 (c) −0.4 < Z < 1.5 (d) |Z| > 2 2.26 Area under the curve, Part II. What percent of a standard normal distribution N (µ = 0, σ = ... |
), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups: • The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds... |
cutoff time for the fastest 5% of athletes in the men’s group, i.e. those who took the shortest 5% of time to finish. (b) The cutoff time for the slowest 10% of athletes in the women’s group. 2.31 LA weather, Part I. deviation of 5◦F. Suppose that the temperatures in June closely follow a normal distribution. (a) What is... |
.34 Find the SD. Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter (mg/dl). Women with cholesterol levels above 220 mg/dl are considered to have high cholesterol and about 18.5% of women fall into this category. What is the standard deviation o... |
segmented bar chart, or mosaic plot. 3. Calculate marginal and joint frequencies for two-way tables. 2.4.1 Contingency tables and bar charts Figure 2.34 summarizes two variables: app type and homeownership. A table that summarizes data for two categorical variables in this way is called a contingency table. Each value... |
by its row total, 8505. So what does 0.411 represent? It corresponds to the proportion of individual applicants who rent. individual joint Total rent mortgage 0.451 0.635 0.479 0.411 0.242 0.386 own Total 1.000 1.000 1.000 0.138 0.122 0.135 Figure 2.37: A contingency table with row proportions for the app type and hom... |
?57 EXAMPLE 2.77 Data scientists use statistics to filter spam from incoming email messages. By noting specific characteristics of an email, a data scientist may be able to classify some emails as spam or not spam with high accuracy. One such characteristic is whether the email contains no numbers, small numbers, or big ... |
For instance, in the email example, the email format was seen as a possible explanatory variable of whether the message was spam, so we would find it more interesting to compute the relative frequencies (proportions) for each email format. 56(a) 0.451 represents the proportion of individual applicants who have a mortga... |
20.40.60.81.0jointindividualProportionrentmortgageown0.00.20.40.60.81.0Proportionjointindividual 120 CHAPTER 2. SUMMARIZING DATA EXAMPLE 2.79 Examine the four bar charts in Figure 2.40. When is the segmented, side-by-side, standardized segmented bar chart, or standardized side-by-side the most useful? The segmented bar... |
Figure 2.41(a). Each column represents a level of homeownership, and the column widths correspond to the proportion of loans in each of those categories. For instance, there are fewer loans where the borrower is an owner than where the borrower has a mortgage. In general, mosaic plots use box areas to represent the nu... |
it is also difficult to decipher details in a pie chart. For example, it takes a couple seconds longer to recognize that there are more loans where the borrower has a mortgage than rent when looking at the pie chart, while this detail is very obvious in the bar chart. While pie charts can be useful, we prefer bar charts... |
political ideology are shown below.58 Political ideology Conservative Moderate Response (i) Apply for citizenship (ii) Guest worker (iii) Leave the country (iv) Not sure Total 57 121 179 15 372 120 113 126 4 363 Liberal Total 278 262 350 20 910 101 28 45 1 175 (a) What percent of these Tampa, FL voters identify themse... |
supportNot sureDemocratRepublicanIndep / OtherRaise taxes on the richRaise taxes on the poorNot sure 2.5. CASE STUDY: MALARIA VACCINE (SPECIAL TOPIC) 125 2.5 Case study: malaria vaccine (special topic) How large does an observed difference need to be for it to provide convincing evidence that something real is going on... |
ively. The results are summarized in Figure 2.44, where 9 of the 14 treatment patients remained free of signs of infection while all of the 6 patients in the control group patients showed some baseline signs of infection. 61We would be assuming that these two variables are independent. 126 CHAPTER 2. SUMMARIZING DATA t... |
-A”: H0: Independence model. The variables treatment and outcome are independent. They have no relationship, and the observed difference between the proportion of patients who developed an infection in the two groups, 64.3%, was due to chance. HA: Alternative model. The variables are not independent. The difference in in... |
infections were independent of the vaccine and we were able to rewind back to when the researchers randomized the patients in the study. If we happened to randomize the patients differently, we may get a different result in this hypothetical world where the vaccine doesn’t influence the infection. Let’s complete another ... |
distribution of these simulated differences is centered around 0. We simulated these differences assuming that the independence model was true, and under this condition, we expect the difference to be near zero with some random fluctuation, where near is pretty generous in this case since the sample sizes are so small in ... |
%, though it can depend upon the situation. Using the 5% cutoff, we would reject the independence model in favor of the alternative. That is, we are concluding the data provide strong evidence that the vaccine provides some protection against malaria in this clinical setting. When there is strong enough evidence that th... |
(a) Determine if each of the following statements is true or false. If false, explain why. Be careful: The In such cases, the statement reasoning may be wrong even if the statement’s conclusion is correct. should be considered false. i. Since more patients on pioglitazone had cardiovascular problems (5,386 vs. 2,593),... |
do the simulation results suggest about the relationship between taking rosiglitazone and having cardiovascular problems in diabetic patients? 64D.J. Graham et al. “Risk of acute myocardial infarction, stroke, heart failure, and death in elderly Medicare patients treated with rosiglitazone or pioglitazone”. In: JAMA 3... |
(treatment - control) and record this value. We repeat this 100 times to build a. Lastly, we calculate the fraction of simulations distribution centered at. If this fraction is low, where the simulated differences in proportions are we conclude that it is unlikely to have observed such an outcome by chance and that the... |
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a frequency table and with a graph, such as a stem-and-leaf plot or dot plot for small data sets, or a histogram for larger data sets. If only a summary is desired, a box plot may be used. • The distribution of a variable can be described and summarized with center (mean or median), spread (SD or IQR), and shape (righ... |
the distribution of estimated infant death rates for 224 countries for which such data were available in 2014.66 (a) Estimate Q1, the median, and Q3 from the histogram. (b) Would you expect the mean of this data set to be smaller or larger than the median? Explain your reasoning. 2.45 TV watchers. Students in an AP St... |
exam scores of twenty introductory statistics students. 57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94 Create a box plot of the distribution of these scores. The five number summary provided below may be useful. Min Q1 Q2 (Median) Q3 Max 57 72.5 78.5 82.5 94 2.50 Marathon winners. The hi... |
650. The article also states that 25% of California residents pay more than $1,800. (a) What is the Z-score that corresponds to the top 25% (or the 75th percentile) of the standard normal distribution? (b) What is the mean insurance cost? What is the cutoff for the 75th percentile? (c) Identify the standard deviation of... |
familiar to most people. For videos, slides, and other resources, please visit www.openintro.org/ahss 3.1. DEFINING PROBABILITY 137 3.1 Defining probability What is the probability of rolling an even number on a die? Of getting 5 heads in row when tossing a coin? Of drawing a Heart or an Ace from a deck of cards? The s... |
5/6. Alternatively, we could have noticed that not rolling a 2 is the same as getting a 1, 3, 4, 5, or 6, which makes up five of the six equally likely outcomes and has probability 5/6. 138 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.5 Consider rolling two dice. If 1/6th of the time the first die is a... |
relative frequency tends to get closer to the probability 1/6 ≈ 0.167 as the number of rolls increases. LAW OF LARGE NUMBERS As more observations are collected, the observed proportion ˆpn of occurrences with a particular outcome after n trials converges to the true probability p of that outcome. n (number of rolls)11... |
exclusive are equivalent and interchangeable. Calculating the probability of disjoint outcomes is easy. When rolling a die, the outcomes 1 and 2 are disjoint, and we compute the probability that one of these outcomes will occur by adding their separate probabilities: P (1 or 2) = P (1) + P (2) = 1/6 + 1/6 = 1/3 What a... |
joint outcomes to determine P (1 or 4 or 5).2 GUIDED PRACTICE 3.8 In the email data set in Chapter 2, the number variable described whether no number (labeled none), only one or more small numbers (small), or whether at least one big number appeared in an email (big). Of the 3,921 emails, 549 had no numbers, 2,827 had ... |
0.139. (c) 123456ABD 3.1. DEFINING PROBABILITY 141 GUIDED PRACTICE 3.9 (a) Verify the probability of event A, P (A), is 1/3 using the Addition Rule. (b) Do the same for event B.4 GUIDED PRACTICE 3.10 (a) Using Figure 3.2 as a reference, what outcomes are represented by event D? (b) Are events B and D disjoint? (c) Are... |
card, it will be in part of the left circle that is not in the right circle (and so on). The total number of cards that are diamonds is given by the total number of cards in the diamonds circle: 10 + 3 = 13. The probabilities are also shown (e.g. 10/52 = 0.1923). 4(a) P (A) = P (1 or 2) = P (1) + P (2) = 1 5(a) Outcom... |
B)? Events A and B are not disjoint – the cards J♦, Q♦, and K♦ fall into both categories – so we cannot use the Addition Rule for disjoint events. Instead we use the Venn diagram. We start by adding the probabilities of the two events: P (A) + P (B) = P (♦) + P (face card) = 13/52 + 12/52 However, the three cards that... |
.2308 3.1. DEFINING PROBABILITY 143 GUIDED PRACTICE 3.14 (a) If A and B are disjoint, describe why this implies P (A and B) = 0. (b) Using part (a), verify that the General Addition Rule simplifies to the simpler Addition Rule for disjoint events if A and B are disjoint.10 GUIDED PRACTICE 3.15 In the email data set with... |
0 (see part (a)) and we are left with the Addition Rule for disjoint events. 11Both the counts and corresponding probabilities (e.g. 2659/3921 = 0.678) are shown. Notice that the number of emails represented in the left circle corresponds to 2659 + 168 = 2827, and the number represented in the right circle is 168 + 19... |
of getting at least one 6 in 10 rolls of the die? (b) What is the complement of getting at most three 6’s in 10 rolls of the die?15 3.1.6 Independence Just as variables and observations can be independent, random processes can be independent, too. Two processes are independent if knowing the outcome of one provides no... |
) The complement of getting at most three 6’s in 10 rolls is getting four, five,..., nine, or ten 6’s in 10 rolls. All rolls1/6th of the firstrolls are a 1.1/6th of those times wherethe first roll is a 1 thesecond roll is also a 1. 3.1. DEFINING PROBABILITY 145 EXAMPLE 3.20 What if there was also a blue die independent ... |
-handed? 16(a) The probability the first person is left-handed is 0.09, which is the same for the second person. We apply the Multiplication Rule for independent processes to determine the probability that both will be left-handed: 0.09×0.09 = 0.0081. (b) It is reasonable to assume the proportion of people who are ambid... |
and right-handed? (b) What is the probability that the first two people are male and right-handed?. (c) What is the probability that the third person is female and left-handed? (d) What is the probability that the first two people are male and right-handed and the third person is female and left-handed? Sometimes we won... |
inclusive. • The probability of an event and the probability of its complement add up to 1. Sometime we use P (A) = 1 − P (not A) when P (not A) is easier to calculate than P (A). • A and B are disjoint, i.e. mutually exclusive, if they cannot happen together. In this case, the events do not overlap and P (A and B) = ... |
cards are mutually exclusive events. (c) Drawing a face card and drawing an ace from a full deck of playing cards are mutually exclusive events. 3.2 Roulette wheel. The game of roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a s... |
Your friend claims that you are cheating, because rolling double 6s twice in a row is very unlikely. Using probability, show that your rolls were just as likely as his. 3.5 Coin flips. If you flip a fair coin 10 times, what is the probability of (a) getting all tails? (b) getting all heads? (c) getting at least one tail... |
the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home? 3.9 Disjoint vs. independent. In parts (a) and (b), identify whether the events are disjoint, independent, or neither (events cannot be both disjoint and independent). (a) You and a randomly ... |
x P(teen went to college immediately after high school)? Explain why this is or is not the case. 3.12 School absences. Data collected at elementary schools in DeKalb County, GA suggest that each year roughly 25% of students miss exactly one day of school, 15% miss 2 days, and 28% miss 3 or more days due to sickness.24... |
of joint events. 4. Determine whether two events are independent and whether they are mutually exclusive, based on the definitions of those terms. 5. Draw a tree diagram with at least two branches to organize possible outcomes and their probabilities. Understand that the second branch represents conditional probabiliti... |
learn is pred not) = 112 1603 = 0.070 ML Predicts FashionFashion Photos0.110.010.06Neither: 0.82 3.2. CONDITIONAL PROBABILITY 153 3.2.2 Marginal and joint probabilities Figure 3.8 includes row and column totals for each variable separately in the photo classify data set. These totals represent marginal probabilities f... |
joint, all probabilities are non-negative, and the probabilities sum to 1.25 25Each of the four outcome combination are disjoint, all probabilities are indeed non-negative, and the sum of the probabilities is 0.1081 + 0.0121 + 0.0615 + 0.8183 = 1.00. 154 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS We can compu... |
, we computed the probability a photo was about fashion based on the condition that the ML algorithm predicted it was about fashion as a fraction: P (truth is fashion | mach learn is pred fashion) = = # cases where truth is fashion and mach learn is pred fashion # cases where mach learn is pred fashion 197 219 = 0.900 ... |
prediction was correct, if the photo was about fashion”. Here the condition is now based on the photo’s truth status, not the ML algorithm. (b) Determine the probability from part (a). Figure 3.10 on page 153 may be helpful.26 GUIDED PRACTICE 3.29 (a) Determine the probability that the algorithm is incorrect if it is ... |
and 3.13. result lived died Total inoculated yes 238 6 244 5136 844 5980 no Total 5374 850 6224 Figure 3.12: Contingency table for the smallpox data set. result lived died Total inoculated yes 0.0382 0.0010 0.0392 no 0.8252 0.1356 0.9608 Total 0.8634 0.1366 1.0000 Figure 3.13: Table proportions for the smallpox data, ... |
5 General multiplication rule Section 3.1.6 introduced the Multiplication Rule for independent processes. Here we provide the General Multiplication Rule for events that might not be independent. GENERAL MULTIPLICATION RULE If A and B represent two outcomes or events, then P (A and B) = P (A|B) × P (B) For the term P (... |
people who were inoculated died. 34The samples are large relative to the difference in death rates for the “inoculated” and “not inoculated” groups, so it seems there is an association between inoculated and outcome. However, as noted in the solution to Guided Practice 3.32, this is an observational study and we cannot... |
P (Q1 = not picked, Q2 = not picked, Q3 = not picked.) = 14 15 × 14 15 × 14 15 = 0.813 You have a slightly higher chance of not being picked compared to when she picked a new person for each question. However, you now may be picked more than once. 35The three probabilities we computed were actually one marginal probab... |
all three questions) = 1 15 37(a) First determine the probability of not winning. The tickets are sampled without replacement, which means the probability you do not win on the first draw is 29/30, 28/29 for the second,..., and 23/24 for the seventh. The probability you win no prize is the product of these separate pro... |
the chances of getting black six times in a row is very small (about 1/64) and puts his paycheck on red. What is wrong with his reasoning?40 3.2.8 Checking for independent and mutually exclusive events If A and B are independent events, then the probability of A being true is unchanged if B is true. Mathematically, th... |
whether equality holds in this equation. P (teen college | parent degree)?= P (teen college) 0.83 = 0.56 The value 0.83 came from a probability calculation using Figure 3.14: 231 280 ≈ 0.83. Because the sides are not equal, teenager college attendance and parent degree are dependent. That is, we estimate the probabili... |
Are teen college attendance and parent college degrees mutually exclusive? Looking in the table, we see that there are 231 instances where both the teenager attended college and parents have a degree, indicating the probability of both events occurring is greater than 0. Since we have found an example where both of th... |
may (and usually do) construct joint probabilities at the end of each branch in our tree by multiplying the numbers we come across as we move from left to right. These joint probabilities are computed using the General Multiplication Rule: P (inoculated and lived) = P (inoculated) × P (lived | inoculated) = 0.0392 × 0... |
probability: P (statement about variable 2 | statement about variable 1) For example, instead of wanting to know P (lived | inoculated), we might want to know P (inoculated | lived). This is more challenging because it cannot be read directly from the tree diagram. In these instances we use Bayes’ Theorem. Let’s begin... |
breast cancer, and “mammogram+” means the mammogram screening was positive, which in this case means the test suggests the patient has breast cancer. (Watch out for the non-intuitive medical language: a positive test result suggests the possible presence of cancer in a mammogram screening.) We can use the conditional ... |
side was broken down into a product of a conditional probability and marginal probability using the tree diagram. P (mammogram+) = P (mammogram+ and no BC) + P (mammogram+ and has BC) = P (mammogram+| no BC)P (no BC) + P (mammogram+| has BC)P (has BC) We can see an application of Bayes’ Theorem by substituting the res... |
visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35% of evenings, sporting events on 20% of evenings, and no events on 45% of evenings. When there is an academic event, the garage fills up about 25% of the time, and it fills up... |
a conditional probability, we are given some information. In an unconditional probability, such as P (A), we are not given any information. • Sometimes P (A|B) can be deduced. For example, when drawing without replacement from a deck of cards, P (2nd draw is an Ace | 1st draw was an Ace) = 3 51. When this is not the c... |
that P(A and B) = 0.1, are the random variables giving rise to events A and B indepen- dent? (d) If we are given that P(A and B) = 0.1, what is P(A|B)? 3.14 PB & J. Suppose 80% of people like peanut butter, 89% like jelly, and 78% like both. Given that a randomly sampled person likes peanut butter, what’s the probabil... |
distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance. Health Coverage No Yes Total Excellent Very good 0.0364 0.3123 0.3486 0.0230 0.2099 0.2329 Health Status Good 0.0427 0.2410 0.2838 Fair 0.0192 0.0817 0.1009 Poor 0.0050 0.... |
draw a chip and it is orange, and then you draw a second chip without replacement. What is the probability this second chip is blue? (c) If drawing without replacement, what is the probability of drawing two blue chips in a row? (d) When drawing without replacement, are the draws independent? Explain. 3.2. CONDITIONAL... |
) Calculate the probability that a student is able to construct a box plot if it is known that he passed. 3.24 Predisposition for thrombosis. A genetic test is used to determine if people have a predisposition for thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the flow of blood I... |
probability. With modern computing power, simulations have become an important and powerful tool for data scientists. In this section, we will look at the concepts that underlie simulations. Learning objectives 1. Understand the purpose of a simulation and recognize the application of the long-run relative frequency i... |
Appendix C.1 on page 509. 3.3. SIMULATIONS 173 EXAMPLE 3.54 Mika’s favorite brand of cereal is running a special where 20% of the cereal boxes contain a prize. Mika really wants that prize. If her mother buys 6 boxes of the cereal over the next few months, what is the probability Mika will get a prize? To solve this p... |
one has a 20% chance of containing a prize.45 GUIDED PRACTICE 3.56 In the previous example, the probability that a box of cereal contains a prize is 20%. The question presented is equivalent to asking, what is the probability of getting at least one prize in six randomly selected boxes of cereal. This probability ques... |
trial 1 we see 461, so we record 3. For trial 2 we see 3395641, so we record 7. For trial 3, we see 0, so we record 1. The rest of this exercise is left as a Guided Practice problem for you to complete. GUIDED PRACTICE 3.58 Finish the simulation above and report your estimate for the average number of boxes of cereal ... |
in this section to properly set up those simulations. The difference is in implementation after the setup. Rather than use a random number table, a data scientist will write a program that uses a pseudo-random number generator in a computer to run the simulations very quickly – often times millions of trials each secon... |
ulated) fleet passed? (a) Flip a coin seven times where each toss represents a car. A head means the car passed and a tail means it failed. If all cars passed, we report PASS for the fleet. If at least one car failed, we report FAIL. (b) Read across a random number table starting at line 5. If a number is a 0 or 1, let i... |
not 100% accurate. Suppose it is known that a lie detector has a failure rate of 25%. A thief will slip by the test 25% of the time and an honest employee will only pass 75% of the time. (a) Describe how you would simulate whether an employee is honest or is a thief using a random number table. Write your simulation v... |
or difference of random variables when those variables are independent. 3.4.1 Introduction to expected value EXAMPLE 3.62 Two books are assigned for a statistics class: a textbook and its corresponding study guide. The university bookstore determined 20% of enrolled students do not buy either book, 55% buy the textbook... |
distribution is a table of all disjoint outcomes and their associated probabili- ties. Figure 3.19 shows the probability distribution for the sum of two dice. RULES FOR PROBABILITY DISTRIBUTIONS A probability distribution is a list of the possible outcomes with corresponding probabilities that satisfies three rules: 1.... |
probability in (b) is negative. This leaves (c), which sure enough satisfies the requirements of a distribution. One of the three was said to be the actual distribution of US household incomes, so it must be (c). Dice sum234567891011120.000.050.100.15Probability 180 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS ... |
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