Statistics: The Art and Science of Learning From Data, Books a la Carte Edition (4th Edition)
4th Edition
ISBN: 9780133860825
Author: Alan Agresti, Christine A. Franklin, Bernhard Klingenberg
Publisher: PEARSON
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Chapter 5, Problem 83CP
To determine
Compute the
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Chapter 5 Solutions
Statistics: The Art and Science of Learning From Data, Books a la Carte Edition (4th Edition)
Ch. 5.1 - Probability Explain what is meant by the long-run...Ch. 5.1 - Testing a coin Your friend decides to flip a coin...Ch. 5.1 - Vegetarianism You randomly sample 10 people in...Ch. 5.1 - Airline accident deaths Airplane safety has been...Ch. 5.1 - World Cup 2014 The powerrank.com website...Ch. 5.1 - Random digits Consider a random number generator...Ch. 5.1 - Polls and sample size A pollster agency wants to...Ch. 5.1 - Heart transplant Before the first human heart...Ch. 5.1 - Nuclear war You are asked to use your best...Ch. 5.1 - Simulate coin flips Use the web app Random Numbers...
Ch. 5.1 - Unannounced pop quiz A teacher announces a pop...Ch. 5.2 - Student union poll Part of a student opinion poll...Ch. 5.2 - Random digit A single random digit (from 09) is...Ch. 5.2 - Pop quiz A teacher gives a four-question...Ch. 5.2 - More true-false questions Your teacher gives a...Ch. 5.2 - Rolling two dice Some friends are playing a game...Ch. 5.2 - Two girls A couple plans to have two children....Ch. 5.2 - Three children A couple plans to have three...Ch. 5.2 - Wrong sample space A couple plans to have four...Ch. 5.2 - Insurance Every year the insurance industry spends...Ch. 5.2 - Cell phone and case At the local cell phone store,...Ch. 5.2 - Seat belt use and auto accidents Based on records...Ch. 5.2 - Protecting the environment When the General Social...Ch. 5.2 - Prob. 25PBCh. 5.2 - Newspaper sales You are the director of newspaper...Ch. 5.2 - Prob. 27PBCh. 5.3 - Recidivism rates A 2014 article from Business...Ch. 5.3 - Prob. 29PBCh. 5.3 - Audit and low income Table 5.3 on audit status and...Ch. 5.3 - Religious affiliation The 2012 Statistical...Ch. 5.3 - Prob. 32PBCh. 5.3 - Prob. 33PBCh. 5.3 - Go Celtics! Larry Bird, who played pro basketball...Ch. 5.3 - Prob. 35PBCh. 5.3 - Homeland security According to an article in The...Ch. 5.3 - Down syndrome again Example 8 discussed the Triple...Ch. 5.3 - Prob. 38PBCh. 5.3 - Happiness in relationship Are people happy in...Ch. 5.3 - Prob. 40PBCh. 5.3 - Prob. 41PBCh. 5.3 - Prob. 42PBCh. 5.3 - Prob. 43PBCh. 5.3 - Prob. 44PBCh. 5.3 - Family with two children For a family with two...Ch. 5.3 - Prob. 46PBCh. 5.4 - Birthdays of presidents Of the first 44 presidents...Ch. 5.4 - Prob. 48PBCh. 5.4 - Lots of pairs Show that with 25 students, there...Ch. 5.4 - Prob. 50PBCh. 5.4 - Prob. 51PBCh. 5.4 - Prob. 52PBCh. 5.4 - Coincidence in your life State an event that has...Ch. 5.4 - Monkeys typing Shakespeare Since events of low...Ch. 5.4 - A true coincidence at DisneyWorld Wisconsin has...Ch. 5.4 - Prob. 56PBCh. 5.4 - Mammogram diagnostics Breast cancer is the most...Ch. 5.4 - Prob. 58PBCh. 5.4 - Prob. 59PBCh. 5.4 - Convicted by mistake In criminal trials (e.g.,...Ch. 5.4 - DNA evidence compelling? DNA evidence can be...Ch. 5.4 - Prob. 62PBCh. 5.4 - Simulating donations to local blood bank The...Ch. 5.4 - Probability of winning In Example 16, we estimated...Ch. 5 - Peyton Manning completions As of the end of the...Ch. 5 - Due for a boy? A couple has five children, all...Ch. 5 - Prob. 68CPCh. 5 - Choices for lunch For the set lunch at Amelias...Ch. 5 - Caught doctoring the books After the major...Ch. 5 - Life after death In a General Social Survey, in...Ch. 5 - Death penalty jury In arguing against the death...Ch. 5 - Drivers exam Three 15-year-old friends with no...Ch. 5 - Prob. 74CPCh. 5 - Health insurance According to a 2006 census bureau...Ch. 5 - Prob. 76CPCh. 5 - Prob. 77CPCh. 5 - Prob. 78CPCh. 5 - Prob. 79CPCh. 5 - Roll two more dice Refer to the previous exercise....Ch. 5 - Prob. 81CPCh. 5 - Waste dump sites A federal agency is deciding...Ch. 5 - Prob. 83CPCh. 5 - Prob. 84CPCh. 5 - Prob. 85CPCh. 5 - Death penalty and false positives For the decision...Ch. 5 - Prob. 87CPCh. 5 - Screening for heart attacks Biochemical markers...Ch. 5 - Prob. 89CPCh. 5 - Color blindness For genetic reasons, color...Ch. 5 - Prob. 91CPCh. 5 - Prostate cancer A study of the PSA blood test for...Ch. 5 - Prob. 95CPCh. 5 - Illustrate probability terms with scenarios a....Ch. 5 - Short term versus long run Short-term aberrations...Ch. 5 - Prob. 98CPCh. 5 - Mrs. Test Mrs. Test (see www.mrstest.com) sells...Ch. 5 - Prob. 100CPCh. 5 - Prob. 101CPCh. 5 - Prob. 102CPCh. 5 - Protective bomb Before the days of high security...Ch. 5 - Streak shooter Sportscaster Maria Coselli claims...Ch. 5 - Multiple choice Choose ALL correct responses. For...Ch. 5 - Prob. 106CPCh. 5 - Prob. 107CPCh. 5 - Multiple choice: Dream come true You have a dream...Ch. 5 - Prob. 109CPCh. 5 - True or false Answer true of false for each part....Ch. 5 - True or false When you flip a balanced coin twice,...Ch. 5 - Prob. 112CPCh. 5 - Prob. 113CPCh. 5 - Prob. 114CPCh. 5 - Prob. 115CPCh. 5 - Prob. 116CP
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- 1 M&Ms colors come in the following percent- ages: 13 percent brown, 14 percent yellow, 13 percent red, 24 percent blue, 20 percent orange, and 16 percent green. Reach into a bag of M&Ms without looking. a. What's the chance that you pull out a brown or yellow M&M? b. What's the chance that you won't pull out a blue? swarrow_forward11. Prove or disprove: (a) If is a characteristic function, then so is ²; (b) If is a non-negative characteristic function, then so is √√4.arrow_forward4. Suppose that P(X = 1) = P(X = -1) = 1/2, that Y = U(-1, 1) and that X and Y are independent. (a) Show, by direct computation, that X + Y = U(-2, 2). (b) Translate the result to a statement about characteristic functions. (c) Which well-known trigonometric formula did you discover?arrow_forward
- 9. The concentration function of a random variable X is defined as Qx(h) = sup P(x ≤ X ≤x+h), h>0. x (a) Show that Qx+b (h) = Qx(h). (b) Is it true that Qx(ah) =aQx(h)? (c) Show that, if X and Y are independent random variables, then Qx+y (h) min{Qx(h). Qy (h)). To put the concept in perspective, if X1, X2, X, are independent, identically distributed random variables, and S₁ = Z=1Xk, then there exists an absolute constant, A, such that A Qs, (h) ≤ √n Some references: [79, 80, 162, 222], and [204], Sect. 1.5.arrow_forward29 Suppose that a mound-shaped data set has a must mean of 10 and standard deviation of 2. a. About what percentage of the data should lie between 6 and 12? b. About what percentage of the data should lie between 4 and 6? c. About what percentage of the data should lie below 4? 91002 175/1 3arrow_forward2,3, ample and rical t? the 28 Suppose that a mound-shaped data set has a mean of 10 and standard deviation of 2. a. About what percentage of the data should lie between 8 and 12? b. About what percentage of the data should lie above 10? c. About what percentage of the data should lie above 12?arrow_forward
- 27 Suppose that you have a data set of 1, 2, 2, 3, 3, 3, 4, 4, 5, and you assume that this sample represents a population. The mean is 3 and g the standard deviation is 1.225.10 a. Explain why you can apply the empirical rule to this data set. b. Where would "most of the values" in the population fall, based on this data set?arrow_forward30 Explain how you can use the empirical rule to find out whether a data set is mound- shaped, using only the values of the data themselves (no histogram available).arrow_forward5. Let X be a positive random variable with finite variance, and let A = (0, 1). Prove that P(X AEX) 2 (1-A)² (EX)² EX2arrow_forward
- 6. Let, for p = (0, 1), and xe R. X be a random variable defined as follows: P(X=-x) = P(X = x)=p. P(X=0)= 1-2p. Show that there is equality in Chebyshev's inequality for X. This means that Chebyshev's inequality, in spite of being rather crude, cannot be improved without additional assumptions.arrow_forward4. Prove that, for any random variable X, the minimum of EIX-al is attained for a = med (X).arrow_forward8. Recall, from Sect. 2.16.4, the likelihood ratio statistic, Ln, which was defined as a product of independent, identically distributed random variables with mean 1 (under the so-called null hypothesis), and the, sometimes more convenient, log-likelihood, log L, which was a sum of independent, identically distributed random variables, which, however, do not have mean log 1 = 0. (a) Verify that the last claim is correct, by proving the more general statement, namely that, if Y is a non-negative random variable with finite mean, then E(log Y) log(EY). (b) Prove that, in fact, there is strict inequality: E(log Y) < log(EY), unless Y is degenerate. (c) Review the proof of Jensen's inequality, Theorem 5.1. Generalize with a glimpse on (b).arrow_forward
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