778 APPENDIX D 5. a. Discrete random variable b. Continuous random variable c. Discrete random variable d. Not a random variable e. Discrete random variable 7. Not a probability distribution because the causes are not values of a numerical random variable. 9. Probability distribution with m = 1.6, s = 0.9. 11. Not a probability distribution because the responses are not values of a numerical random variable. 13. Probability distribution with m = 2.4, s = 1.0. 15. m = 0.4, s = 0.6 17. Significantly high numbers of matches are greater than or equal to m + 2s, and m + 2s = 0.4 + 210.62 = 1.6 matches. Because 4 matches is greater than or equal to 1.6 matches, it is a significantly high number of matches. 19. a. 0.004 b. 0.004 c. Part (b) d. Yes, because the probability of 0.004 is low (less than or equal to 0.05). 21. m = 2.1, s = 1.1 23. Significantly low numbers of drivers who say that they text while driving is less than or equal to m-2s = 2.1-211.12 = -0.1. Because 1 driver is not less than or equal to -0.1, 1 is not a significantly low number of drivers who say that they text while driving. 25. a. 0.344 b. 0.648 c. Part (b) d. No, because the probability of 2 or fewer drivers who say that they text while driving is not low (less than or equal to 0.05). 27. Because the probability of 270 or more saying that we should use biometrics is 0.0995, which is not low (less than or equal to 0.05), 270 is not significantly high. Given that 270 is not significantly greater than 50%, there is not sufficient evidence to conclude that the majority of the population says that we should replace passwords with biometric security. 29. a. 1000 b. 1>1000 c. $499 d. -50. e. The $1 bet on the pass line in craps is better because its expected value of -1.4. is much greater than the expected value of -50. for the Florida Pick 3 lottery. 31. a. -39. b. The bet on the number 27 is better because its expected value of -26. is greater than the expected value of -39. for the other bet. Section 5-2 1. The given calculation assumes that the first two speaking characters are females and the last three are not females, but there are other arrangements consisting of two females and three males. The probabilities corresponding to those other arrangements should also be included in the result. 3. Because the 50 selections are made without replacement, they are dependent, not independent. Based on the 5% guideline for cumbersome calculations, the 50 selections can be treated as being independent. (The 50 selections constitute 3.33% of the population of 1500 speaking characters, and 3.33% is not more than 5% of the population.) 5. Not binomial. Each of the ages has more than two possible outcomes. 13. a. 1>365 b. 31>365 c. Answer varies, but it is probably quite small, such as 0.01 or less. d. Yes 14. 0.0422. No. 15. 1>1,832,600. No, the jackpot seems disproportionately small given the probability of winning it. But then, no lotteries are fair. Chapter 4: Cumulative Review Exercises 1. a. 6.919 rnfl b. 6.485 rnfl c. 9.485 rnfl d. 12.090 rnfl e. 3.400 rnfl f. 11.558 rnfl2 2. a. 3.44 rnfl, 4.480 rnfl, 6.485 rnfl, 7.845 rnfl, 15.53 rnfl b. c. The amount of 15.53 rnfl appears to be an outlier. 3. a. 46% b. 0.460 c. Stratified sample 4. a. Convenience sample b. If the students at the college are mostly from a surrounding region that includes a large proportion of one ethnic group, the results might not reflect the general population of the United States. c. 0.75 d. 0.64 5. Based on the scatterplot, it is reasonable to conclude that there is no association between heights of presidents and the heights of the presidential candidates who were runners-up. It is also reasonable to conclude that there is a very weak association with increasing heights of winners corresponding to decreasing heights of runners-up. (More objective criteria will be introduced in Chapter 10.) Chapter 5 Answers Section 5-1 1. The random variable is x, which is the number of unlicensed software packages. The possible values of x are 0, 1, 2, 3, and 4. The values of the random variable x are numerical. 3. ΣP1x2 = 0.008 + 0.076 + 0.265 + 0.412 + 0.240 = 1.001. The sum is not exactly 1 because of a round-off error. The sum is close enough to 1 to satisfy the requirement. Also, the variable x is a numerical random variable and its values are associated with probabilities, and each of the probabilities is between 0 and 1 inclusive, as required. The table does describe a probability distribution.
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