Age Distribution and Probability Analysis of Americans

3.2.2 The following table shows the distribution of ages of Americans.’ Age distribution in reference population Age Proportion 0-19 0.27 20-29 0.14 30-39 0.13 4049 0.14 50-64 0.19 65+ 0.13 Find the probability that the age of a randomly chosen American (a) is less than 20. (b) is between 20 and 49. (c) is greater than 49. (d) is greater than 29. Answer (2.5pt): 3.2.2 (a) 0.27 (b) 0.14 + 0.13 + 0.14 = 0.41 () 0.19+0.13 = 0.32 (d)1-027-0.14=0.59 3.2.3 Inacertain college,55% of the students are women. Suppose we take a sample of two students. Use a proba- bility tree to find the probability (a) that both chosen students are women. (b) that at least one of the two students is a woman. Answer (2.5pt): 3.2.3 (a) (0.55)(0.55) = 0.3025 (b) (0.55)(0.55) + (0.55)(0.45) + (0.45)(0.55) = 0.7975 or 1 - (0.45)(0.45) = 1 - 0.2025 = 0.7975 0 058 woman < 04y probability 0.55 X5 = 0. Y0y 0.yI X 04y = 0, )Y Wy x 0.y : o‘W'lS 045y 0¥ = 0,905

3.2.6 If a woman takes an early pregnancy test, she will either test positive, meaning that the test says she is preg- nant, or test negative, meaning that the test says she is not pregnant. Suppose that if a woman really is pregnant, there is a 98% chance that she will test positive. Also, sup- pose that if a woman really is not pregnant, there is a 99% chance that she will test negative. (a) Suppose that 1,000 women take early pregnancy tests and that 100 of them really are pregnant. What is the probability that a randomly chosen woman from this group will test positive? (b) Suppose that 1,000 women take early pregnancy tests and that 50 of them really are pregnant. What is the probability that a randomly chosen woman from this group will test positive? Answer (2.5pt): * 3.2.6 (a) There are two ways to test positive. A true positive happens with probability (0.1)(0.98) = 0.098. A false positive happens with probability (0.9)(0.01) = 0.009. Thus, Pr{test positive} = 0.098 + 0.009 = 0.107. (b) Using the same reasoning as in part (a), Pr{test positive} = (0.05)(0.98) + (0.95)(0.01) = 0.049 + 0.095 = 0.0585. 3.3.3 The following data table is taken from the study reported in Exercise 3.3.1. Here “stressed” means that the person reported that most days are extremely stressful or quite stressful; “not stressed” means that the person reported that most days are a bit stressful, not very stress- ful, or not at all stressful. Income Low Medium High Total Stressed 526 274 216| 1,016 Not stressed |1,954 1,680 1,899 5,533 Total 2,480 1,954 2,115 6,549 (a) What is the probability that someone in this study is stressed? (b) Given that someone in this study is from the high income group, what is the probability that the person 1s stressed? (c) Compare your answers to parts (a) and (b). Is being stressed independent of having high income? Why or why not?

Answer (2.5pt): 3.3.3 (a) 1016/6549 = 0.1551 = 0.155 (b) 216/2115=0.1021 = 0.102 (c) No; the probability of a person being stressed depends on whether or not the person has high income, since the answers to (a) and (b) differ. 3.3.4 Consider the data table reported in Exercise 3.3.3. (a) What is the probability that someone in this study has low income? (b) What is the probability that someone in this study either is stressed or has low income (or both)? (c) What is the probability that someone in this study is stressed and has low income? Answer (2.5pt): 3.3.4 (a) 2480/6549 = 0.3787 = 0.379 (b) (1016 + 1954)/6549 = 2970/6549 = 0.4535 =~ 0.454 (c) 526/6549 = 0.0803 = 0.08 3.4.3 Inacertain population of the parasite Trypanosoma, the lengths of individuals are distributed as indicated by the density curve shown here. Areas under the curve are shown in the figure.!’ Consider the length of an individual trypanosome chosen at random from the population. Find (a) Pr{20 < length < 30} (b) Pr{length = 20} (c) Pr{length < 20} 0.01 0.03 0.34 0.41 0.21 10 15 20 25 30 35

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Answer (2.5pt): *«3.4.3 (a) Pr{20<Y <30} =0.41 +0.21 =0.62 (b) 0.41 +0.21 + 0.03 = 0.65 () 0.01 +0.34=0.35

HW3-answer

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