soln%207 - X X X X X X X X X X X X X X X X X AP Statistics...

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X X X X X X X X X X X X X X X X X AP Statistics Solutions to Packet 7 X Random Variables Discrete and Continuous Random Variables Means and Variances of Random Variables X X X X X X X X X X X X X
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2 HW #44 2, 3, 6 – 8, 13 – 17 7.2 THREE CHILDREN A couple plans to have three children. There are 8 possible arrangements of girls and boys. For example, GGB means the first two children are girls and the third child is a boy. All 8 arrangements are (approximately) equally likely. (a) Write down all 8 arrangements of the sexes of three children. What is the probability of any one of these arrangements? BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG. Each has probability 1/8. (b) Let X be the number of girls the couple has. What is the probability that X = 2? Three of the eight arrangements have two (and only two) girls, so P( X = 2) = 3/8 = 0.375. (c) Starting from your work in (a), find the distribution of X . That is, what values can X take, and what are the probabilities for each value? Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 7.3 SOCIAL CLASS IN ENGLAND A study of social mobility in England looked at the social class reached by the sons of lower-class fathers. Social classes are numbered from 1 (low) to 5 (high). Take the random variable X to be the class of a randomly chosen son of a father in Class I. The study found that the distribution of X is: Son’s class: Probability: 1 0.48 2 0.38 3 0.08 4 0.05 5 0.01 (a) What percent of the sons of lower-class fathers reach the highest class, Class 5? 1% (b) Check that this distribution satisfies the requirements for a discrete probability distribution. All probabilities are between 0 and 1; the probabilities add to 1. (c) What is P( X 3)? P( X 3) = 0.48 + 0.38 + 0.08 = 1 - 0.01 - 0.05 = 0.94. (d) What is P( X < 3)? P( X < 3) = 0.48 + 0.38 = 0.86. (e) Write the event “a son of a lower-class father reaches one of the two highest classes” in terms of X . What is the probability of this event? Write either X 4 or X > 3. The probability is 0.05 + 0.01 = 0.06. (f) Briefly describe how you would use simulation to answer the question in (c). Read two random digits from Table B. Here is the correspondence: 01 to 48 Class 1, 49 to 86 Class 2, 87 to 94 Class 3, 95 to 99 Class 4, and 00 Class 5. Repeatedly generate 2 digit random numbers. The proportion of numbers in the range 01 to 94 will be an estimate of the required probability.
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3 7.6 CONTINUOUS RANDOM VARIABLE, I Let X be a random number between 0 and 1 produced by the idealized uniform random generator described in Example 7.3 and Figure 7.5 (text p. 398). Find the following probabilities: (a) P(0 X 0.4) = 0.4 (b) P(0.4 X 1) = 0.6 (c) P(0.3 X 0.5) = 0.2 (d) P(0.3 < X < 0.5) = 0.2 (e) P(0.226 X 0.713) = 0.713 – 0.226 = 0.487 (f) What important fact about continuous random variables does comparing your answer to (c) and (d) illustrate? A continuous distribution assigns probability 0 to every individual outcome. In this case, the probabilities in (c) and (d) are the same because the events differ by 2 individual values, 0.3 and 0.5, each of which has probability 0.
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soln%207 - X X X X X X X X X X X X X X X X X AP Statistics...

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