soln%207

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 18

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online