Homework_5_Solutions - Homework 5 Solutions 8.1 8.7 8.8 a....

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Homework 5 Solutions 8.1 a. Continuous b. Discrete c. Continuous d. Discrete e. Discrete 8.7 1 – (.05 + .20 + .50 + .15) = .10. (The sum of all probabilities must equal 1.) 8.8 a. Condition 1: Sum = .1 + .1 + .3 + .5 = 1. Condition 2: Each of the four probabilities is between 0 and 1. b. Figure for Exercise 8.8c c. k 0 1 2 3 P(X k) .1 .2 .5 1 As an example, P(X 1 ) = P ( X =0) + P ( X =1) = .1+.1=.2. 8.17 a. .06 + .13 = .19 b. P ( X > 0) = 1 – P ( X = 0) = 1− .73 = .27. Another method is to add the probabilities for X = 1, 2, 3, 4. c. k 0 1 2 3 4 P ( X k ) .73 .89 .95 .98 1 8.22 a. E( X ) = - 2(.25) + 0(.50) + 2(.25) = 0. b. 2 σ = ( - 2 - 0) 2 (.25) + (0 - 0) 2 (.50) + (2 - 0) 2 (.25) = 2. c. 414 . 1 2 = = σ 8.23 An intuitive solution (and a correct one) is that the expected value for the number of girls among three children is 3 × .5 = 1.5. A more elaborate solution is to determine the probability distribution for X =number of girls, and then use the formula for E( X ). The distribution given in Example 8.6
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This homework help was uploaded on 04/07/2008 for the course STAT 200 taught by Professor Barroso,joaor during the Spring '08 term at Pennsylvania State University, University Park.

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Homework_5_Solutions - Homework 5 Solutions 8.1 8.7 8.8 a....

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