# HW 16 Answers (5.1) - X is binomial with parameters 6 1 10...

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Math331, Spring 2008 Instructor: David Anderson Section 5.1 Homework Answers Homework: pgs. 196 - 197, #’s 1, 3, 4, 8, 13. 1. If we think of each draw as a Bernoulli RV with success deFned as drawing a spade, the p = . 25. Therefore, we see that if X is the number of spades in the Frst 8 draws, X is a binomial RV with parameters 8 and .25. Therefore, P { X = 4 } = ( 8 4 ) ( . 25) 4 ( . 75) 4 = . 0865 . 3. The event of any one of them being born in either April or October can be modeled as a Bernoulli RV with p = 1 / 6. Therefore, if X is the number of students out of the 6 who were born in either April or October, X is binomial with parameters 6 and 1/6. Therefore, P { X = 3 } = ( 6 3 ) (1 / 6) 3 (5 / 6) 3 = . 0536 . 4. The event that any given digit is a 9 can be modeled as a Bernoulli RV with p = 1 / 10. Therefore, if X is the random variable that gives the number of 9’s on a given licence plate,
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Unformatted text preview: X is binomial with parameters 6, 1 / 10. Therefore, P { X = 2 } = ( 6 2 ) (1 / 10) 2 (9 / 10) 4 = . 0984. 8. Let X be the number of the 15 that will be audited. Then X is binomial with parameters 15 and .20. Therefore, P { X ≤ 4 } = p 15 P . 2 . 8 15 + p 15 1 P . 2 1 . 8 14 + p 15 2 P . 2 2 . 8 13 + p 15 3 P . 2 3 . 8 12 + p 15 4 P . 2 4 . 8 11 = . 8358 . 13. Let X represent the number of days that the price of the stock increased. Then X is a binomial RV with parameters 6 and 1 / 3. ±or the price of the stock to be unchanged after 6 days, we must have X = 3. Therefore, P { X = 3 } = ( 6 3 ) (1 / 3) 3 (2 / 3) 3 = . 2195 . 1...
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## This homework help was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.

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