hw5F05solns

# hw5F05solns - Solutions Homework#5 1 Chapter 4 Problems 1...

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Unformatted text preview: Solutions - Homework #5 1 Chapter 4 Problems 1. Problem 32 : To determine whether or not they have a certain disease, 100 people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to group the people in groups of 10. The blood samples of the 10 people in each group will be pooled and analyzed together. If the test is negative, one test will suffice for the 10 people; whereas, if the test is positive each of the 10 people will also be individually tested and, in all, 11 tests will be made on this group. Assume the probability that a person has the disease is . 1 for all people, independently of each other, and compute the expected number of tests necessary for each group. (Note that we are assuming that the pooled test will be positive if at least one person in the pool has the disease.) Within each group of 10, each person has probability 0 . 10 of having the disease, independently. If the combined sample from the group of 10 is tested, it will not be diseased with probability: P ( P c 1 P c 2 P c 10 ) = P ( P c 1 ) P ( P c 2 ) P ( P c 10 ) (independence of P i s) = ( . 9)( . 9) ( . 9) = ( . 9) 10 , where P i = event that the i th person has the disease. Let X = the number of tests required per group of 10. Then X takes on the values 1 &amp;amp; 11 with probabilities: P ( X = 1) = P ( P c 1 P c 10 ) = ( . 9) 10 , P ( X = 11) = 1- ( . 9) 10 . So E ( X ) = X x xP ( X = x ) = (1)( . 9 10 ) + (11)(1- . 9 10 ) = 11- 10( . 9) 10 . Letting Y = 10 X be the total number of tests required for the group of 100 people, E ( Y ) = E (10 X ) = 10 E ( X ) (by Corollary 5.1) = 10[11- 10( . 9) 10 ] = 110- 100( . 9) 10 . The number of tests necessary for each group of 10 is 1 / 10 of this value, namely 11- 10( . 9) 10 . 2. Problem 33 : A newsboy purchases papers at 10 cents and sells them at 15 cents. However, he is not allowed to return unsold papers. If his daily demand is a binomial random variable with n = 10 and p = 1 / 3, approximately how many papers should he purchase so as to maximize his expected profit? Let: Y = the number of papers purchased and X = the number of papers sold. Then X Bin( n = 10 ,p = . 3) assuming at least 10 papers were purchased. The profit gained is: P = 15 X- 10 Y . Since E ( P ) = 15 E ( X )- 10 Y , we need E ( X ) for different values of Y . Computing: If Y = 1 , E ( X ) = (0) P ( X = 0) + (1) P ( X = 1) = (0)(2 / 3) 10 + (1)(1- 2 / 3) 10 = (0)( . 0173) + (1)( . 9827) = . 9827 , If Y = 2 , E ( X ) = (0) P ( X = 0) + (1) P ( X = 1) + (2) P ( X = 2) = (0)(2 / 3) 10 + 1 &amp;quot; 10 1 ! (1 / 3) 1 (2 / 3) 9 # + 2 1- (2 / 3) 10- 10(1 / 3)(2 / 3) 9 / 1 = 0 + (1)( . 0867) + (2)( . 8960) = 1 . 8787 , If Y = 3 , E ( X ) = (0)( . 0173) + (1)( . 0867) + (2)( . 1951) + (3)( . 7009) = 2 . 5796 , If Y = 4 , E ( X ) = (0)( . 0173) + (1)( . 0867) + (2)( . 1951) + (3)( . 2601) + (4)( . 4408) = 3 . 0204 ,......
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hw5F05solns - Solutions Homework#5 1 Chapter 4 Problems 1...

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