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Unformatted text preview: AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY THEORY Homework Set # 4 – Solution Notes (1). (25 points) The cdf of X is given by F ( x ) = x < 4 3 / 10 4 ≤ x < 1 7 / 10 1 ≤ x < 4 1 x ≥ 4 (a). (15 points) Find the variance and the standard deviation of X . The probability mass function is given by p ( x ) = 3 / 10 if x = 4 4 / 10 if x = 1 3 / 10 if x = 4 otherwise Thus, E ( X ) = ( 4)(3 / 10) + (1)(4 / 10) + (4)(3 / 10) = 4 / 10, E ( X 2 ) = ( 4) 2 (3 / 10) + (1) 2 (4 / 10) + (4) 2 (3 / 10) = 10, var ( X ) = E ( X 2 ) [ E ( X )] 2 = 10 (4 / 10) 2 = 9 . 84, and the standard deviation is σ X = √ 9 . 84. (b). (10 points) Find the variance of Y = 3 p ( X ) + P ( X > 2), where p ( · ) is the pmf of X . First, we note that P ( X > 2) = p (4) = 3 / 10. We compute E ( Y 2 ) = E ((3 p ( X ) + 3 / 10) 2 ) = (3 p ( 4) + 3 / 10) 2 (3 / 10) + (3 p (1) + 3 / 10) 2 (4 / 10) + (3 p (4) + 3 / 10) 2 (3 / 10) = (3(3 / 10) + 3 / 10) 2 (3 / 10) + (3(4 / 10) + 3 / 10) 2 (4 / 10) + (3(3 / 10) + 3 / 10) 2 (3 / 10) = a (some number I could evaluate if I really wanted to), Similarly, we compute E ( Y ) = E (3 p ( X ) + 3 / 10) = (3 p ( 4) + 3 / 10)(3 / 10) + (3 p (1) + 3 / 10)(4 / 10) + (3 p (4) + 3 / 10)(3 / 10) = (3(3 / 10) + 3 / 10)(3 / 10) + (3(4 / 10) + 3 / 10)(4 / 10) + (3(3 / 10) + 3 / 10)(3 / 10) = b (some number I could evaluate if I really wanted to), Finally, we get var ( Y ) = E ( Y 2 ) [ E ( Y )] 2 = a b 2 . (2). (25 points) Problem 4.46, Ross 8th edition (it is problem 46, page 193, Ross 7th edition). I will help you set up the problem. Let C be the event that the defendent is convicted (found to be guilty by at least 9 of the 12 jurors). Let G be the event that the defendent actually is guilty. Let E be the event that the jury renders a correct decision. The problem asks you to compute P ( E ) and P ( C ) ; however, you only need to do the first part: compute P ( E ) . We want to compute P ( E ). Since the probability of correctness of the decision depends on whether or not the defendent is guilty, we condition on this information, breaking into two cases: G and G c . We get, by conditioning on G , P ( E ) = P ( E  G ) P ( G ) + P ( E  G c ) P ( G c ) We know P ( G ) = . 65 and thus P ( G c ) = . 35....
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This note was uploaded on 02/07/2010 for the course AMS 311 taught by Professor Tucker,a during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Tucker,A

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