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Unformatted text preview: AMS 311 (Fall, 2010) 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 The cdf is a staircase (piecewiseconstant), so it corresponds to the cdf of a discrete r.v. The corresponding probability mass function is given by p ( x ) = 3 / 10 if x = − 4 4 / 10 if x = 1 3 / 10 if x = 4 otherwise (a). (10 points) Find the variance and the standard deviation of X . From the pmf of X , we readily get 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). (5 points) Find the variance of Y = X 3 + 12 . We compute E ( Y 2 ) = E (( X 3 + 12) 2 ) = (( − 4) 3 + 12) 2 (3 / 10) + (1 3 + 12) 2 (4 / 10) + (4 3 + 12) 2 (3 / 10) = a (some number I could evaluate if I really wanted to), and E ( Y ) = E ( X 3 + 12) = (( − 4) 3 + 12)(3 / 10) + (1 3 + 12)(4 / 10) + (4 3 + 12)(3 / 10) = b (some number I could evaluate if I really wanted to), so that we get var ( Y ) = E ( Y 2 ) − [ E ( Y )] 2 = a − b 2 . Altrnatively, I could use the following property of variance (easy to prove!): var ( cX + d ) = c 2 var ( X ). Then, var ( Y ) = var ( X 3 + 12) = var ( X 3 ) = E ( X 6 ) − [ E ( X 3 )] 2 = ( − 4) 6 (3 / 10) + (1) 6 (4 / 10) + (4) 6 (3 / 10) − [( − 4) 3 (3 / 10) + (1) 3 (4 / 10) + (4) 3 (3 / 10)] 2 . (c). (10 points) Find the cdf of W = 2 X 2 − 3 . We know that X takes on values 4, 1, or 4, with probabilities 3/10, 4/10, or 3/10, respectively. Thus, W takes on values 29 (if X = − 4 or X = 4) or 1 (if X = 1). Thus, P ( W = 29) = 3 / 10 + 3 / 10 = 6 / 10 and P ( W = − 1) = 4 / 10. So the mass function of W is p W ( w ) = P ( W = w ) = braceleftBigg 4 / 10 if w = − 1 6 / 10 if w = 29 otherwise and the cdf is F W ( w ) = P ( W ≤ w ) = braceleftBigg if w < − 1 4 / 10 if − 1 ≤ w < 29 1 if w ≥ 29 (2). Suppose it takes at least 9 votes from a 12member jury to convict a defendant. Suppose the probability that a juror votes a guilty person innocent is 0.2, whereas the probability that the juror votes an innocent person guilty is 0.1. If each juror acts independently and if 65 percent of the defendants are guilty, find the probability that the jury renders a correct decision. 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 ) ....
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This note was uploaded on 07/15/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Tucker,A

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