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# hw4-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...

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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 ) = 0 x < 4 3 / 10 4 x < 1 7 / 10 1 x < 4 1 x 4 The cdf is a staircase (piecewise-constant), 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 0 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 0 otherwise and the cdf is F W ( w ) = P ( W w ) = braceleftBigg 0 if w < 1 4 / 10 if 1 w < 29 1 if w 29 (2). Suppose it takes at least 9 votes from a 12-member 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 ) . 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 )

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We know P ( G ) = . 65 and thus P ( G c ) = . 35. We need to find P ( E | G ). Let X be the number of jurors that decide “guilty”. Then, given that the defendent is guilty , X is Binomial(12,0.8), since we are told that, if the defendent is guilty, each juror (independently) makes a “mistake” with probability 0.2 (and thus makes the correct, “guilty” decision with probability 0.8). Event E
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hw4-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...

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