P-Manual-9thEd-Errata.pdf

The first line of the formula in the solution of

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The first line of the formula in the solution of Problem 6 in Practice Examination 19 should be Pr 2 X ! X 2 > 0 ( ) = Pr X 2 ! X ( ) > 0 ( ) = Pr X X ! 2 ( ) < 0 ( ) = instead of Pr 2 X ! X 2 > 0 ( ) = Pr X 2 ! X ( ) > 0 ( ) = Pr X X ! 2 ( ) > 0 ( ) = Posted July 28, 2011 In the formula in the solution of Problem 26 in Practice Examination 17, the formula should be f X x ( ) = ! F X x ( ) = " d dx x k # e " x k ! k = 0 3 \$ = " d dx e " x + xe " x + 1 2 x 2 # e " x + 1 6 x 3 # e " x % & ' ( ) * = = " " e " x + e " x " xe " x ( ) + xe " x " 1 2 x 2 # e " x % & ' ( ) * + 1 2 x 2 # e " x " 1 6 x 3 # e " x % & ' ( ) * % & ' ( ) * = 1 6 x 3 # e " x . The formula was missing a minus sign in the second line just after the first parenthesis. The rest of the solution is unaffected. Posted July 28, 2011

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In the solution of Problem 16 in Practice Examination 17, the word “ bad” in the first sentence should be replaced by “bag”. Posted July 28, 2011 The second sentence of the solution of Problem 12 in Practice Examination 16 should be: Box 1 contains 1 blue and 4 red marbles, box 2 contains 2 blue and 3 red marbles and box 3 contains 3 blue and 2 red marbles. instead of Box 1 contains 1 blue and 4 red marbles, box 2 contains 2 blue and 3 red marbles and box 3 contains 3 red and 2 blue marbles. The rest of the solution is unaffected by this typo. Posted July 26, 2011 In Problem 9 in Practice Examination 14, the third condition should be: (ii) The future lifetimes follow a Weibull distribution with ! = 1.5 and ! = 2.0 for smokers, and ! = 2.0 and ! = 2.0 for nonsmokers. Also, the survival function of the Weibull distribution should be given as s T t ( ) = e ! t " # \$ % & ' ( ) . Posted June 26, 2011 In Problem 29 in Practice Examination 19, the last sentence should be: Find the variance of the conditional distribution of X , given Y = y . Posted March 10, 2011 The second sentence of the solution of Problem 10 in Practice Examination 1 should be: As the policy has a deductible of 1 (thousand), the claim payment is Y = 0, when there is no damage,with probability 0.94, max 0, X ! 1 ( ) , when 0 < X < 15, with probability 0.04, 14, in the case of total loss,with probability 0.02. " # \$ % \$ \$ Posted January 25, 2011 The last two sentences of the solution of Problem 11 in Practice Examination 16 should be replaced by But the memoryless property of the exponential distribution tells us that Y and Y ! 3 Y > 3 ( ) have the same distribution. Note, however, that
Y Y > 3 ( ) = 3 + Y ! 3 Y > 3 ( ) , so that Var Y Y > 3 ( ) = Var Y ! 3 Y > 3 ( ) = Var Y ( ) . This implies that Var Y X > 3 { } ! Y > 3 { } ( ) = Var Y Y > 3 ( ) = Var Y ( ) = 1 2 2 = 0.25. Answer A. Posted January 15, 2011 In Problem 16 in Practice Examination 6, the calculation of the second moment of X should be: E X 2 ( ) = 1 4 ! 0 2 + 3 4 ! 1 + 1 ( ) Second moment of T ! = 3 2 . instead of E X 2 ( ) = E X ( ) = 1 4 ! 0 2 + 3 4 ! 1 + 1 ( ) Second moment of T ! = 3 2 . Posted July 24, 2010 In the solution of Problem 21 in Practice Examination 6, the statement under the first expression on the right-hand side of the third to last formula should be: number of ways to pick ordered samples of size 2 from population of size n instead of number of ways to pick ordered samples of size n ! 2 from population of size n Posted July 17, 2010 Practice Examinations: An Introduction on page 109, the third sentence of the last section should be: Practice examinations 6-18 are meant to be more challenging.

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• Fall '16
• daniel

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