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hmwk10sol2027

# hmwk10sol2027 - Solutions to Homework 10 ISyE 2027 Fall...

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Solutions to Homework 10, ISyE 2027 Fall 2006 Problem 1 (a) geometric (b) Poisson (c) binomial (d) normal (e) exponential Problem 2 (a) P ( Z 1 . 44) = 0 . 9251. (b) We see that P ( | Z | ≤ 1 . 44) = P ( - 1 . 44 Z 1 . 44) = P ( Z 1 . 44) - P ( Z ≤ - 1 . 44) = 2 P ( Z 1 . 44) - 1 = 0 . 8502 . (c) P ( Z > 1 . 44) = 1 - P ( Z 1 . 44) = 0 . 0749. (d) P ( Z = 1 . 44) = 0 ( Z is continuous). (e) We see that 0 . 05 = P ( | Z | > x ) = P ( Z x ) + P ( Z ≤ - x ) = 2 P ( Z x ) = 2(1 - P ( Z x )) . This implies that P ( Z x ) = 0 . 975. After using a normal table, we see that x = 1 . 96. Problem 3 Now suppose X N (2 , 100). Then (a) P ( X 9) = P (10 Z + 2 9) = P ( Z . 7) = 0 . 7580. (b) P ( | X | > 9) = 1 - P ( | X | ≤ 9) = 1 - P ( - 9 X 9) = 1 - P ( - 9 10 Z + 2 9) = 1 - P ( - 1 . 1 Z 0 . 7) = 2 - P ( Z 0 . 7) - P ( Z 1 . 1) = 0 . 3777 . (c) Now, 0 . 05 = P ( | X - 2 | ≥ 3 x ) = P ( X - 2 3 x ) + P ( X - 2 ≤ - 3 x ) = P (10 Z 3 x ) + P (10 Z 3 x ) = 2 P ( Z 0 . 3 x ) so now we find that P ( Z 0 . 3 x ) = 0 . 975 1

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2 which implies that 0 . 3 x = 1 . 96, or x = 6 . 533. Problem 4 Here P ( μ - kσ < Y < μ + ) = P ( μ - kσ < μ + σZ < μ + ) = P ( - k < Z < k ) = 2 P ( Z k ) - 1 . If k = 0, the probability is 0. If k = 1, the probability is 0.6826. If k = 2, the probability is 0 . 9544. Finally, if k = 3, the probability is 0 . 9974. Problem 5 Using the formula given for the even moments of a standard normal r.v., we see that E [ Z 2 ] = 1 and E [ Z 4 ] = 3 (we’ll use these later). (a) The mean of X n is just E [ X n ] = E [ n k =1 Z 2 k ] = n k =1 E [ Z 2 k ] = nE [ Z 2 1 ] = n. The variance of X n is V ar ( X n ) = V ar ( n k =1 Z 2 k ) = n k =1 V ar ( Z 2 k ) = nV ar ( Z 2 1 ) . But V ar ( Z 2 1 ) = E [ Z 4 1 ] - E [ Z 2 1 ] 2 = 3 - 1 = 2, so V ar ( X n ) = 2 n . (b) Now P ( X 1 1 . 44) = P ( Z 2 1 1 . 44) = P ( - 1 . 2 Z 1 . 2) = 2 P ( Z 1 . 2) - 1 = 0 . 7698 . Furthermore,
3 E [ X n | Z 1 = . 2] = n k =1 E [ Z 2 k | Z 1 = . 2] = 0 . 04 + n k =2 E [ Z 2 k ] = 0 . 04 + ( n - 1) . Problem 6 Let T i denote the time to pick the i th item, for 1 i 100. Then the total time to pick 100 items is just S 100 = n k =1 T k Using the CLT, we see that P ( S 100 550) = P S 100 - 5(100) (3) 100 5 / 3 P ( Z 5 / 3) = 0 . 9521 .

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