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Dr. Hackney STA Solutions pg 79

Dr. Hackney STA Solutions pg 79 - Second Edition 5-13 b...

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Second Edition 5-13 b. Using the normal approximation, we have μ v = r (1 - p ) /p = 20( . 3) /. 7 = 8 . 57 and σ v = r (1 - p ) /p 2 = (20)( . 3) /. 49 = 3 . 5 . Then, P ( V n = 0) = 1 - P ( V n 1) = 1 - P V n - 8 . 57 3 . 5 1 - 8 . 57 3 . 5 = 1 - P ( Z ≥ - 2 . 16) = . 0154 . Another way to approximate this probability is P ( V n = 0) = P ( V n 0) = P V - 8 . 57 3 . 5 0 - 8 . 57 3 . 5 = P ( Z ≤ - 2 . 45) = . 0071 . Continuing in this way we have P ( V = 1) = P ( V 1) - P ( V 0) = . 0154 - . 0071 = . 0083, etc. c. With the continuity correction, compute P ( V = k ) by P ( k - . 5) - 8 . 57 3 . 5 Z ( k + . 5) - 8 . 57 3 . 5 , so P ( V = 0) = P ( - 9 . 07 / 3 . 5 Z ≤ - 8 . 07 / 3 . 5) = . 0104 - . 0048 = . 0056, etc. Notice that the continuity correction gives some improvement over the uncorrected normal approximation. 5.39 a. If h is continuous given > 0 there exits δ such that | h ( x n ) - h ( x ) | < for | x n - x | < δ . Since X 1 , . . . , X n converges in probability to the random variable X , then lim n →∞ P ( | X n - X | < δ ) = 1. Thus lim n →∞ P ( | h ( X n ) - h ( X ) | < ) = 1. b. Define the subsequence X j ( s ) = s + I [ a,b ] ( s ) such that in I [ a,b ] , a is always 0, i.e, the subse- quence X 1 , X 2 , X 4 , X 7 , . . . . For this subsequence X j ( s ) s if s >
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