Mth562-HW7-5

Mth562-HW7-5 - MTH/STA 562 Exercise 7.44 . ( a ) p + 3 r pq...

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Unformatted text preview: MTH/STA 562 Exercise 7.44 . ( a ) p + 3 r pq n < 1 if and only if 3 r pq n < 1 & p if and only if 3 r pq n < q if and only if 9 pq n < q 2 if and only if 9 p n < q if and only if n 9 p > 1 q if and only if n > 9 p q ! ( b ) < p & 3 r pq n if and only if 3 r pq n < p if and only if 9 pq n < p 2 if and only if 9 q n < p if and only if n 9 q > 1 p if and only if n > 9 q p ! ( c ) From ( a ) and ( b ), we have n > 9 p q ! and n > 9 q p ! if and only if n > 9 larger of p and q smaller of p and q ! : Exercise 7.45 . ( a ) If p = 0 : 5, n > 9 & : 5 : 5 = 9 : ( b ) If p = 0 : 6, n > 9 & : 6 : 4 = 13 : 5 : If p = 0 : 4, n > 9 & : 6 : 4 = 13 : 5 : If p = 0 : 8, n > 9 & : 8 : 2 = 18 : If p = 0 : 2, n > 9 & : 8 : 2 = 18 : If p = 0 : 99, n > 9 & : 99 : 01 = 891 : If p = 0 : 001, n > 9 & : 999 : 001 = 8991 : Exercise 7.46 . The quantity Y & np p np (1 & p ) converges in distribution to a standard normal random variable. Hence, with n = 100 and p = 0 : 1, we have P f Y & 15 g P 8 < : Z & 14 : 5 100 (0 : 1) q 100 (0 : 1) (0 : 9) 9 = ; = P f Z & 1 : 5 g = 0 : 0668 : Exercise 7.47 . Let Y be the number of persons not showing up for a given &ight. Using the normal approximation to the binomial distribution with n = 190 and p = 0 : 05, the desired probability is P f Y & 5 g P 8 < : Z & 4 : 5 160 (0 : 05) q 160 (0 : 05) (0 : 95) 9 = ; = P f Z & 1 : 27 g = 1 P f Z & 1 : 27 g = 1 : 1020 = 0 : 8980 : Exercise 7.48 . Let Y be the number of lawyers in the sample. ( a ) This is a binomial random variable with n = 1500 and p = 1 = 410. The desired probability is P f Y & 1 g P 8 > > < > > : Z & : 5 1500 1 410 r 1500 1 410 409 410 9 > > = > > ; = P f Z & 1 : 65 g = 1 P f Z & 1 : 65 g = 1 : 0495 = 0 : 9505 : ( b...
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Mth562-HW7-5 - MTH/STA 562 Exercise 7.44 . ( a ) p + 3 r pq...

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