STAT 400 Lecture AL1 Spring 2015 Dalpiaz Answers for 5 7 Normal Approximation to Binomial Distribution Normal Binomial mean n p standard deviation n p 1 p 1 Binomial distribution n 25 p 0 50 Normal approximation mean n p 25 0 50 12 5 n p 1 p 25 0 50 0 50 6 25 SD 6 25 2 5 a P X 17 PMF 17 0 0322 b 16 5 12 5 17 5 12 5 Z P X 17 P 16 5 X 17 5 P 2 5 2 5 P 1 60 Z 2 00 0 9772 0 9452 0 0320 c P X 11 1 CDF 10 1 0 2122 0 7878 d 10 5 12 5 P X 11 P X 10 5 P Z 2 5 P Z 0 80 1 0 2119 0 7881 e P 10 X 14 CDF 14 CDF 9 0 7878 0 1148 0 6730 f 9 5 12 5 14 5 12 5 Z P 10 X 14 P 9 5 X 14 5 P 2 5 2 5 P 1 20 Z 0 80 0 7881 0 1151 0 6730 2 Let X number of passengers who do not cancel their reservations Then X has Binomial distribution n 100 p 0 85 Normal approximation 100 0 85 85 2 100 0 85 0 15 12 75 3 57 92 5 85 P Z 2 10 0 9821 P X 92 P X 92 5 P Z 3 57 Binomial P X 92 0 9878 2 5 A fair 6 sided die is rolled 180 times The sum of the outcomes is likely to be around give or take or so The average of the outcomes is likely to be around give or take or so 1 2 3 4 5 6 21 3 5 6 6 1 3 5 2 2 3 5 2 3 3 5 2 4 3 5 2 5 3 5 2 6 3 5 2 6 6 25 2 25 0 25 0 25 2 25 6 25 6 17 5 1 708 6 E Sum n 180 3 5 630 SD Sum n 180 1 708 22 9 The sum of the outcomes is likely to be around 630 give or take 23 or so E Average 3 5 SD Average n 1 708 180 0 1273 The average of the outcomes is likely to be around 3 5 give or take 0 13 or so A fair 6 sided die is rolled 180 times The number of 6 s is likely to be around give or take or so Let X denote the number of 6 s p 1 6 n 180 Binomial distribution E X n p 180 1 6 30 SD X n p 1 p 180 1 6 5 6 5 The number of 6 s is likely to be around 30 give or take 5 or so 3 Binomial distribution n 180 p 1 6 a 35 1 P X 35 180 C 35 6 5 6 145 0 0464 Normal approximation mean n p 180 1 6 30 n p 1 p 180 1 6 5 6 25 b 25 5 SD P X 35 P 34 5 X 35 5 P 0 90 Z 1 10 0 8643 0 8159 0 0484 c P X 35 P X 34 5 P Z 0 90 1 0 8159 0 1841 180 Binomial P X 35 k 35 d 1 180 C k 6 k 5 6 180 k 0 18283 P 20 X 40 P 19 5 X 40 5 P 2 10 Z 2 10 0 9642 40 Binomial 1 P 20 X 40 180 C k 6 k 20 k 5 6 180 k 0 965 e Use Normal approximation to find the probability that the sum of the results is between 600 and 640 both inclusive Recall 3 5 2 17 5 6 35 12 P 600 Sum 640 P 599 5 Sum 640 5 599 5 180 3 5 640 5 180 3 5 P Z 180 35 180 35 12 12 P 1 33 Z 0 46 0 5854 4 Poisson distribution 1 4 52 72 8 a P X 68 72 868 e 72 8 68 0 0411 Normal approximation 72 8 b 72 8 8 5323 P X 68 P 67 5 X 68 5 P 0 62 Z 0 50 0 3085 0 2676 0 0409 c P X 70 P X 70 5 P Z 0 27 0 3936 Poisson d P X 70 70 72 8 k e 72 8 k 0 k 0 40078 P 65 X 80 P 64 5 X 80 5 P 0 97 Z 0 90 0 8159 0 1660 0 6499 Poisson P 65 X 80 80 72 8 k e 72 8 k 65 k 0 65218 Hype for 0 5 continuity correction Let X be a Binomial n 400 p 0 80 random variable P X 312 0 1738 A 1 B BINOMDIST 312 400 0 8 1 A 2 1 B 0 173821 2 Without 0 5 continuity correction P X 312 z 312 400 0 80 400 0 80 0 20 With 0 5 continuity correction P X 312 P X 312 5 1 00 P Z 1 00 0 1587 z 312 5 400 0 80 400 0 80 0 20 0 9375 P Z 0 94 0 1736 P Z 0 9375 0 17425
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