jim1174
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find the standard deviation for the binomial distribution which has stated values of n=50 and p=0.3. use normal distribution to approximate the binomial distribution and find probability of (a) x> 13 and (b) 13<x 17.
Answer
standard deviation = sqrt[np(1-p)] = sqrt[50*.3*.7] = sqrt(10.5) = 3.24037
(a) First, change x = 13 to a z-score using a continuity correction:
(13.5 - .3*50) / sqrt(.3*.7*50) = -0.46
P(x > 13) = P(z > -0.46) = .5000 + .1772 = .6772
(b) Change 17 to a z-score:
(17.5 - 15) / sqrt(.3*.7*50) = 0.77
P(13 < x < 17) = P(-0.77 < z < 0.77) = 2 * .2794 = .5588
Answer
standard deviation = sqrt[np(1-p)] = sqrt[50*.3*.7] = sqrt(10.5) = 3.24037
(a) First, change x = 13 to a z-score using a continuity correction:
(13.5 - .3*50) / sqrt(.3*.7*50) = -0.46
P(x > 13) = P(z > -0.46) = .5000 + .1772 = .6772
(b) Change 17 to a z-score:
(17.5 - 15) / sqrt(.3*.7*50) = 0.77
P(13 < x < 17) = P(-0.77 < z < 0.77) = 2 * .2794 = .5588