Question

A binomial distribution has p​ = 0.27 and n​ = 94

Use the normal approximation to the binomial distribution to answer parts ​(a) through ​(d) below.

​a) What are the mean and standard deviation for this​ distribution?

​b) What is the probability of exactly 18 ​successes?

​c) What is the probability of 20 to 27 ​successes?

​d) What is the probability of 13 to 22 ​successes?

Answer 1

Solution :

Given that p​ = 0.27 and n​ = 94

=> q = 1 - p = 0.73

=> n*p = 94*0.27 = 25.38 , n*q = 94*0.73 = 68.62

=> Therefore,np > 5 and nq > 5 then we use normal approximation

a) => mean μ = n*p = 94*0.27 = 25.38

=> standard deviation σ = sqrt(n*p*q) = sqrt(94*0.27*0.73) = 4.3043

b) => P(x = 18) = P(18 - 0.5 < x < 18 + 0.5) (using continuity correction)

= P(17.5 < x < 18.5)

= P((17.5 - 25.38)/4.3043 < (x - μ)/σ < (18.5 - 25.38)/4.3043)

= P(-1.8307 < Z < -1.5984)

= 0.0212

c) P(20 <= x <= 27) = P(20 - 0.5 < x < 27 + 0.5) (using continuity correction)

= P(19.5 < x < 27.5)

= P((19.5 - 25.38)/4.3043 < (x - μ)/σ < (27.5 - 25.38)/4.3043)

= P(-1.3661 < Z < 0.4925)

= 0.6026

d) P(13 <= x <= 22) = P(13 - 0.5 < x < 22 + 0.5) (using continuity correction)

= P(12.5 < x < 22.5)

= P((12.5 - 25.38)/4.3043 < (x - μ)/σ < (22.5 - 25.38)/4.3043)

= P(-2.9924 < Z < -0.6691)

= 0.25