Question

Find the​ mean, variance, and standard deviation of the binomial distribution with the given values of n and p.

n equals 124n=124​,

p equals 0.72

Answer 1

(i)

n = 124

p = 0.72

q = 1 - p = 0.28

Mean = np = 124 X 0.72 = 89.28

(ii)

Variance = npq = 124 X 0.72 X 0.28 = 24.9984

(iii)

Standard Deviation = $\sqrt{24.9984}=4.9998$

So,

Answers are:

Mean = np = 89.28

Variance = npq = 24.9984

Standard Deviation =