Question

an airplane manufacturer sells at most 1 airplanes per month. on any given month, the company sells no airplanes with probability 0.91. if the company is expected to sell 0.69 airplanes per month, what is the variance of the number of airplanes the company sells per month?

Answer 1

Let X be the number of airplanes the company sells per month

Probability that the company sells no airplanes = 0.91

As, X can be at most 1

Probability that the company sells 1 airplanes = 1 - Probability that the company sells no airplanes = 1 - 0.91 = 0.09

Clearly X follows binomial distribution with p = 0.09 (p is Probability that the company sells 1 airplanes)

q = 0.91 (1 - p)

Expected value of X = E(X) = 0.69 (given)

For binomial distribution E(X) = np (Here n is the number of trial)

so, np = 0.69

Here we need to find variance of X = Var(X) = npq (for binomial distribution)

Var(X) = 0.69 0.91 (as np = 0.69 and q = 0.91)

Var(X) = 0.6279

Answer: The variance of the number of airplanes the company sells per month is 0.6279