Question

Acme Electronics makes calculators. Consumer satisfaction is one the top priorities of the company’s management. The company guarantees a refund or a replacement for any calculator that malfunctions within 2 years from the date of purchase. It is known from past data that despite all efforts, 5% of the calculators manufactured by Acme Electronics malfunction within a 2-year period. The company mailed a package of 10 randomly selected calculators to a store.

a. Let x denote the number of calculators in this package of 10 than will be returned for a refund or replacement within a 2-year period. Write the probability distribution for x.

b. Find the probability that at least 2 calculators will be returned or replaced.

c. Find the probability that no calculators will be returned or replaced.

d. Determine the mean and standard deviation of x.

Answer 1

a)

B(10, 0.05)

b)

Here, n = 10, p = 0.05, (1 - p) = 0.95 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 2).
P(X >= 2) = (10C2 * 0.05^2 * 0.95^8) + (10C3 * 0.05^3 * 0.95^7) + (10C4 * 0.05^4 * 0.95^6) + (10C5 * 0.05^5 * 0.95^5) + (10C6 * 0.05^6 * 0.95^4) + (10C7 * 0.05^7 * 0.95^3) + (10C8 * 0.05^8 * 0.95^2) + (10C9 * 0.05^9 * 0.95^1) + (10C10 * 0.05^10 * 0.95^0)
P(X >= 2) = 0.0746 + 0.0105 + 0.001 + 0.0001 + 0 + 0 + 0 + 0 + 0
P(X >= 2) = 0.0862

c)

Here, n = 10, p = 0.05, (1 - p) = 0.95 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 10C0 * 0.05^0 * 0.95^10
P(X = 0) = 0.5987

d)

mean = np
= 10 * 0.05
= 0.5

std,dev= sqrt(npq)
=sqrt(10 * 0.05 *(1-0.05))
= 0.6892