A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames?

a) 0.3585

b) 0.6415

c) 0.07548

Step-by-step explanation:

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 20

x = Number of successes required

p = probability of window frame with minor defect = 0.05

q = probability of window frame with no defect = 1 - 0.05 = 0.95

a) Probability that none of the frames picked will have defects

x = 0

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

P(X = 0) = ²⁰C₀ 0.05⁰ 0.95²⁰⁻⁰ = 0.3585

b) Probability that At least one will need adjustment = P(X ≥ 1) = 1 - P(X < 1) = 1 - P(X = 0) = 1 - 0.3585 = 0.6415

c) Probability that More than two will need adjustment = P(X > 2)

P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

So, we evaluate each of that using the binomial distribution formula, slotting in the appropriate variables, and sum it all up.

P(X > 2) = 1 - 0.9245 = 0.07548