Question

A manufacturer of window frames knows from experience that 6% of the production will have some type of defect that will require an adjustment. What is the probability that in a sample of 21 window frames, more than 2 will need an adjustment?

a. 0.8716

b. 0.2727

c. 0.2333

d. 0.1284

e. None of the above

Answer 1

Solution

Given that ,

p = 0.06

q = 1 - p = 1 - 0.06 = 0.94

n = 21

Using binomial probability formula ,

P(X = x) = (n C x) * p x * (1 - p)n - x

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

= 1 - P(X = 0) - P(X = 1)- P(X = 2)

= 1 -  (21 C 0) * 0.06 0 * (0.94)21-  (21 C 1) * 0.06 1 * (0.94)20-  (21 C 2) * 0.06 2 * (0.94)19

= 1 - 0.8716

= 0.1284

Probability = 0.1284