Question

A manufacturer of window frames knows from long experience that 5 percent 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. None will need adjustment

Answer 1

0.385 is the probability under given conditions.

Lets simplify the above problem,

Given n=20, p=0.05, q=0.95

Probability that in a sample of 20 window frames:

None will need adjustments.

P(x=0)= 0.95^20 = 0.3585

At least one will need adjustment?

P(x>=1) = 1 - P(x=0)

= 1 - 0.3585

= 0.6415

More than two will need adjustment?

P(x>2) = 1 - P(x=0) -P(x=1) -P(x=2)

= 1 - 0.3585 - 20C1(0.05)^1*(0.95)^19 - 20C2(0.05)^2*(0.95)^18

= 0.6415 - 20*0.05*0.3774 - 190*0.0025*0.3972

= 0.2641-0.18868

= 0.0755

Hence 0.385 is the solution of the given problem.

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