Question

Suppose a large shipment of telephones contained 22% defectives.

If a sample of size 433 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%? Round your answer to four decimal places.

Answer 1

Solution

Given that,

=  [p ( 1 - p ) / n] = [(0.22 * 0.78) / 433 ] = 0.0199

= P[(-0.03) /0.0199 < ( - ) / < (0.03) / 0.0199]

= P(-1.51 < z < 1.51)

= P(z < 1.51) - P(z < -1.51)

= 0.9345 - 0.0655

= 0.8690

Probability = 0.8690