Question

Suppose a large shipment of microwave ovens contained 20% defectives.

If a sample of size 336 is selected, what is the probability that the sample proportion will be greater than 16%? Round your answer to four decimal places.

Answer 1

Solution

Given that,

p = 0.20

1 - p = 0.80

n = 336

= p = 0.20

=  [p( 1 - p ) / n] = [(0.20 * 0.80) / 336 ] = 0.0218

P( > 0.16) = 1 - P( < 0.16)

= 1 - P(( - ) / < (0.16 - 0.20) / 0.0218)

= 1 - P(z < -1.83)

= 0.9664