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.
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
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