Suppose a large shipment of microwave ovens contained 22% defectives. If a sample of size 499 is selected, what is the probability that the sample proportion will be less than 17%? Round your answer to four decimal places.
Solution
Given that,
p = 0.22
1 - p = 1 - 0.22 = 0.78
n = 499
= p = 0.22
= [p ( 1 - p ) / n] = [(0.22 * 0.78) / 499 ] = 0.0185
P( < 0.17)
= P[( - ) / < (0.17 - 0.22) / 0.0185 ]
= P(z < - 2.70)
Using z table,
= 0.0035
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