Question

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.

Answer #1

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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