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sergij07 [2.7K]
3 years ago
6

Suppose a large shipment of telephones contained 21% defectives. If a sample of size 498 is selected, what is the probability th

at the sample proportion will differ from the population proportion by less than 3%
Mathematics
1 answer:
grandymaker [24]3 years ago
6 0

Answer:

P(-3\% < x < 3\%) = 0.901

Step-by-step explanation:

Given

p = 21\%

n = 498

Required

P(-3\% < x < 3\%)

First, we calculate the z score

z = \sqrt{p * (1 - p)/n}

z = \sqrt{21\% * (1 - 21\%)/498}

z = \sqrt{21\% * (79\%)/498}

z = \sqrt{0.1659/498}

z = \sqrt{0.000333}

z = 0.0182

So:

P(-3\% < x < 3\%) = P(-3\%/0.0182 < z

P(-3\% < x < 3\%) = P(1.648 < z

From z probability, we have:

P(-3\% < x < 3\%) = 0.901

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