Question

Suppose a large shipment of telephones contained 21% defectives. If a sample of size 498 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

Answer 1

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:

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

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

From z probability, we have:

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