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

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

Explanation:

Given

`p = 21\%`

`n = 498`

Required

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

First, we calculate the z score

`z = √(p * (1 - p)/n)`

`z = √(21\% * (1 - 21\%)/498)`

`z = √(21\% * (79\%)/498)`

`z = √(0.1659/498)`

`z = √(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`