Question

Suppose a large shipment of microwave ovens contained 7% defectives. If a sample of size 300 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

Answer 1

95.86% probability that the sample proportion will differ from the population proportion by less than 3%

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`p = 0.07`

For a proportion, we have that:

`\mu = p = 0.07`

`\sigma = \sqrt{(p(1-p))/(n)} = \sqrt{(0.07*0.93)/(300)} = 0.0147`

What is the probability that the sample proportion will differ from the population proportion by less than 3%

This is the pvalue of Z when X = 0.07 + 0.03 = 0.1 subtracted by the pvalue of Z when X = 0.07 - 0.03 = 0.04. So

X = 0.1

`Z = (X - \mu)/(\sigma)`

`Z = (0.1 - 0.07)/(0.0147)`

`Z = 2.04`

`Z = 2.04` has a pvalue of 0.9793

X = 0.04

`Z = (X - \mu)/(\sigma)`

`Z = (0.04 - 0.07)/(0.0147)`

`Z = -2.04`

`Z = -2.04` has a pvalue of 0.0207

0.9793 - 0.0207 = 0.9586

95.86% probability that the sample proportion will differ from the population proportion by less than 3%