Question

Suppose a large shipment of laser printers contained 22% defectives. If a sample of size 276 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 6%

Answer 1

Answer:

The probability that the sample proportion will differ from the population proportion by less than 6% is 0.992.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

The information provided is:

$p=0.22\\n=276$

As the sample size is large, i.e. n = 276 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion.

Compute the value of $P(\hat p-p$ as follows:

$P(\hat p-p$

Thus, the probability that the sample proportion will differ from the population proportion by less than 6% is 0.992.