The manager of a customer support center takes an SRS of 40 reported issues and finds that 22% of the sampled issues required more than one call to resolve. The manager may take an SRS like this each month. Suppose that it is really an average of 25% of the approximately 1000 issues reported per month that require more than one call.

Let represent the proportion of a sample of 40 reported issues that require more than one call to resolve. What are the mean and standard deviation of the sampling distribution of p?

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

25% of the approximately 1000 issues reported per month that require more than one call.

This means that [tex]p = 0.25[/tex]

What are the mean and standard deviation of the sampling distribution of p?

Sample of 40 means that [tex]n = 40[/tex].

By the Central Limit Theorem,

The mean is [tex]\mu = p = 0.25[/tex]

The standard deviation is [tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.25*0.75}{40}} = 0.0685[/tex]

The mean of the sampling distribution of p is 0.25 and the standard deviation is 0.0685.