Question

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?

Answer 1

Answer:

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

Step-by-step explanation:

Central Limit Theorem

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

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 $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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

This means that $p = 0.25$

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

Sample of 40 means that $n = 40$.

By the Central Limit Theorem,

The mean is $\mu = p = 0.25$

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

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