The safety officers of a mining company take an SRS of 500500500 employees and finds that 15\, percent of the sampled employees wear contact lenses. The safety officers may take several more samples like this. Suppose it is really 12\, percent of the approximately 34{,}00034,00034, comma, 000 employees in the company who wear contact lenses. What are the mean and standard deviation of the sampling distribution of the proportion of employees who wear contact lenses

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]

12% of the employees wear contact lenses.

This means that [tex]p = 0.12[/tex]

Samples of 500:

This means that [tex]n = 500[/tex]

What are the mean and standard deviation of the sampling distribution of the proportion of employees who wear contact lenses?

Mean:

[tex]\mu = p = 0.12[/tex]

Standard deviation:

[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.12*0.88}{500}} = 0.0145[/tex]

The mean of the sampling distribution of the proportion of employees who wear contact lenses is 0.12 and the standard deviation is 0.0145.