Answer:
5
Step-by-step explanation:
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.
In this problem, we have that:
[tex]\mu = 80, \sigma = 20[/tex]
On average, how much error would be expected between the sample mean and the population mean?
This is the standard deviation of the sample. We have that [tex]n = 16[/tex]. So
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{16}} = 5[/tex]
The answer is 5.
