Question

A population has a mean of 200 and a standard deviation of 50. Suppose a random sample of 100 people is selected from this population. What is the probability that the sample mean will be within +/- 5 of the population mean? Hint: use the z-score.

Answer 1

a)

Given:

Population mean \(\mu=200\)

Standard deviation \(\sigma=50\)

Size of the sample \(\mathrm{n}=100\)

The expected value of \(\bar{x}\) is given by:

\(E(\bar{x})=\mu\)

\(=200\)

b)

Standard deviation of \(\bar{x}\) is given by:

$$ \begin{aligned} \sigma_{\bar{x}} &=\frac{\sigma}{\sqrt{n}} \\ &=\frac{50}{\sqrt{100}} \\ &=5 \end{aligned} $$

c)

In selecting simple random samples of size \(n\) from a population, the sampling distribution of the

sample mean \(\bar{x}\) can be approximated by a normal distribution as the sample size becomes

large.

So the sampling distribution of \(\bar{x}\) is normal with mean \(E(\bar{x})=200\) and standard deviation

\(\sigma_{\bar{x}}=5\)

d)

The sampling distribution of \(\bar{x}\) provides the probability information about how close the sample

mean \(\bar{x}\) is to the population mean \(\mu\)