A population distribution has mean 50 and standard deviation 20. For a random sample of size 100, what is the mean and standard error of the sampling distribution of the sample mean?

Sep 30, 2017

${\mu}_{m} = 50$

${\sigma}_{m} = \sqrt{{20}^{2} / 100} = \frac{20}{10} = 2$

Explanation:

we are not told what the background distribution is. But since the sample size is $> 30$ we can use the Central Limit Theorem.

if$\text{ } \overline{X}$ is the sample mean the sampling distribution is approximately Normal given by

barX~N(mu, sigma^2/n)

in this case we have

barX~N(50, 20^2/100)

the mean will be ${\mu}_{m} = 50$

standard error ${\sigma}_{m} = \sqrt{{20}^{2} / 100} = \frac{20}{10} = 2$