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?

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Answer 1 · 2017-09-30T08:11:19

$μ_{m} = 50$ $mu_m=50$

$σ_{m} = \sqrt{\frac{20^{2}}{100}} = \frac{20}{10} = 2$ $sigma_m=sqrt(20^2/100)=20/10=2$

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

if $¯ ¯ ¯ X$ $" "barX$ is the sample mean the sampling distribution is approximately Normal given by

$¯ ¯ ¯ X ~ N (μ, \frac{σ^{2}}{n})$ $barX~N(mu, sigma^2/n)$

in this case we have

$¯ ¯ ¯ X ~ N (50, \frac{20^{2}}{100})$ $barX~N(50, 20^2/100)$

the mean will be $μ_{m} = 50$ $mu_m=50$

standard error $σ_{m} = \sqrt{\frac{20^{2}}{100}} = \frac{20}{10} = 2$ $sigma_m=sqrt(20^2/100)=20/10=2$