Describe the sampling distribution of x-bar for samples of size 20 drawn from a normal population with a mean of 50 and standard deviation of 4.

If the original population has a Normal distribution, then the distribution of the sample mean is also Normal.

If we extract random samples of size n from a population with half a μ and standard deviation σ, then the distribution of the sample mean will have the following parameters:

μ_Xbar = μ

σ_Xbar = σ / √ (n).

Where:

μ = 50

σ = 4

n = 20

So:

μ_Xbar = 50

σ_Xbar = 0.8944