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