Answer to Question #305888 in Statistics and Probability for lester

Question

Consider a population consisting of 2,6,8,0 and 1. Suppose samples of size 2 are drawn from this

population. Find the Mean and Variance of the sampling distribution of sample means.

Answer 1

Let X be the random variable representing the possible values of the sample means

Then

Sample space Mean (X) P(X)

{2,6} (2+6)/2 = 8 1/10

{2,8} (2+8)/2 = 5 1/10

{2,0} (2+0)/2 = 1 1/10

{6,8} (6+8)/2 = 7 1/10

{6,1} (6+1)/2 = 3.5 1/10

{8,0} (8+0)/2 = 4 1/10

{8,1} (8+1)/2 = 4.5 1/10

{0,1} (0+1)/2 = 0.5 1/10

{0,6} (0+6)/2 = 3 1/10

{2,1} (2+1)/2 = 1.5 1/10

The mean of X = $\frac{1}{10}$ (8+5+1+7+3.5+4+4.5+0.5+3+1.5)

= 3.8

Answer: Mean X = 3.8

The variance of X

σ_X² = Σ(x – μ)²⋅ P(x)

=$\frac{1}{10}$(8-3.8)² + $\frac{1}{10}$(5-3.8)² + $\frac{1}{10}$(1-3.8)²+ $\frac{1}{10}$(7-3.8)² + $\frac{1}{10}$(3.5-3.8)² + $\frac{1}{10}$(4-3.8)² +

$\frac{1}{10}$(4.5-3.8)²+ $\frac{1}{10}$(0.5-3.8)²+ $\frac{1}{10}$(3-3.8)² + $\frac{1}{10}$(1.5-3.8)² 

= 5.46

Answer: Variance of X = 5.46