Correct. The answer is true

(a) μ = (2+3+6+9)/4 = 5

(b) s² = (2-5)²+(3-5)²+(6-5)²+(9-5)²=9+4+1+16=30 and s=5.48

(c) There are 4*4=16 samples of size two which can be drawn with replacement. These are

(2,2) (2,3) (2,6) (2,9)

(3,2) (3,3) (3,6) (3,9)

(6,2) (6,3) (6,6) (6,9)

(9,2) (9,3) (9,6) (9,9)

The corresponding sample means are:

2.0 2.5 4.0 5.5

2.5 3.0 4.5 6.0

4.0 4.5 6.0 7.5

5.5 6.0 7.5 9.0

and the mean of the sampling distribution of means is

μ _mean = sum of all sample means in (1) above/16 = 5.0 illustrating the fact that m=m_mean.

(d) The variance s²_mean of the sampling distribution of means is obtained by subtracting the mean 5 from each number (1), squaring the result, adding all 16 numbers thus obtained, and dividing by 16. The final result is

s²_mean=120/16=7.5 and s_mean=2.74

This illustrates the fact that for finite populations involving sampling with replacement (or infinite populations), s²_mean= s²/n since the right-hand side is 5.48/2=2.74, agreeing with the above value.