Correct. The answer is true

(a) μ = (2+3+6+9)/4 = 5

(b) s^{2} = (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 s^{2}_{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

s^{2}_{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), s^{2}_{mean}=
s^{2}/n
since the right-hand side is 5.48/2=2.74, agreeing with the above value.