Answer to Question #306331 in Calculus for Jeremiah Macapobre Ramos

Question

Our hypothetical population contains the scores 4, 6, 7, and 9. Determine

the mean and variance of the sampling distribution of the sample mean,

given that samples contain two scores drawn from the population with

replacement?

Answer 1

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

Given the  the scores 4, 6, 7, and 9, the possible samples and their means are:

Then,

Sample space Mean (X)

{4,4} (4+4)/2 = 4

{4,6} (4+6)/2 = 5

{6,4} (6+4)/2 = 5

{4,7} (4+7)/2 = 5.5

(7,4} (7+4)/2 = 5.5

{4,9} (4+9)/2 = 6.5

{9,4} (9+4)/2 = 6.5

{6,6) (6+6)/2 = 6

{6,7} (6+7)/2 = 6.5

{7,6} (7+6)/2 = 6.5

{6,9} (6+9)/2 = 7.5

{9,6} (9+6)/2 = 7.5

{7,9} (7+9)/2 = 8

{9,7) (9+7)/2 = 8

{7,7) (7+7)/2 = 7

{9,9} (9+9)/2 = 9

Therefore, the possible values for X with respective probabilities are:

X P(X)

4 1/16

5 2/16

5.5 2/16

6 1/16

6.5 4/16

7 1/16

7.5 2/16

8 2/16

9 1/16

Then, the mean of X = ΣX*P(X)

=4(1/16)+5(2/16)+5.5(2/16)+6(1/16)+6.5(4/16)+7(1/6)+7.5(2/16)+8(2/16)+9(1/16)

=7.23

Answer: Mean of X = 7.23

Variance of X

σ_X² = Σ(x – μ)²⋅ P(x)

=$\frac{1}{6}$(4-7.23)² + $\frac{2}{16}$(5-7.23)² + $\frac{2}{16}$(5.5-7.23)²+ $\frac{1}{16}$(6-7.23)² + $\frac{4}{16}$(6.5-7.23)² + $\frac{1}{16}$(7-7.23)² +$\frac{2}{16}$(7.5-7.23)²+ $\frac{2}{16}$(8-7.23)²+ $\frac{1}{16}$(9-7.23)²

= 2.1579

Answer: Variance of X = 2.1579