Question

A population of PWD learners in 5 schools in a certain municipality of Quezon Province are (𝒙) 9, 5, 6, 12, and 15. Suppose that two schools were selected as samples, determine the mean and variance of the sampling distribution of sample mean. 

Answer 1

The mean of the sampling distribution of the sample means is the mean of the population from which the scores were sampled:

$\mu_{\bar x} =\mu=\\ =\cfrac{9+5+6+12+15} {5} =9.4.$

Population variance:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\\ =\begin{Bmatrix} 9-9.4, 5-9.4, 6-9.4, 12-9.4, 15-9.4 \end{Bmatrix}=$

$=\begin{Bmatrix} -0.4, -4.4, - 3.4, 2.6, 5.6 \end{Bmatrix},$

$\sigma^2=(-0.4)^2\cdot \cfrac{1}{5}+(-4.4)^2\cdot \cfrac{1}{5}+(-3.4)^2\cdot \cfrac{1}{5}+\\ +2.6^2\cdot \cfrac{1}{5}+5.6^2\cdot \cfrac{1}{5}=13.84.$

Variance of the sampling distribution of sample means:

$\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{13.84}{2}=6.92.$