Question #323880

A population consist of 2,4,5,9 and 10, with sample size of 3. Compute the mean and variance of the

sampling distribution of the sample mean.


Expert's answer

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

μxˉ=μ==2+4+5+9+105=6.\mu_{\bar x} =\mu=\\ =\cfrac{2+4+5+9+10} {5} =6.

Population variance:

σ2=(xiμ)2P(xi),\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),

Xμ=={26,46,56,96,106}=X-\mu=\\ =\begin{Bmatrix} 2-6, 4-6, 5-6, 9-6, 10-6 \end{Bmatrix}=

={4,2,1,3,4},=\begin{Bmatrix} -4, -2, - 1, 3, 4 \end{Bmatrix},

σ2=(4)215+(2)215+(1)215++3215+4215=9.2.\sigma^2=(-4)^2\cdot \cfrac{1}{5}+(-2)^2\cdot \cfrac{1}{5}+(-1)^2\cdot \cfrac{1}{5}+\\ +3^2\cdot \cfrac{1}{5}+4^2\cdot \cfrac{1}{5}=9.2.


Variance of the sampling distribution of sample means:

σxˉ2=σ2n=9.23=3.07.\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{9.2}{3}=3.07.

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