Question #327708

Random sample of size N = 2 are drawn from a finite population consisting of


the numbers 1, 3, 8, 9.


Find the following:


Mean, Variance and Standard deviation of the Population.


Mean, Variance and Standard deviation of the Sample means.

1
Expert's answer
2022-04-16T04:12:56-0400

The population mean:

μ=1+3+8+94=5.25.\mu=\cfrac{1+3+8+9}{4}=5.25.


The population variance:

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

Xμ=={15.25,35.25,85.25,95.25}=X-\mu=\\ =\begin{Bmatrix} 1-5.25,3-5.25,8-5.25,9-5.25 \end{Bmatrix}=

={4.25,2.25,2.75,3.75},=\begin{Bmatrix} -4.25, -2.25, 2.75, 3.75 \end{Bmatrix},

σ2=(4.25)214+(2.25)214+2.75214++3.75214=11.189.\sigma^2=(-4.25)^2\cdot \cfrac{1}{4}+(-2.25)^2\cdot \cfrac{1}{4}+2.75^2\cdot \cfrac{1}{4}+\\ +3.75^2\cdot \cfrac{1}{4}=11.189.


The population standard deviation:

σ=11.189=3.345.\sigma=\sqrt{11.189}=3.345.


The mean of the sampling distribution of sample means:

μxˉ=μ=5.25.\mu_{\bar x} =\mu=5.25.


The variance of the sampling distribution of sample means:

σxˉ2=σ2n=11.1892=5.594.\sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{11.189}{2}=5.594.


The standard deviation of the sampling distribution of sample means:

σxˉ=σn=3.3452=2.365.\sigma_{\bar x}=\cfrac{\sigma}{\sqrt n}=\cfrac{3.345}{\sqrt 2}=2.365.



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