Question #325911

Random samples of size 3 are taken from a population of the numbers 3, 4, 5, 6 , 7 and 8. Construct the sampling distribution of the sample means and determine its standard deviation.

1
Expert's answer
2022-04-11T09:52:07-0400

All the possible samples of sizes n=3 wich can be drawn without replacement from the population:

{(3,4,5),(3,4,6),(3,4,7),(3,4,8),(3,5,6),(3,5,7),(3,5,8),(3,6,7),(3,6,8),(3,7,8),(4,5,6),(4,5,7),(4,5,8),(4,6,7),(4,6,8),(4,7,8),(5,6,7),(5,6,8),(5,7,8),(6,7,8)}.\{ (3,4,5), (3,4,6),(3,4,7),(3,4,8),(3,5,6),\\ (3,5,7),(3,5,8),(3,6,7),(3,6,8),(3,7,8),\\ (4,5,6),(4,5,7),(4,5,8),(4,6,7),(4,6,8),\\ (4,7,8),(5,6,7),(5,6,8),(5,7,8),(6,7,8)\}.


Population mean:

μ=3+4+5+6+7+86=5.5.\mu=\cfrac{3+4+5+6+7+8}{6}=5.5.


Population variance:

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

Xμ={35.5,45.5,55.5,65.5,75.5,85.5}=X-\mu=\begin{Bmatrix} 3-5.5,4-5.5, 5-5.5, 6-5.5,7-5.5,8-5.5 \end{Bmatrix}=

={2.5,1.5,0.5,0.5,1.5,2.5},=\begin{Bmatrix} -2.5, -1.5, -0.5,0.5, 1.5,2.5 \end{Bmatrix},

σ2=(2.5)216+(1.5)216+(0.5)216++0.5216+1.5216+2.5216=2.917.\sigma^2=(-2.5)^2\cdot \cfrac{1}{6}+(-1.5)^2\cdot \cfrac{1}{6}+(-0.5)^2\cdot \cfrac{1}{6}+\\ +0.5^2\cdot \cfrac{1}{6}+1.5^2\cdot \cfrac{1}{6}+2.5^2\cdot \cfrac{1}{6}=2.917.


Population standard deviation:

σ=2.917=1.708.\sigma=\sqrt{2.917}=1.708.


The standard deviation of the sampling distribution of sample means:

σxˉ=σn=1.7083=0.986.\sigma_{\bar x}=\cfrac{\sigma}{\sqrt n}=\cfrac{1.708}{\sqrt 3}=0.986.

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