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)\}. {( 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 + 8 6 = 5.5. \mu=\cfrac{3+4+5+6+7+8}{6}=5.5. μ = 6 3 + 4 + 5 + 6 + 7 + 8 = 5.5.
Population variance:
σ 2 = ∑ ( x i − μ ) 2 ⋅ P ( x i ) , \sigma^2=\sum(x_i-\mu)^2\cdot P(x_i), σ 2 = ∑ ( x i − μ ) 2 ⋅ P ( x i ) ,
X − μ = { 3 − 5.5 , 4 − 5.5 , 5 − 5.5 , 6 − 5.5 , 7 − 5.5 , 8 − 5.5 } = X-\mu=\begin{Bmatrix}
3-5.5,4-5.5, 5-5.5, 6-5.5,7-5.5,8-5.5
\end{Bmatrix}= X − μ = { 3 − 5.5 , 4 − 5.5 , 5 − 5.5 , 6 − 5.5 , 7 − 5.5 , 8 − 5.5 } =
= { − 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}, = { − 2.5 , − 1.5 , − 0.5 , 0.5 , 1.5 , 2.5 } ,
σ 2 = ( − 2.5 ) 2 ⋅ 1 6 + ( − 1.5 ) 2 ⋅ 1 6 + ( − 0.5 ) 2 ⋅ 1 6 + + 0. 5 2 ⋅ 1 6 + 1. 5 2 ⋅ 1 6 + 2. 5 2 ⋅ 1 6 = 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. σ 2 = ( − 2.5 ) 2 ⋅ 6 1 + ( − 1.5 ) 2 ⋅ 6 1 + ( − 0.5 ) 2 ⋅ 6 1 + + 0. 5 2 ⋅ 6 1 + 1. 5 2 ⋅ 6 1 + 2. 5 2 ⋅ 6 1 = 2.917.
Population standard deviation:
σ = 2.917 = 1.708. \sigma=\sqrt{2.917}=1.708. σ = 2.917 = 1.708.
The standard deviation of the sampling distribution of sample means:
σ x ˉ = σ n = 1.708 3 = 0.986. \sigma_{\bar x}=\cfrac{\sigma}{\sqrt n}=\cfrac{1.708}{\sqrt 3}=0.986. σ x ˉ = n σ = 3 1.708 = 0.986.
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