The population mean:
μ = 1 + 3 + 8 + 9 4 = 5.25. \mu=\cfrac{1+3+8+9}{4}=5.25. μ = 4 1 + 3 + 8 + 9 = 5.25.
The 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 = = { 1 − 5.25 , 3 − 5.25 , 8 − 5.25 , 9 − 5.25 } = X-\mu=\\
=\begin{Bmatrix}
1-5.25,3-5.25,8-5.25,9-5.25
\end{Bmatrix}= X − μ = = { 1 − 5.25 , 3 − 5.25 , 8 − 5.25 , 9 − 5.25 } =
= { − 4.25 , − 2.25 , 2.75 , 3.75 } , =\begin{Bmatrix}
-4.25, -2.25, 2.75, 3.75
\end{Bmatrix}, = { − 4.25 , − 2.25 , 2.75 , 3.75 } ,
σ 2 = ( − 4.25 ) 2 ⋅ 1 4 + ( − 2.25 ) 2 ⋅ 1 4 + 2.7 5 2 ⋅ 1 4 + + 3.7 5 2 ⋅ 1 4 = 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. σ 2 = ( − 4.25 ) 2 ⋅ 4 1 + ( − 2.25 ) 2 ⋅ 4 1 + 2.7 5 2 ⋅ 4 1 + + 3.7 5 2 ⋅ 4 1 = 11.189.
The population standard deviation:
σ = 11.189 = 3.345. \sigma=\sqrt{11.189}=3.345. σ = 11.189 = 3.345.
The mean of the sampling distribution of sample means:
μ x ˉ = μ = 5.25. \mu_{\bar x} =\mu=5.25. μ x ˉ = μ = 5.25.
The variance of the sampling distribution of sample means:
σ x ˉ 2 = σ 2 n = 11.189 2 = 5.594. \sigma^2_{\bar x}=\cfrac{\sigma^2}{n}=\cfrac{11.189}{2}=5.594. σ x ˉ 2 = n σ 2 = 2 11.189 = 5.594.
The standard deviation of the sampling distribution of sample means:
σ x ˉ = σ n = 3.345 2 = 2.365. \sigma_{\bar x}=\cfrac{\sigma}{\sqrt n}=\cfrac{3.345}{\sqrt 2}=2.365. σ x ˉ = n σ = 2 3.345 = 2.365.
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