I.
m e a n = μ = 8 + 10 + 12 + 14 + 16 5 = 12 mean=\mu=\dfrac{8+10+12+14+16}{5}=12 m e an = μ = 5 8 + 10 + 12 + 14 + 16 = 12
v a r i a n c e = σ 2 = 1 5 ( ( 8 − 12 ) 2 + ( 10 − 12 ) 2 variance=\sigma^2=\dfrac{1}{5}((8-12)^2+(10-12)^2 v a r ian ce = σ 2 = 5 1 (( 8 − 12 ) 2 + ( 10 − 12 ) 2
+ ( 12 − 12 ) 2 + ( 14 − 12 ) 2 + ( 16 − 12 ) 2 ) = 8 +(12-12)^2+(14-12)^2+(16-12)^2)=8 + ( 12 − 12 ) 2 + ( 14 − 12 ) 2 + ( 16 − 12 ) 2 ) = 8
σ = σ 2 = 8 = 2 2 \sigma=\sqrt{\sigma^2}=\sqrt{8}=2\sqrt{2} σ = σ 2 = 8 = 2 2
II. There are ( 5 2 ) = 10 \dbinom{5}{2}=10 ( 2 5 ) = 10 samples of size two which can be drawn without replacement:
S a m p l e S a m p l e m e a n ( 8 , 10 ) 9 ( 8 , 12 ) 10 ( 8 , 14 ) 11 ( 8 , 16 ) 12 ( 10 , 12 ) 11 ( 10 , 14 ) 12 ( 10 , 16 ) 13 ( 12 , 14 ) 13 ( 12 , 16 ) 14 ( 14 , 16 ) 15 \begin{matrix}
Sample & Sample\ mean \\
(8,10) & 9 \\
(8,12) & 10 \\
(8,14) & 11 \\
(8,16) & 12 \\
(10,12) & 11 \\
(10,14) & 12 \\
(10,16) & 13\\
(12,14) & 13 \\
(12,16) & 14 \\
(14,16) & 15 \\
\end{matrix} S am pl e ( 8 , 10 ) ( 8 , 12 ) ( 8 , 14 ) ( 8 , 16 ) ( 10 , 12 ) ( 10 , 14 ) ( 10 , 16 ) ( 12 , 14 ) ( 12 , 16 ) ( 14 , 16 ) S am pl e m e an 9 10 11 12 11 12 13 13 14 15
III.
X ˉ P ( X ˉ ) 9 0.1 10 0.1 11 0.2 12 0.2 13 0.2 14 0.1 15 0.1 \begin{matrix}
\bar{X} & P(\bar{X}) \\
9 & 0.1 \\
10 & 0.1 \\
11 & 0.2 \\
12 & 0.2 \\
13 & 0.2 \\
14 & 0.1\\
15 & 0.1\\
\end{matrix} X ˉ 9 10 11 12 13 14 15 P ( X ˉ ) 0.1 0.1 0.2 0.2 0.2 0.1 0.1 IV.
μ X ˉ = 9 ( 0.1 ) + 10 ( 0.1 ) + 11 ( 0.2 ) + 12 ( 0.2 ) \mu_{\bar{X}}=9(0.1)+10(0.1)+11(0.2)+12(0.2) μ X ˉ = 9 ( 0.1 ) + 10 ( 0.1 ) + 11 ( 0.2 ) + 12 ( 0.2 )
+ 13 ( 0.2 ) + 14 ( 0.1 ) + 15 ( 0.1 ) = 12 +13(0.2)+14(0.1)+15(0.1)=12 + 13 ( 0.2 ) + 14 ( 0.1 ) + 15 ( 0.1 ) = 12
∑ i X ˉ i 2 P ( X i ˉ ) = 9 2 ( 0.1 ) + 1 0 2 ( 0.1 ) + 1 1 2 ( 0.2 ) \sum_i\bar{X}_i^2P(\bar{X_i})=9^2(0.1)+10^2(0.1)+11^2(0.2) ∑ i X ˉ i 2 P ( X i ˉ ) = 9 2 ( 0.1 ) + 1 0 2 ( 0.1 ) + 1 1 2 ( 0.2 )
+ 1 2 2 ( 0.2 ) + 1 3 2 ( 0.2 ) + 1 4 2 ( 0.1 ) + 1 5 2 ( 0.1 ) = 147 +12^2(0.2)+13^2(0.2)+14^2(0.1)+15^2(0.1)=147 + 1 2 2 ( 0.2 ) + 1 3 2 ( 0.2 ) + 1 4 2 ( 0.1 ) + 1 5 2 ( 0.1 ) = 147
σ X ˉ 2 = ∑ i X ˉ i 2 P ( X i ˉ ) − μ X ˉ 2 = 147 − 1 2 2 = 3 \sigma_{\bar{X}}^2=\sum_i\bar{X}_i^2P(\bar{X_i})-\mu_{\bar{X}}^2=147-12^2=3 σ X ˉ 2 = ∑ i X ˉ i 2 P ( X i ˉ ) − μ X ˉ 2 = 147 − 1 2 2 = 3
σ X ˉ = σ X ˉ 2 = 3 \sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{3} σ X ˉ = σ X ˉ 2 = 3
μ X ˉ = 12 , σ X ˉ = 3 \mu_{\bar{X}}=12, \sigma_{\bar{X}}=\sqrt{3} μ X ˉ = 12 , σ X ˉ = 3
V.
The mean μ X ˉ \mu_{\bar{X}} μ X ˉ and standard deviation σ X ˉ \sigma_{\bar{X}} σ X ˉ of the sample mean X ˉ \bar{X} X ˉ satisfy
μ X ˉ = 12 = μ , \mu_{\bar{X}}=12=\mu, μ X ˉ = 12 = μ ,
σ X ˉ = 3 = 2 2 2 5 − 2 5 − 1 = σ n N − n N − 1 \sigma_{\bar{X}}=\sqrt{3}=\dfrac{2\sqrt{2}}{\sqrt{2}}\sqrt{\dfrac{5-2}{5-1}}=\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}} σ X ˉ = 3 = 2 2 2 5 − 1 5 − 2 = n σ N − 1 N − n
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