A.
We have population values 23, 26, 29, 32, 35, population size N=5 and sample size n=3.
Mean of population ( μ ) (\mu) ( μ ) = 23 + 26 + 29 + 32 + 35 5 = 29 \dfrac{23+26+29+32+35}{5}=29 5 23 + 26 + 29 + 32 + 35 = 29
Variance of population
σ 2 = Σ ( x i − x ˉ ) 2 N = 36 + 9 + 0 + 9 + 36 5 = 18 \sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{36+9+0+9+36}{5}=18 σ 2 = N Σ ( x i − x ˉ ) 2 = 5 36 + 9 + 0 + 9 + 36 = 18
σ = σ 2 = 18 = 3 2 ≈ 4.24264 \sigma=\sqrt{\sigma^2}=\sqrt{18}=3\sqrt{2}\approx4.24264 σ = σ 2 = 18 = 3 2 ≈ 4.24264
The number of possible samples which can be drawn without replacement is N C n = 5 C 3 = 10. ^{N}C_n=^{5}C_3=10. N C n = 5 C 3 = 10.
n o S a m p l e S a m p l e m e a n ( x ˉ ) 1 23 , 26 , 29 26 2 23 , 26 , 32 27 3 23 , 26 , 35 28 4 23 , 29 , 32 28 5 23 , 29 , 35 29 6 23 , 32 , 35 30 7 26 , 29 , 32 29 8 26 , 29 , 35 30 9 26 , 32 , 35 31 10 29 , 32 , 35 32 \def\arraystretch{1.5}
\begin{array}{c:c:c:c:c}
no & Sample & Sample \\
& & mean\ (\bar{x})
\\ \hline
1 & 23,26,29 & 26 \\
\hdashline
2 & 23,26,32 & 27 \\
\hdashline
3 & 23,26,35 & 28\\
\hdashline
4 & 23,29,32 & 28 \\
\hdashline
5 & 23,29,35 & 29 \\
\hdashline
6 & 23,32,35 & 30 \\
\hdashline
7 & 26,29,32 & 29 \\
\hdashline
8 & 26,29,35 & 30 \\
\hdashline
9 & 26,32,35 & 31 \\
\hdashline
10 & 29,32,35 & 32 \\
\hdashline
\end{array} n o 1 2 3 4 5 6 7 8 9 10 S am pl e 23 , 26 , 29 23 , 26 , 32 23 , 26 , 35 23 , 29 , 32 23 , 29 , 35 23 , 32 , 35 26 , 29 , 32 26 , 29 , 35 26 , 32 , 35 29 , 32 , 35 S am pl e m e an ( x ˉ ) 26 27 28 28 29 30 29 30 31 32
X ˉ f ( X ˉ ) X ˉ f ( X ˉ ) X ˉ 2 f ( X ˉ ) 26 1 / 10 26 / 10 676 / 10 27 1 / 10 27 / 10 729 / 10 28 2 / 10 56 / 10 1568 / 10 29 2 / 10 58 / 10 1682 / 10 30 2 / 10 60 / 10 1800 / 10 31 1 / 10 31 / 10 961 / 10 32 1 / 10 32 / 10 1024 / 10 \def\arraystretch{1.5}
\begin{array}{c:c:c:c:c}
\bar{X} & f(\bar{X}) &\bar{X} f(\bar{X})& \bar{X}^2f(\bar{X})
\\ \hline
26 & 1/10 & 26/10 & 676/10 \\
\hdashline
27 & 1/10 & 27/10 & 729/10 \\
\hdashline
28 & 2/10 & 56/10 & 1568/10 \\
\hdashline
29 & 2/10 & 58/10 & 1682/10 \\
\hdashline
30 & 2/10 & 60/10 & 1800/10 \\
\hdashline
31 & 1/10 & 31/10 & 961/10 \\
\hdashline
32 & 1/10 & 32/10 & 1024/10 \\
\hdashline
\end{array} X ˉ 26 27 28 29 30 31 32 f ( X ˉ ) 1/10 1/10 2/10 2/10 2/10 1/10 1/10 X ˉ f ( X ˉ ) 26/10 27/10 56/10 58/10 60/10 31/10 32/10 X ˉ 2 f ( X ˉ ) 676/10 729/10 1568/10 1682/10 1800/10 961/10 1024/10
Mean of sampling distribution
μ X ˉ = E ( X ˉ ) = ∑ X ˉ i f ( X ˉ i ) = 29 = μ \mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=29=\mu μ X ˉ = E ( X ˉ ) = ∑ X ˉ i f ( X ˉ i ) = 29 = μ
The variance of sampling distribution
V a r ( X ˉ ) = σ X ˉ 2 = ∑ X ˉ i 2 f ( X ˉ i ) − [ ∑ X ˉ i f ( X ˉ i ) ] 2 Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2 Va r ( X ˉ ) = σ X ˉ 2 = ∑ X ˉ i 2 f ( X ˉ i ) − [ ∑ X ˉ i f ( X ˉ i ) ] 2
= 8440 10 − ( 29 ) 2 = 3 = σ 2 n ( N − n N − 1 ) =\dfrac{8440}{10}-(29)^2=3= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1}) = 10 8440 − ( 29 ) 2 = 3 = n σ 2 ( N − 1 N − n )
σ X ˉ = 3 ≈ 1.73205 \sigma_{\bar{X}}=\sqrt{3}\approx1.73205 σ X ˉ = 3 ≈ 1.73205
B.
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