Solution
Population size N = 5 N= 5 N = 5
Sample size n = 4 n=4 n = 4
Possible samples
= C 5 4 = ( 5 4 ) = 5 =C_5^4= \binom{5}{4}=5 = C 5 4 = ( 4 5 ) = 5 Samples
n o . S a m p l e M e a n 1. 5 , 6 , 9 , 12 8.00 2. 5 , 6 , 9 , 13 8.25 3. 5 , 6 , 12 , 13 9.00 4. 5 , 9 , 12 , 13 9.75 5. 6 , 9 , 12 , 13 10.00 \begin {matrix}no.& Sample &Mean\\1.&5,6,9,12&8.00\\2.&5,6,9,13&8.25\\3.&5,6,12,13&9.00\\4.&5,9,12,13&9.75\\5.&6,9,12,13&10.00\end{matrix} n o . 1. 2. 3. 4. 5. S am pl e 5 , 6 , 9 , 12 5 , 6 , 9 , 13 5 , 6 , 12 , 13 5 , 9 , 12 , 13 6 , 9 , 12 , 13 M e an 8.00 8.25 9.00 9.75 10.00
1. Mean = ∑ X ˉ f ( X ˉ ) =\sum \bar Xf(\bar X) = ∑ X ˉ f ( X ˉ )
X ˉ f f ( X ˉ ) X ˉ f ( X ˉ ) X ˉ 2 f ( X ˉ ) 8.00 1 1 / 5 1.60 12.8000 8.25 1 1 / 5 1.65 13.6125 9.00 1 1 / 5 1.80 16.2000 9.75 1 1 / 5 1.95 19.0125 10.00 1 1 / 5 2.00 20.0000 ∑ 5 1 9.00 81.625 \begin {matrix}\bar X&f&f(\bar X)&\bar Xf(\bar X)&\bar X^2f(\bar X)\\8.00&1&1/5&1.60&12.8000\\8.25&1&1/5&1.65&13.6125\\9.00&1&1/5&1.80&16.2000\\9.75&1&1/5&1.95&19.0125\\10.00&1&1/5&2.00&20.0000\\\sum&5&1&9.00&81.625\end{matrix} X ˉ 8.00 8.25 9.00 9.75 10.00 ∑ f 1 1 1 1 1 5 f ( X ˉ ) 1/5 1/5 1/5 1/5 1/5 1 X ˉ f ( X ˉ ) 1.60 1.65 1.80 1.95 2.00 9.00 X ˉ 2 f ( X ˉ ) 12.8000 13.6125 16.2000 19.0125 20.0000 81.625
Mean = ∑ X ˉ f ( X ˉ ) =\sum \bar Xf(\bar X) = ∑ X ˉ f ( X ˉ )
Mean = 9.000 =9.000 = 9.000
2. Standard deviation
σ = ∑ X ˉ 2 f ( X ˉ ) − ( ∑ X ˉ f ( X ˉ ) ) 2 \sigma=\sqrt {\sum \bar X^2 f(\bar X)-(\sum \bar X f(\bar X))^2 } σ = ∑ X ˉ 2 f ( X ˉ ) − ( ∑ X ˉ f ( X ˉ ) ) 2
σ = 81.625 − 81.000 \sigma=\sqrt{81.625-81.000} σ = 81.625 − 81.000
σ = 0.791 \sigma =0.791 σ = 0.791
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