Solution
1.
For a population;
mean μ = ∑ X i N \mu=\dfrac{\sum X_i}{N} μ = N ∑ X i
Variance σ 2 = ∑ ( X i − μ ) 2 N \sigma^2 =\dfrac{\sum (X_i-\mu)^2}{N} σ 2 = N ∑ ( X i − μ ) 2
Standard deviation σ = ∑ ( X i − μ ) 2 N \sigma=\sqrt\dfrac{\sum(X_i-\mu)^2}{N} σ = N ∑ ( X i − μ ) 2
mean μ = 1 + 9 + 2 + 12 + 8 + 7 + 1 7 \mu = \dfrac{1+9+2+12+8+7+1}{7} μ = 7 1 + 9 + 2 + 12 + 8 + 7 + 1
μ = 5.71 \mu=5.71 μ = 5.71
Variance
X i X i − μ ( X i − μ ) 2 1 − 4.71 22.18 1 − 4.71 22.18 2 − 3.71 13.76 7 1.29 1.66 8 2.29 5.24 9 3.29 10.82 12 6.29 39.56 ∑ 115.40 \def\arraystretch{1.5}\begin{array}{c:c:c} X_i&X_i -\mu&(X_i-\mu)^2\\\hline1&-4.71&22.18\\\hdashline1&-4.71&22.18\\\hdashline2&-3.71&13.76\\\hdashline7&1.29&1.66\\\hdashline8&2.29&5.24\\\hdashline9&3.29&10.82\\\hdashline12&6.29&39.56\\\hline \sum&&115.40\\\hline\end{array} X i 1 1 2 7 8 9 12 ∑ X i − μ − 4.71 − 4.71 − 3.71 1.29 2.29 3.29 6.29 ( X i − μ ) 2 22.18 22.18 13.76 1.66 5.24 10.82 39.56 115.40
σ 2 = 115.4 7 = 16.49 \sigma^2=\dfrac{115.4}{7}=16.49 σ 2 = 7 115.4 = 16.49
Standard deviation
σ = 115.4 7 = 4.06 \sigma=\sqrt\dfrac{115.4}{7} =4.06 σ = 7 115.4 = 4.06
2.
Population size N = 7 N=7 N = 7
Sample size n = 6 n =6 n = 6
Possible number of samples
= C 7 6 = ( 7 6 ) = 7 =C_7^6= \binom{7}{6}=7 = C 7 6 = ( 6 7 ) = 7 Samples
n o . S a m p l e S a m p l e m e a n 1 1 , 1 , 2 , 7 , 8 , 9 4.67 2 1 , 1 , 2 , 7 , 8 , 12 5.17 3 1 , 1 , 2 , 7 , 9 , 12 5.33 4 1 , 1 , 2 , 8 , 9 , 12 5.50 5 1 , 1 , 7 , 8 , 9 , 12 6.33 6 1 , 2 , 7 , 8 , 9 , 12 6.50 7 1 , 2 , 7 , 8 , 9 , 12 6.50 \def\arraystretch{1.5}\begin{array}{c:c:c}no.& Sample & Sample mean\\\hline1&1,1,2,7,8,9&4.67\\\hdashline2&1,1,2,7,8,12&5.17\\\hdashline3&1,1,2,7,9,12&5.33\\\hdashline4&1,1,2,8,9,12&5.50\\\hdashline5&1,1,7,8,9,12&6.33\\\hdashline6&1,2,7,8,9,12&6.50\\\hdashline7&1,2,7,8,9,12&6.50\\\hline\end{array} n o . 1 2 3 4 5 6 7 S am pl e 1 , 1 , 2 , 7 , 8 , 9 1 , 1 , 2 , 7 , 8 , 12 1 , 1 , 2 , 7 , 9 , 12 1 , 1 , 2 , 8 , 9 , 12 1 , 1 , 7 , 8 , 9 , 12 1 , 2 , 7 , 8 , 9 , 12 1 , 2 , 7 , 8 , 9 , 12 S am pl e m e an 4.67 5.17 5.33 5.50 6.33 6.50 6.50
Sampling distribution of the sample means
M e a n P r o b a b i l i t y 4.67 1 7 5.17 1 7 5.33 1 7 5.50 1 7 6.33 1 7 6.50 2 7 \def\arraystretch{1.5}\begin{array}{c:c} Mean& Probability \\\hline4.67&\frac17\\\hdashline5.17&\frac17\\\hdashline5.33&\frac17\\\hdashline5.50&\frac17\\\hdashline6.33&\frac17\\\hdashline6.50&\frac27\\\hline\end{array} M e an 4.67 5.17 5.33 5.50 6.33 6.50 P ro babi l i t y 7 1 7 1 7 1 7 1 7 1 7 2