We have population values 8, 10, 12, 14, 16 and 18
Mean and Standard Deviation and Variance of population is computed as follows:
1.) Mean= μ = ∑ X N \mu = \frac{{\sum X}}{N} μ = N ∑ X =8 + 10 , + 12 + 14 + 16 + 18 6 = 13 \frac{8+ 10,+12+ 14+16 +18}{6}=13 6 8 + 10 , + 12 + 14 + 16 + 18 = 13
2.) Standard Deviation σ = ∑ i = 1 n ( x i − μ ) 2 n \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2} {n}} σ = n ∑ i = 1 n ( x i − μ ) 2 =
σ = ( 8 − 13 ) 2 + ( 10 − 13 ) 2 + ( 12 − 13 ) 2 + ( 14 − 13 ) 2 + ( 16 − 13 ) 2 + ( 18 − 13 ) 2 6 \sigma = \sqrt{\frac{(8-13)^2+(10-13)^2+(12-13)^2+(14-13)^2+(16-13)^2+(18-13)^2} {6}} σ = 6 ( 8 − 13 ) 2 + ( 10 − 13 ) 2 + ( 12 − 13 ) 2 + ( 14 − 13 ) 2 + ( 16 − 13 ) 2 + ( 18 − 13 ) 2 =70 6 = 11.667 = 3.42 \sqrt{\frac{70}{6}}=\sqrt{11.667}=3.42 6 70 = 11.667 = 3.42
3.) Variance = s d 2 = 3.4 2 2 = 11.67 sd^2= 3.42^2= 11.67 s d 2 = 3.4 2 2 = 11.67
⟹ σ n N – n N – 1 = 3.42 3 6 – 3 6 – 1 = 1.529 \implies \dfrac{\sigma }{{\sqrt n }}\sqrt {\dfrac{{N – n}}{{N – 1}}} = \dfrac{{3.42}}{{\sqrt 3 }}\sqrt {\dfrac{{6 – 3}}{{6 – 1}}} = 1.529 ⟹ n σ N –1 N – n = 3 3.42 6–1 6–3 = 1.529 .
Comments