Question #202043

solve for the mean, variance, and standard deviation of the given set of data of a population.


1
Expert's answer
2021-06-03T15:38:34-0400

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=XN\mu = \frac{{\sum X}}{N} =8+10,+12+14+16+186=13\frac{8+ 10,+12+ 14+16 +18}{6}=13



2.) Standard Deviation σ=i=1n(xiμ)2n\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2} {n}} =


σ=(813)2+(1013)2+(1213)2+(1413)2+(1613)2+(1813)26\sigma = \sqrt{\frac{(8-13)^2+(10-13)^2+(12-13)^2+(14-13)^2+(16-13)^2+(18-13)^2} {6}} =706=11.667=3.42\sqrt{\frac{70}{6}}=\sqrt{11.667}=3.42


3.) Variance= sd2=3.422=11.67sd^2= 3.42^2= 11.67


    σnNnN1=3.4236361=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 .


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