solve for the mean, variance, and standard deviation of the given set of data of a population.
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}μ=N∑X =8+10,+12+14+16+186=13\frac{8+ 10,+12+ 14+16 +18}{6}=1368+10,+12+14+16+18=13
2.) Standard Deviation σ=∑i=1n(xi−μ)2n\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2} {n}}σ=n∑i=1n(xi−μ)2 =
σ=(8−13)2+(10−13)2+(12−13)2+(14−13)2+(16−13)2+(18−13)26\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 =706=11.667=3.42\sqrt{\frac{70}{6}}=\sqrt{11.667}=3.42670=11.667=3.42
3.) Variance= sd2=3.422=11.67sd^2= 3.42^2= 11.67sd2=3.422=11.67
⟹ σnN–nN–1=3.4236–36–1=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⟹nσN–1N–n=33.426–16–3=1.529 .
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