If for distribution of 18 observations `sum(x_i-5)=3a n dsum(x_i-5)^2=43 ,` find the mean and standard deviation.

Answer

N/a

Explanation

Given, n=18,`Sigma(x-5)=3 " and " Sigma(x-5)^(2)=43`
`therefore " Mean " =A+(Sigma(x-5))/(18)`
`=5+(3)/(18)=5+0.1666=5.1666=5.17`
and `SD=sqrt((Sigma(x-5)^(2))/n)-((Sigma(x-5))/(n))^(2)`
`=sqrt((43)/(18)-((3)/(18))^(2))`
`=sqrt(2.3944-(0.166)^(2))=sqrt(2.3944-0.2755)=1.59`