Mean and standard deviation of 100 observations of are found to be 40 and 10. If at the time of calculation two observations are wrongly taken as 30 and 70 instead of 3 and 27 respectively. Find the correct standard deviation.

Explanation

Given , n=100,` barx=40,sigma=10 ans bar x =40 `
` therefore (Sigma x_(i))/n=40 `
`rArr (Sigmax_(i))/(100)=40`
`rArr Sigma x_(i)=4000`
Now , Corrected `Sigmax_(i)=4000-30-70-+3+27`
` therefore =4030-100=3930`
Corrected mean `=(2930)/(100)=39.3`
Now `sigma^(2)=(Sigmax_(i)^(2))/(n)-(40)^(2)`
`rArr 100=(Sigmax_(i)^(2))/(100)-1600`
`rArr Sigmax_(i)^(2)=17000`
Now Corrected `Sigmax_(i)^(2)=17000-(30)^(2)-(70)^(2)+3^(2)+(27)^(2)`
` rArr Corrected sigma=sqrt((164939)/(100)-(39.3)^(2))`
`=sqrt(1649.39-39.3xx39.3)`
`=sqrt(1649.39-15.44.49)`
`sqrt(104.9)=10.24`