The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean sand standard deviation?

Explanation

`"Mean "=40rArr(overset(100)underset(i=1)Sigmax_(i))/(10)=40`
`rArr" "overset(100)underset(i=1)Sigmax_(i)=4000.`
`SD=5.1rArrsigma^(2)=(5.1)^(2)`
`rArr" "(overset(100)underset(i=1)Sigmax_(i)^(2))/(100)-(40)^(2)=26.01`
`rArr" "overset(100)underset(i=1)Sigmax_(i)^(2)=162601.`
Thus, incorrect `(overset(100)underset(i=1)Sigmax_(i))=4000 and"incorrect "(overset(100)underset(i=1)Sigmax_(i)^(2))=162601`
Now, incorrect `(overset(100)underset(i=1)Sigmax_(i))=4000`
`rArr" correct "(overset(100)underset(i=1)Sigmax_(i))=(4000-50+40)=3990`
`rArr" correct mean"=(3990)/(100)=39.9." ...(i)"`
`"And, correct "(overset(100)underset(i=1)Sigmax_(i)^(2))=162601`
`rArr" correct"(overset(100)underset(i=1)Sigmax_(i)^(2))={162601-(50)^(2)+(40)^(2)}=161701`
`rArr" correct variance "=("correct "(overset(100)underset(i=1)Sigmax_(i)^(2)))/(100)-("correct mean")^(2)`
`={(161701)/(100)-(39.9)^(2)}={1617.01-(40-0.1)^(2)}`
`=(1617.01)-{1600+0.01-8}`
`=(1617.01-1592.01)=25`
`rArr" correct "SD=sqrt(25)=5.`
Hence, correct mean =39.9 and correct SD=5.