The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which are recorded as 21, 21 and 18. Find the mean and standard deviation if the

Explanation

Here `n=100`
Mean `barx=20impliessumx_(i)=20xx100=2000`
and standard deviation `=3`
`impliessqrt((sum(x_(i)-barx)^(2))/100)=3impliessum(x_(i)-barx)^(2)=900` ………….. 1
Omitting the incorrect observation `21,21,18`
Remaining observations `=97`
Their sum `sumx_(i)=sumx_(i)-(21+21+18)`
`=2000-60=1940`
New mean `=(sumX_(i))/97=1940/97=20`
Now from equation (1) `sum(x_(i)-barx)^(2)=900`
`implies sumx_(i)^(2)-2barxsumx_(i)+sum(barx)^(2)=900`
`implies sumx_(i)^(2)-2xx20xx2000+100(20)^(2)=900`
`implies sumx_(i)^(2)=40900`
`:. sumX_(i)^(2)=sumx_(i)^(2)-(21)^(2)+21^(2)+18^(2)`
`=4090-(441+441+324)=39394`
New standard deviation
`=sqrt((sumX_(i)^(2))/97-("new mean")^(2))=sqrt(39694/97-(20)^(2))`
`=sqrt(409.2165-400)=sqrt(9.2165)=3.036`