The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 marks each. The mean and the variance obtained from this distribution are found to be 40 and 49, respectively. It was later discovered that two observations belonging to the class interval (21 - 30) were included in the class interval (31 - 40) by mistake. Find the mean and the variance after correcting the error.

Answer

49.25

Explanation

Given , `n=40, bar(x)=40, var(x) =49`
`rArr bar(x)=(Sigmaf_(i)x_(i))/(40)=40 rArrSigmaf_(i)x_(i)=1600`
Also, var`(x)=49`
`rArr (1)/(40)Sigma f_(i)(x_(i)-40)^(2)=49`
`therefore 49=(1)/(40) (Sigmax_(i)^(2)f_(i))-2Sigma x_(i)f_(i)+40Sigma f_(i)`
`rArr 49=(1)/(40) (Sigma x_(i)^(2)f_(i))-2(1600)+40xx40`
`therefore Sigma x_(i)^(2) f_(i)=1649 xx40`
Let (21-30) and (31-40) denote the kth and (k + 1)th class intervals, respectively.
Then, if before correction `f_(k) and f_(k+1)` are frequencies of those intervals, then after correction (2 observations are shifted form (31-40) to (21-30), frequency of kth interval becomes `f_(k+2)` and frequency of (k+1)th interval becomes `f_(k+1)-2`
`therefore bar(x)_("new")=(1)/(40){underset(i ne k,k+1)(sum_(i=1)^(40)f_(i)x_(i))+(f_(k)+2)x_(k)+(f_(k+1)-2)x_(k+1)}`
`bar(x)_("new")=(1)/(40){sum_(i=1)^(40)f_(i)x_(i)}+(2)/(40){(x_(k)-x_(k+1))}`
`=(1)/(40)sum_(i=1)^(40)f_(i)x_(i) +(1)/(20)(-10)=39.5`
`=(1)/(40)[sum_(i=1)^(40)f_(i)(x_(i)-39.5)^(2) +f_(k)(x_(k)-39.5)^(2)+f_(k+1)(x_(k+1)-39.5)^(2)" where " i ne k, k+1]`
`=(1)/(40)[Sigma(f_(i)x_(i)^(2)-79f_(i)x_(i)+(39.5)^(2)f_(i))]`
`=(1)/(40) sum_(i=1)^(40)f_(i)*x_(i)^(2)-79((1)/(40)) sum_(i=1)^(40)f_(i)x_(i)+(39.5)^(2)*(1)/(40) sum_(i=1)^(40)f_(i)`
`=1649-3160+1560.25=49.25`