Question

The weighted arithmetic mean of 10 observations was 36. However a particular observation was recorded as 60 instead of 40. In what ratio should be the weights of correct and incorrect reading so as to have no change in AM.

Answer 1

The weighted mean of a non-empty set of data $x_1{x}_2{x}_3.......x_n$ with non negative weights $w_1,w_2,......w_n$

$\Rightarrow \overline{x}=\frac{{\sum }_{i=1}^n{w}_i{x}_i}{{\sum }_{i=1}^n{w}_i}=\frac{w_1{x}_1+w_2{x}_2+w_3{x}_3+.......+w_n{x}_n}{w_1+w_2+w_3+......w_n}$

Given, the weighted mean of 10 onservations is 36 and also an observation was wrongly written as 60 instead of 40. Say the true weight associated with that observation be 'w' and incorrect be ${}^{\prime }{w}_1^{\prime }.$

Also let the sum of weights of rest of the 9 observations be 'x'. and the sum of rest of weighted observations be 'y'. We have to keep the Am same in either case.

$\therefore 36=\frac{40w+y}{w+x}......\left(1\right)$

and $36=\frac{60w_1+y}{w_1+x}.......\left(2\right)$

From $\left(1\right)$ we have,

$\Rightarrow 36w+36x=40w+y$

$36x-4w=\mathrm{4.......}\left(3\right)$

Using $\left(3\right)$ in $\left(2\right)$ we have

$\Rightarrow 36w_1+36x=36x-4w+60w_1$

$\Rightarrow 4w=24w_1$

$\Rightarrow \frac{w}{w_1}=\frac{6}{1}$

$\therefore$ The ratio of the weights of correct and incorrect reading so as to have no change in AM is $6:1.$

Hence, solved.