The sum of deviations of a set of values `x_(1), x_(2), x_(3), …… x_(n)` measured from `50` is `-10` and the sum of deviations of the values from `46` is `70.` Find the value of `n` and the mean.

Options

A. `40.5`
B. `39.5`
C. `47.5`
D. `49.5`

Answer

D

Explanation

Here, `(x_(1)-50)+(x_(2)-50)+ .. + (x_(n)-50) = -10`
`rArr (x_(1)+x_(2)+x_(3)+...+ x_(n)) = 50n-10 " " ....(1)`
`Further, (x_(1)-46)+(x_(2)-46)+(x_(3)-46)+...+(x_(n)-46) = 70`
`rArr (x_(1) +x_(2) +x_(3)+...+x_(4)) = 46n +70 " "....(2)`
From eqs. (1) and (2), we get `50n-10 = 46n+70`
`rArr n=(80)/(4) = 20`
`therefore " The required mean"`
`(bar(x)) = (x_(1) + x_(2) + ..+x_(n))/(n) = (46 xx 20+70)/(20) = 49.5`