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The sums of the deviations of a set of $n$ values $x_1, x_2,… x_n$ measured from $15$ and $-3$ are $-90$ and $54$ respectively. Find the value of $n$ and mean.
Given:
The sums of the deviations of a set of $n$ values $x_1, x_2,… x_n$ measured from $15$ and $-3$ are $-90$ and $54$ respectively.
To do:
We have to find the value of $n$ and mean.
Solution:
We know that,
Mean $\overline{X}=\frac{Sum\ of\ the\ observations}{Number\ of\ observations}$
Therefore,
In the first case,
$(x_1 - 15) + (x_2 - 15) + (x_3 - 15) + … + (x_n - 15) = -90$
$x_1 + x_2 + x_3 + … + x_n - 15 \times n = -90$
$n \bar{x}-15 n=-90$........(i)
In the second case,
$(x_1 +3) + (x_2 + 3) + (x_3 + 3) + … + (x_n + 3) = 54$
$x_1 + x_2 + x_3 + … + x_n + 3 \times n = 54$
$n \bar{x}+3 n=54$..............(ii)
Subtracting (ii) from (i), we get,
$-18 n=-144$
$n=\frac{-144}{-18}$
$n=8$
$n \bar{x}-15 \times 8=-90$
$8 \bar{x}-120=-90$
$8 \bar{x}=-90+120$
$8 \bar{x}=30$
$\bar{x}=\frac{30}{8}$
$=\frac{15}{4}$