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,

$-18 n=-144$

$n=\frac{-144}{-18}$

$n=8$

$n \bar{x}-15 \times 8=-90$

$8 \bar{x}-120=-90$

$8 \bar{x}=-90+120$

$8 \bar{x}=30$

$\bar{x}=\frac{30}{8}$

$=\frac{15}{4}$

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Updated on: 10-Oct-2022

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