# The sum of the deviations of a set of n values

Question:

The sum of the deviations of a set of $n$ values $x_{1}, x_{2}, x_{3}, \cdots, x_{n}$ measured from 15 and $-3$ are $-90$ and 54 respectively. Find the value of $n$ and mean.

Solution:

Given: $\sum_{\mathrm{n}}^{\mathrm{i}=1}\left(\mathrm{x}_{\mathrm{i}}-15\right)=-90$

$\Rightarrow\left(x_{1}-15\right)+\left(x_{2}-15\right)+\cdots \cdot \cdot+\left(x_{n}-15\right)=-90$

$\Rightarrow\left(x_{1}+x_{2}+\cdots+n\right)-(15+15+15+\cdots \cdots+15)=-90$

$\Rightarrow \sum x-15 n=-90 \cdots(1)$

And $\sum_{n}^{i=1}\left(x_{i}+3\right)=54$

$\Rightarrow\left(x_{1}+3\right)+\left(x_{2}+3\right)+\cdots \cdot+\left(x_{n}+3\right)=54$

$\Rightarrow\left(x_{1}+x_{2}+\cdots+n\right)+(3+3+3+\cdots \cdots+3)=54$

$\Rightarrow \sum x+3 n=54 \cdots(2)$

By subtracting equation (1) from equation (2), we get

$\sum x+3 n-\sum x+15 n=54+90$

⇒18n=144

⇒ n = 144/18 = 8

Put value of n in equation (1)

$\sum x-15 \times 8=-90$

$\sum x-120=-90$

$\sum x=120-90=30$

$\therefore \mathrm{x}=\frac{\sum \mathrm{x}}{\mathrm{n}}=\frac{30}{8}=3.75$