Question 15 Marks
The sum of the deviations of a set of n values $x_1, x_2, x_3,..., x_n$ measured from $15$ and $-3$ are $-90$ and $54$ respectively. Find the value of n and mean.
Answer
View full question & answer→Given $\sum\limits^{\text{i}=1}_{\text{n}}(\text{x}_\text{i}-15)=-90$$\Rightarrow(\text{x}_1-15)+(\text{x}_2-15)+\dots+(\text{x}_\text{n}-15)=-90$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})\\-(15+15+15+\dots+15)=-90$
$\Rightarrow\sum\text{x}-\text{15n}=-90\dots(1)$
And $\sum\limits^{\text{i}-1}_{\text{n}}(\text{x}_\text{i}+3)=54$
$\Rightarrow(\text{x}_1+3)+(\text{x}_2+3)+\dots+(\text{x}_\text{n}+3)=54$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(3+3+3+\dots+3)=54$
$\Rightarrow\sum\text{x}+\text{3n}=54\dots(2)$
By subtracting equation (1) from equation (2), we get
$\sum\text{x}+\text{3n}-\sum\text{x}+\text{15n}=54+90$
$\Rightarrow\text{18n}=144$
$\Rightarrow\text{n}=\frac{144}{18}=8$
Put value of n in equation (1)
$\sum\text{x}-15\times8=-90$
$\Rightarrow\sum\text{x}-120=-90$
$\Rightarrow\sum\text{x}=120-90=30$
$\therefore\text{x}=\frac{\sum\text{x}}{\text{n}}=\frac{30}{8}=3.75$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})\\-(15+15+15+\dots+15)=-90$
$\Rightarrow\sum\text{x}-\text{15n}=-90\dots(1)$
And $\sum\limits^{\text{i}-1}_{\text{n}}(\text{x}_\text{i}+3)=54$
$\Rightarrow(\text{x}_1+3)+(\text{x}_2+3)+\dots+(\text{x}_\text{n}+3)=54$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(3+3+3+\dots+3)=54$
$\Rightarrow\sum\text{x}+\text{3n}=54\dots(2)$
By subtracting equation (1) from equation (2), we get
$\sum\text{x}+\text{3n}-\sum\text{x}+\text{15n}=54+90$
$\Rightarrow\text{18n}=144$
$\Rightarrow\text{n}=\frac{144}{18}=8$
Put value of n in equation (1)
$\sum\text{x}-15\times8=-90$
$\Rightarrow\sum\text{x}-120=-90$
$\Rightarrow\sum\text{x}=120-90=30$
$\therefore\text{x}=\frac{\sum\text{x}}{\text{n}}=\frac{30}{8}=3.75$