Question 13 Marks
There are two factories employing 100 and 80 men, respectively. If the arithmetic mean of their monthly salaries are Rs.575 and Rs.625, then find the arithmetic mean of the salaries of both the factories together.
Answer
View full question & answer→Let $n_1$ be the no. of persons in the first factory and $\bar{X}_1$ be the mean of the first factory workers, and $n_2$ be the number of persons in the second factory and their mean be $\bar{X}_2$
$\begin{array}{l}
\because n_1=100 \text { and } \bar{X}_1=575 \text { and } n_2=80 \text { and } \bar{X}_2=625 \\
\therefore \text { Combined Mean }\left(\bar{X}_{1,2}\right)=\frac{n_1 \bar{X}_1+n_2 \bar{X}_2}{n_4+n_2} \\
\Rightarrow \bar{X}_{1,2}=\frac{575 \times 100+625 \times 80}{100+80}=\frac{57500+50000}{180} \\
=\frac{107500}{\frac{180}{X_{1,2}}=597.2} \\
\therefore 597.2
\end{array}$
$\begin{array}{l}
\because n_1=100 \text { and } \bar{X}_1=575 \text { and } n_2=80 \text { and } \bar{X}_2=625 \\
\therefore \text { Combined Mean }\left(\bar{X}_{1,2}\right)=\frac{n_1 \bar{X}_1+n_2 \bar{X}_2}{n_4+n_2} \\
\Rightarrow \bar{X}_{1,2}=\frac{575 \times 100+625 \times 80}{100+80}=\frac{57500+50000}{180} \\
=\frac{107500}{\frac{180}{X_{1,2}}=597.2} \\
\therefore 597.2
\end{array}$