Question

The daily wages of two firms A,B are given. If combined mean is $49.87$ and SD is $10.67$, find $a$.

Firm	A	B
No. of employees	$586$	$647$
Mean Salary	$52.5$	$47.5$
S.D.of salary	$10$	$a$

• A. $4$
• B. $12$
• C. $11$
• D. $20$

Answer 1

Answer: (C)

$11$

$n_1=586,{\overline{x}}_1=52.5,{\sigma }_1=10,$ $n_2=647,{\overline{x}}_2=47.5,{\sigma }_2=a$

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}=\frac{586\times 52.5+647\times 47.5}{586+647}=419.87$

$d_1={\overline{x}}_1-\overline{x}=52.5-49.87=2.62$

$d_2={\overline{x}}_2-\overline{x}=47.5-49.87=-2.37$

${\sigma }^2(n_1+n_2)=n_1({\sigma }_1^2+d_1^2)+n_2({\sigma }_2^2+d_2^2)$

${\sigma }^2\left(1233\right)=586\times (100+6.86)+647(a^2+5.61)$

${\left(10.67\right)}^2\times 1233=62619.96+647a^2+3629.6$

$a^2=114.5$

$a=10.7$

$10.7\approx 11$