Question

Statement I For the following frequency distribution, the standard deviation is $1.38$.

x	2	3	4	5	6	7
f	4	9	16	14	11	6

Statement II The mean life of a sample of 60 bulbs was $650h$ and the standard deviation was $8h$. A second sample of 80 bulbs has a mean life of $660h$ and standard deviation $7h$. Then, the overall standard deviation is $8.9$.

• A. Only Statement I is true
• B. Only Statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (C)

$x_i$	$f_i$	$d_i=x_i-4$	$f_i{d}_i$	$f_i{d}_i^2$
2	4	-2	-8	16
3	9	-1	-9	9
4	16	0	0	0
5	14	1	14	14
6	11	2	22	44
7	6	3	18	54
	$\Sigma f_i=60$		$\Sigma f_i{d}_i=37$	$\Sigma f_i{d}_i^2=137$

Standard deviation $(\sigma )=\sqrt{\frac{\Sigma f{d}_i^2}{\Sigma f_i}-{\left(\frac{\Sigma f{d}_i}{\Sigma f_i}\right)}^2}$
$=\sqrt{\frac{137}{60}-{\left(\frac{37}{60}\right)}^2}$
$=\sqrt{2.2833-(0.616{)}^2}$
$=\sqrt{2.2833-0.3794}$
$=\sqrt{1.9038}=1.38$
II. We have, $n_1=60,{\bar{x}}_1=650,{\sigma }_1=8$
$n_2=80,{\bar{x}}_2=660,{\sigma }_2=7$
Combined standard deviation ( $\sigma$ )
$=\sqrt{\frac{n_1{\sigma }_1^2+n_2{\sigma }_2^2}{n_1+n_2}+\frac{n_1{n}_2{\left({\bar{x}}_1-{\bar{x}}_2\right)}^2}{{\left(n_1+n_2\right)}^2}}$
$=\sqrt{\frac{60\times 64+80\times 49}{60+80}+\frac{60\times 80\times (660-650{)}^2}{(60+80{)}^2}}$
$=\sqrt{\frac{6\times 64+8\times 49}{14}+\frac{60\times 80\times 100}{140\times 140}}$
$=\sqrt{\frac{3\times 64+4\times 49}{7}+\frac{6\times 8\times 100}{14\times 14}}$
$=\sqrt{\frac{388}{7}+\frac{1200}{49}}=\frac{\sqrt{2716+1200}}{49}$
$=\frac{\sqrt{3916}}{7}=\frac{62.58}{7}=8.9$