Question

We have the following results of two samples of polythene bags from two manufacturer $A$ and $B$ tested by a prospective buyer for bursting pressure.

Bursting pressure (in $kg$ )	Number of bags manufactured
	A	B
5 -10	2	9
10-15	9	11
15-20	29	18
20-25	54	32
25-30	11	27
30-35	5	13

Which of the following are true?
I. Bags manufactured by $B$ have higher bursting pressure.
II. Bags manufactured by A have higher bursting pressure.
III. Bags manufactured by $A$ have more uniform pressure.
IV. Bags manufactured by $B$ have more uniform pressure.

• A. I and III are true
• B. III and IV are true
• C. II and IV are true
• D. I and IV are true

Answer 1

Answer: (A)

To find out the correct statements, we compute coefficient of variation.
I. For manufacturer $A$

Bursting pressure	Mid value	$f_i$	$u_i=\frac{x_i-17.5}{5}$	$f_i{u}_i$	$f_i{u}_i^2$
5 -10	7.5	2	-2	-4	8
10-15	12.5	9	-1	-9	9
15-20	17.5	29	0	0	0
20-25	22.5	54	1	54	54
25-30	27.5	11	2	22	44
30-35	32.5	5	3	15	45
		$N=\Sigma f_i=110$	$\Sigma u_i=3$	$\Sigma f_i{u}_i=78$	$\Sigma f_i{u}_i^2=160$

${\bar{x}}_A=a+h\left(\frac{\Sigma f{u}_i}{N}\right)$
$\Rightarrow {\bar{x}}_A=17.5+5\times \frac{78}{110}$
$\Rightarrow {\bar{x}}_A=17.5+3.5$
$\Rightarrow {\bar{x}}_A=\mathrm{21....}$(i)
$\Rightarrow {\sigma }_A^2=h^2\left[\frac{1}{N}\Sigma t{u}_i^2-{\left(\frac{1}{N}\Sigma t_i\right)}^2\right]$
$\Rightarrow {\sigma }_A^2=25\left[\frac{160}{110}-{\left(\frac{78}{110}\right)}^2\right]$
$\Rightarrow {\sigma }_A^2=25\left[\frac{17600-6084}{110\times 110}\right]$
$\Rightarrow {\sigma }_A^2=23.79$
${\sigma }_A=\sqrt{23.79}=4.87$
$\therefore$ Coefficient of variation $=\frac{{\sigma }_A}{{\bar{x}}_A}\times 100=\frac{4.87}{21}\times 100$
$=\mathrm{23.19......}$(ii)
II. For manufacturer B

Bursting pressure	Mid value	$f_i$	$u_i=\frac{x_i-17.5}{5}$	$f_i{u}_i$	$f_i{u}_i^2$
5 -10	7.5	9	-2	-18	36
10-15	12.5	11	-1	-11	36
15-20	17.5	18	0	0	0
20-25	22.5	32	1	32	32
25-30	27.5	27	2	54	108
30-35	32.5	13	3	39	117
		$N=\Sigma f_i=110$	$\Sigma u_i=3$	$\Sigma f_i{u}_i=96$	$\Sigma f_i{u}_i^2=304$

${\bar{x}}_B=a+h\left(\frac{\Sigma f{u}_i}{N}\right)$
$\Rightarrow {\bar{x}}_B=17.5+5\times \frac{96}{110}$
$\Rightarrow {\bar{x}}_B=17.5+4.36$
$\Rightarrow {\bar{x}}_B=\mathrm{21.86.....}$(iii)
$\Rightarrow {\sigma }_B^2=h^2\left[\frac{1}{N}\left(\Sigma f{u}_i^2\right)-{\left(\frac{1}{N}\Sigma f{u}_i\right)}^2\right]$
$\Rightarrow {\sigma }_B^2=25\left[\frac{304}{110}-{\left(\frac{96}{110}\right)}^2\right]$
$\Rightarrow {\sigma }_B^2=25\left[\frac{33440-9216}{110\times 110}\right]$
$\Rightarrow {\sigma }_B^2=50.04$
$\Rightarrow {\sigma }_B=\sqrt{50.04}$
$\Rightarrow {\sigma }_B=7.07$
$\therefore$ Coefficient of variation $=\frac{{\sigma }_B}{{\bar{x}}_B}\times 100$
$=\frac{7.07}{21.86}\times 100=\mathrm{32.34.....}$(iv)
We find that the average bursting pressure is higher for manufacturer B [from Eqs. (i) and (iii)].
Also, the coefficient of variation is less for manufacturer A [from Eqs. (ii) and (iv)]. So, bags manufactured by A have more uniform pressure.