Q.
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.
| 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 |
Statistics
Solution:
To find out the correct statements, we compute coefficient of variation.
I. For manufacturer $A$
Bursting pressure
Mid value
$f_i$
$u_i = \frac{x_i - 17.5}{5}$
$f_iu_i$
$f_iu_i^2$
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
$N = \Sigma f_i = 110$
$\Sigma u_i = 3$
$\Sigma f_iu_i = 78$
$\Sigma f_iu_i^2 = 160$
$ \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=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$
$=23.19......$(ii)
II. For manufacturer B
Bursting pressure
Mid value
$f_i$
$u_i = \frac{x_i - 17.5}{5}$
$f_iu_i$
$f_iu_i^2$
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
$N = \Sigma f_i = 110$
$\Sigma u_i = 3$
$\Sigma f_iu_i = 96$
$\Sigma f_iu_i^2 = 304$
$ \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=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 $
$=\frac{7.07}{21.86} \times 100=32.34 .....$(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.
| Bursting pressure | Mid value | $f_i$ | $u_i = \frac{x_i - 17.5}{5}$ | $f_iu_i$ | $f_iu_i^2$ |
|---|---|---|---|---|---|
| 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 |
| $N = \Sigma f_i = 110$ | $\Sigma u_i = 3$ | $\Sigma f_iu_i = 78$ | $\Sigma f_iu_i^2 = 160$ |
| Bursting pressure | Mid value | $f_i$ | $u_i = \frac{x_i - 17.5}{5}$ | $f_iu_i$ | $f_iu_i^2$ |
|---|---|---|---|---|---|
| 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 |
| $N = \Sigma f_i = 110$ | $\Sigma u_i = 3$ | $\Sigma f_iu_i = 96$ | $\Sigma f_iu_i^2 = 304$ |