Question

Life of bulbs produced by two factories A and B are given below:

Length of life (in hours)	Factory A (Number of bulbs)	Factory B (Number of bulbs)
550 – 650	10	8
650 – 750	22	60
750 – 850	52	24
850 – 950	20	16
950 – 1050	16	12
	120	120

The bulbs of which factory are more consistent from the point of view of length of life?

Answer 1

Here h = 100

Let A (assumed mean) = 800.

Length of life (in hour)	Mid values ${\displaystyle \left(x_i\right)}$	${\displaystyle y_i=\frac{x_i-A}{10}}$	Factory A			Factory B
			${\displaystyle f_i}$	${\displaystyle f_i{y}_i}$	${\displaystyle f_i{y}_i^2}$	${\displaystyle f_i}$	${\displaystyle f_i{y}_i}$	${\displaystyle f_i{y}_i^2}$
550 – 650	600	–2	10	–20	40	8	–16	32
650 – 750	700	–1	22	–22	22	60	– 60	60
750 – 850	800	0	52	0	0	24	0	0
850 – 950	900	1	20	20	20	16	16	16
950 – 1050	1000	2	16	32	64	12	24	48
			120	10	146	120	–36	156

For factory A

Mean ${\displaystyle \left(\overline{x}\right)=800+\frac{10}{120}\times 100}$ = 816.67 hours

S.D. = ${\displaystyle \frac{100}{120}\sqrt{120\left(146\right)-100}}$ = 109.98 

Therefore, Coefficient of variation (C.V.) = ${\displaystyle \frac{S.D.}{\overline{x}}\times 100}$

= ${\displaystyle \frac{109.98}{816.67}\times 100}$

= 13.47

For factory B

Mean = ${\displaystyle 800+\frac{-36}{120}100}$ = 770

S.D. = ${\displaystyle \frac{100}{120}\sqrt{120\left(156\right)-{(-36)}^2}}$ = 110

Therefore, Coefficient of variation = ${\displaystyle \frac{S.D.}{\text{Mean}}\times 100}$

= ${\displaystyle \frac{110}{770}\times 100}$

= 14.29

Since C.V. of factory B > C.V. of factory A

⇒ Factory B has more variability which means bulbs of factory A are more consistent.