Question

Answer the following :

There are two companies U and V which manufacture cars. A sample of 40 cars each from these companies are taken and the average running life (in years) is recorded

Life (in years)	No of Cars
	Company U	Company V
0 - 5	5	14
5 - 10	18	8
10 - 15	17	18

Which company shows greater consistency?

Answer 1

We construct the following table to compute C.V.:

For both companies, U and V, the mid-values of the class-intervals are 2.5, 7.5, 12.5

Company U
xi	fi	ui = `(x_"i" - "A")/"h"` A = 7.5, h = 5	fiui	fiui2 = fiui × ui
2.5	5	– 1	– 5	5
7.5	18	0	0	0
12.5	17	1	17	17
	`sumf_"i"` = 40		`sumf_"i"u_"i"` = 12	`sumf_"i"u_"i"^2` = 22

Company U

`bar(u) = (sumf_"i"u_"i")/(sumf_"i") = 12/40` = 0.3

∴ `bar(x) = "A" + "h"bar(u)`

= 7.5 + 5 × 0.3 = 9

`sigma_u^2 = (sumf_"i"u_"i"^2)/(sumf_"i") - (baru)^2`

= `22/40 - (0.3)^2`

= 0.55 – 0.09 = 0.46

∴ `sigma_u = sqrt(0.46)` = 0.678

∴ `sigma_x = "h"sigma_u`  = 5 × 0.678 = 3.39

C.V. = `sigma_x/bar(x) xx 100`

= `(3.39)/9 xx 100`

= 37.67

Company V
xi	fi	ui = `(x_"i" - "A")/"h"` A = 7.5, h = 5	fiui	fiui2 = fiui × ui
2.5	14	– 1	– 14	14
7.5	8	0	0	0
12.5	18	1	18	18
	`sumf_"i"` = 40		`sumf_"i"u_"i"` = 4	`sumf_"i"u_"i"^2` = 32

Company V

`bar(u) = (sumf_"i"u_"i")/(sumf_"i") = 4/40` = 0.1

∴ `bar(x) = "A" + "h"bar(u)`

= 7.5 + 5 × 0.1 = 8

`sigma_u^2 = (sumf_"i"u_"i"^2)/(sumf_"i") - (baru)^2`

= `32/40 - (0.1)^2`

= 0.8 – 0.01 = 0.79

∴ `sigma_u = sqrt(0.79)` = 0.889

∴ `sigma_x = "h"sigma_u`  = 5 × 0.889 = 4.445

C.V. = `sigma_x/bar(x) xx 100`

= `(4.445)/8 xx 100`

= 55.56

Since `("C.V")_("company U")<("C.V.")_("company V")`

company U shows greater consistency in performance.