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?
Solution
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 | ||||
x_{i} | f_{i} |
u_{i} = `(x_"i" - "A")/"h"` A = 7.5, |
f_{i}u_{i} | f_{i}u_{i}^{2} = f_{i}u_{i} × u_{i} |
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 | ||||
x_{i} | f_{i} |
u_{i} = `(x_"i" - "A")/"h"` A = 7.5, |
f_{i}u_{i} | f_{i}u_{i}^{2} = f_{i}u_{i} × u_{i} |
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.