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प्रश्न
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?
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उत्तर
Here h = 100
Let A (assumed mean) = 800.
| Length of life (in hour) |
Mid values `(x_i)` |
`y_i = (x_i - A)/10` | Factory A | Factory B | ||||
| `f_i` | `f_iy_i` | `f_iy_i^2` | `f_i` | `f_iy_i` | `f_iy_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 `(barx) = 800 + 10/120 xx 100` = 816.67 hours
S.D. = `100/120 sqrt(120(146) - 100)` = 109.98
Therefore, Coefficient of variation (C.V.) = `(S.D.)/barx xx 100`
= `109.98/816.67 xx 100`
= 13.47
For factory B
Mean = `800 + (-36)/120 100` = 770
S.D. = `100/120 sqrt(120(156) - (-36)^2)` = 110
Therefore, Coefficient of variation = `(S.D.)/"Mean" xx 100`
= `110/770 xx 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.
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