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

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

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

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

#### Solution

For factory A
Let the assumed mean A = 800 and h = 100.

 Length of Life(in hours) Mid-Values $\left( x_i \right)$ $u_i = \frac{x_i - 800}{100}$ $u_i^2$ Number of bulbs $\left( f_i \right)$ $f_i u_i$ $f_i u_i^2$ 550–650 600 −2 4 10 −20 40 650–750 700 −1 1 22 −22 22 750–850 800 0 0 52 0 0 850–950 900 1 1 20 20 20 950–1050 1000 2 4 16 32 64 $\sum_{} f_i = 120$ $\sum_{} f_i u_i = 10$ $\sum_{} f_i u_i^2 = 146$

Mean,

$X_A = A + h\left( \frac{\sum_{} f_i u_i}{\sum_{} f_i} \right) = 800 + 100 \times \frac{10}{120} = 808 . 33$
Standard deviation,
$\sigma_A = h\sqrt{\left( \frac{\sum_{} f_i u_i^2}{\sum_{} f_i} \right) - \left( \frac{\sum_{} f_i u_i}{\sum_{} f_i} \right)^2} = 100\sqrt{\left( \frac{146}{120} \right) - \left( \frac{10}{120} \right)^2} = 100 \times 1 . 0998 = 109 . 98$
∴ Coefficient of variation = $\frac{\sigma_A}{X_A} \times 100 = \frac{109 . 98}{808 . 33} \times 100 = 13 . 61$
For factory B
Let the assumed mean A = 800 and h = 100.
 Length of Life(in hours) Mid-Values $\left( x_i \right)$ $u_i = \frac{x_i - 800}{100}$ $u_i^2$ Number of bulbs $\left( f_i \right)$ $f_i u_i$ $f_i u_i^2$ 550–650 600 −2 4 8 −16 32 650–750 700 −1 1 60 −60 60 750–850 800 0 0 24 0 0 850–950 900 1 1 16 16 16 950–1050 1000 2 4 12 24 48 $\sum_{}f_i = 120$ $\sum_{} f_i u_i = - 36$ $\sum_{} f_i u_i^2 = 156$

Mean,

$X_B = A + h\left( \frac{\sum_{} f_i u_i}{\sum_{} f_i} \right) = 800 + 100 \times \frac{\left( - 36 \right)}{120} = 800 - 30 = 770$
Standard deviation,
$\sigma_B = h\sqrt{\left( \frac{\sum_{} f_i u_i^2}{\sum_{} f_i} \right) - \left( \frac{\sum_{} f_i u_i}{\sum_{} f_i} \right)^2} = 100\sqrt{\left( \frac{156}{120} \right) - \left( \frac{- 36}{120} \right)^2} = 100 \times 1 . 1 = 110$
∴ Coefficient of variation = $\frac{\sigma_B}{X_B} \times 100 = \frac{110}{770} \times 100 = 14 . 29$
Since the coefficient of variation of factory B is greater than the coefficient of variation of factory A, therefore, factory B has more variability than factory A. This means bulbs of factory A are more consistent from the point of view of length of life.
Concept: Statistics - Statistics Concept
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.7 | Q 11 | Page 49