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?
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Solution
For factory A
Let the assumed mean A = 800 and h = 100.
Length of Life (in hours) 
MidValues
\[\left( x_i \right)\]

\[u_i = \frac{x_i  800}{100}\]

\[u_i^2\]

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.
Let the assumed mean A = 800 and h = 100.
Length of Life (in hours) 
MidValues
\[\left( x_i \right)\]

\[u_i = \frac{x_i  800}{100}\]

\[u_i^2\]

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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