Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result
\(\begin{array}{c|c} \text{No. defective per box} & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text{No. of boxes} & 2 & 7 & 17 & 10 & 8 & 6\end{array}\)
The mean and the median of the distribution are respectively
\(\begin{array}{c|c} \text{No. defective per box} & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text{No. of boxes} & 2 & 7 & 17 & 10 & 8 & 6\end{array}\)
The mean and the median of the distribution are respectively
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Correct Answer: Option A
Explanation:
\(Mean = \frac{\sum fx}{\sum f}\)
= \(\frac{333}{50} = 6.66 \approxeq 6.7\)
The median is the average of the 25th and 26th position = 6.
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| No of defective(x) | No of boxes (f) | \(fx\) |
| 4 | 2 | 8 |
| 5 | 7 | 35 |
| 6 | 17 | 102 |
| 7 | 10 | 70 |
| 8 | 8 | 64 |
| 9 | 6 | 54 |
| \(\sum\) | 50 | 333 |
\(Mean = \frac{\sum fx}{\sum f}\)
= \(\frac{333}{50} = 6.66 \approxeq 6.7\)
The median is the average of the 25th and 26th position = 6.
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