Fifty boxes, each of 50 balls, were inspected for the number which were defective, with the results below. The mean and the median of the distribution are respectively
No. of defective per box
4
5
6
7
8
9
No. of boxes
2
7
17
10
8
6
Worked solution (try it first)
Total defective: 4(2)+5(7)+6(17)+7(10)+8(8)+9(6)=333, so the mean is 50333=6.66≈6.7.
With 50 boxes the median is halfway between the 25th and 26th values.
Running totals: 2, 9, 26.
The 10th to 26th values are all 6, so the median is 6.