anaik100 wrote:
The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per driver?
A. 0.5
B. 1
C. 1.5
D. 2
E. 4
If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.There are total of \(17+13+21+4+2+2+1=60\) drivers (60 terms) so the median number of accidents per driver would be the average of accidents of 30th and 31st drivers (as we have even # of terms).
30th term equals to 1 and 31st term equals to 2 so \(median=\frac{1+2}{2}\).
Answer: C.
To elaborate more, you can imagine these data points as:
0, ..., 0, 1, ..., 1, 2, ..., 2, 3, ..., 3, 4, ..., 4, 5, ..., 5, 6, ..., 6 --> 17 zeros, 13 ones, 21 twos and so on, total of 60 data points. Median would be the average of 30th and 31st terms: \(median=\frac{1+2}{2}\).
Hope it's clear.
_________________