12.2 Tinkering with the Median
The median is the middle value of the sorted data: as many values fall below it as above it. How does it react to change?
Start with a data set whose median is 9. Now insert a new value 13, which is above 9. There are now more values sitting above 9, so 9 can no longer hold the middle. The median climbs, it becomes the average of the two fresh middle values. If those middle values turn out to be 9 and 12, the new median is
\frac{9+12}{2} = 10.5.
By the same logic, inserting a value below the median nudges the median down.
Math Talk Unlike the mean, the median doesn’t care how far away a new value lands, only whether it falls above or below the centre. Inserting the value 13 or the value 5{,}000{,}000 (both above the median 9) would budge the median by the same amount in the same direction. This makes the median stubborn against extreme values. Where could that stubbornness be handy?
Mean or Median: Which Tells the Truer Tale? The mean uses every value, so a single very large or very small value (an outlier) can tug it far from where most of the data lives. The median shrugs such values off. Compare these two lists of daily auto-rickshaw fares (in rupees):
- 12,\ 14,\ 15,\ 16,\ 18 → mean = 15, median = 15;
- 12,\ 14,\ 15,\ 16,\ 98 → mean = 31, median = 15.
One unusual fare drags the mean from 15 up to 31, but the median holds firm at 15. When the data has a few wild values, the median is often the fairer summary of “a typical value.”
Finding an unknown value
Because the mean locks the sum to the count, a single missing value can be recovered the moment the mean is known.
Worked Example
The blurred stopwatch reading. A swimming coach recorded the lap times (in seconds) of 10 trainees: 58, 55, 61, 49, 52, 57, 60, 54, 50, and one reading w that got blurred. The mean time is 55.4 s. Find w.
\frac{58 + 55 + 61 + 49 + 52 + 57 + 60 + 54 + 50 + w}{10} = 55.4.
The nine known times sum to 496, so 496 + w = 55.4 \times 10 = 554, giving
w = 554 - 496 = 58.
The blurred time is 58 s.
Worked Example
The over-counted mango tree. A gardener noted the average mango yield over 15 trees as 27.6, but later found that one tree’s count had been written 3 higher than the true figure. Find the corrected average.
The individual counts aren’t given, but the recorded total is:
\text{recorded total} = 27.6 \times 15 = 414.
Since one count was 3 too high, the true total is 414 - 3 = 411, so the correct average is
\frac{411}{15} = 27.4.
Notice the shortcut tucked inside: shaving 3 off the total of 15 values lowers the mean by \frac{3}{15} = 0.2, sliding it from 27.6 down to 27.4. You never need the individual counts at all.
Worked Example: Combining two groups
Pooling two sections. Section A has 20 students with a mean test score of 62; Section B has 30 students with a mean of 70. What is the mean score of all 50 students together?
Recover each section’s total from \text{sum} = \text{mean} \times \text{count}:
\text{A's total} = 62 \times 20 = 1240, \qquad \text{B's total} = 70 \times 30 = 2100.
The combined mean is therefore
\frac{1240 + 2100}{20 + 30} = \frac{3340}{50} = 66.8.
Note it is not simply \frac{62+70}{2} = 66, the larger section pulls the combined mean toward its own value.