Tinkering with the median, and finding the unknown

The median is the middle value of the sorted data, with as many values below it as above. What happens to it when the data changes?

Take data whose median is 8, and include a new value 11. That is above the median, so now more values are above 8 than below it. The middle has moved.

With the new count, the median is the average of the two middle values, 8 and 11, which is 9.5. So including a value above the median raises it, and including one below lowers it.

The median usually moves less than the mean, because it only cares about position (where a value sits in the order), not about how far away the value is. One huge value pulls the mean a long way, but hardly moves the median.

Now the reverse problem: using a known mean to find a missing value. The key fact: mean × number of values = total.

Coach Balwan wrote down the weights of ten kushti (wrestling) players and their mean, but one weight got smudged. The nine weights he can read are 42, 40, 39, 33, 48, 38, 42, 35 and 32, and the mean is 39.2 kg.

The total must be 39.2 × 10 = 392. The nine known weights add up to 349. So the missing weight is 392 − 349 = 43 kg.

Venkayya's problem is similar but trickier. His average coconut harvest is 25.6 per tree over 15 trees, so the recorded total is 25.6 × 15 = 384. But one tree's count was written 3 too high, so the true total is 384 − 3 = 381, and the correct average is 381/15 = 25.4.

Neither problem needs all the separate values, only the total, which the mean gives you.