The above table shows a set of scores and their frequencies. If the m

Muammer:

Median is the middle value when all values are arranged in increasing (or decreasing order)

All values are:
10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, ...(x number of times), 13, 13, 13 ... (8 times), 14, 14, 14 (9 times)

So in all there are 5 + 6 + x + 8 + 9 values.

Whenever we have odd number of values, the median is the middle value i.e. if we have 3 values, the median is the 2nd value, if we have 5 values, the median is the 3rd value and so on. So median will be one of the values.
When we have even number of values, then median is the average of the middle two values. So if we have 4 values, median is average of 2nd and 3rd values. Here, since the median does not belong to the list, we must have even number of total values.

Also, the middle two values must be distinct so that their average is different from them. That is, if middle two values are both 12, their average would be 12 too. But if want that the average should not be a part of the list, the middle two values must be something like 12 and 13 so that average and hence median is 12.5.

So what should be the value of x? We already have 5 + 6 = 11 values on the left of x and 8+9 = 17 values on the right of x. So x should be 6 so that we have 17 values (of 10, 11 and 12) and 17 values of (13 and 14). Now the median will be the average of the middle two values, the 17th and the 18th values. Median will be 12 + 13 = 12.5.