Some GRE problems feature a second type of average: the median, or “middle value.” The median is calculated in one of two ways, depending on the number of data points in the set.
For lists containing an odd number of values, the median is the unique middle value when the data are arranged in increasing (or decreasing) order.
For lists containing an even number of values, the median is the average (arithmetic mean) of the two middle values when the data are arranged in increasing (or decreasing) order.
A note on terminology: the GRE is very precise in its use of mathematical terms. If a question refers to a group of numbers as a set, you know that all of the numbers are different from each other; because the mathematical definition of a set says that there can’t be any repeat values. However, if a question refers to a group of numbers as a list, repeat values are allowed, but not required. So {3, 4, 9, 17} is a set and a list, but {3, 4, 9, 9}, is just a list. You can find the median of any list or set of numbers though. You can also find the median of a dataset (or data set), which despite having the word set in its name, is actually just a large list in the language of math.
The median of the set {5, 17, 24, 25, 28} is the unique middle number, 24. The median of the list {3, 4, 9, 17} is the mean of the two middle values (4 and 9), or 6.5. Notice that the median of a list containing an odd number of values must be a value in the set. However, the median of a list containing an even number of values does not have to be in the list—and indeed will not be, unless the two middle values are equal.
Unlike the arithmetic mean, the median of a set depends only on the one or two values in the middle of the ordered set. Therefore, you may be able to determine a specific value for the median of a set even if one or more unknowns are present.
For instance, consider the unordered list {x, 2, 5, 11, 11, 12, 33}. No matter whether x is less than 11, equal to 11, or greater than 11, the median of the resulting set will be 11. (Try substituting different values of x to see that the median does not change.)
By contrast, the median of the unordered list {x, 2, 5, 11, 12, 12, 33} depends on x. If x is 11 or less, the median is 11. If x is between 11 and 12, the median is x. Finally, if x is 12 or more, the median is 12.
What is the median of the set {6, 2, −1, 4, 0}?
What is the median of the set {1, 2, x, 8}, if 2 < x < 8?