Solution: Look at the columns! Notice that 85 is the average of 80 and 90.
If 80 and 90 appeared with equal frequency, then the student’s average
for the course would be 85. However, six of the data points correspond
with the average of 80 and four of the data points correspond with the
average of 90. The average for the course will thus be closer to 80 than to
90 and therefore less than 85. The correct answer is B.
Though it is tested less commonly, you should also know how to calculate a
weighted average. Let’s use the preceding example:
A student took 10 exams for his biology course. His average on 6 of the
exams was 80. His average on the other 4 exams was 90. What was his
average for the 10 exams?
■ Approach 1: Use the average formula. For the 4 courses in which he
averaged 90, the sum is 90(4). For the 6 courses in which he averaged 80, the
sum is 80(6). The average for all 10 courses is thus 80(6) + 90(4) 10 =^84010 = 84.
■ Approach 2: Use the weights.
Note that 80(6) + 90(4) 10 = 80( 106 ) + 90( 104 ). The fractions represent the weight
of each data point as a piece of the total weights of the set. From this,
you can extrapolate the following formula:
weighted average = total number of data pointsfrequency of data point 1 × data point 1 +
frequency of data point
total number of data points × average of data point 2
Note that this approach works when the number of data points is
represented as a ratio or percent as well!
For the first 10 days of a 30-day period, a stock’s average return was $6.00.
For the last 20 days of the 30-day period, the stock’s average return was
$12.00. What was the stock’s average return for the 30-day period?
SOLUTION: The two data points are $6.00 and $12.00. The frequency of the
$6.00 data point is 3010 =^13. The frequency of the $12.00 data point is^2030 =^23.
The weighted average is thus (^13 )6 + (^23 )12 = 2 + 8 = 10.
More on the Median
Earlier, the section defined the median as the middle data point when the data
points are in increasing order. Though calculating the median for small sets simply
requires that you put the data points in increasing order and find the middle value,
it becomes trickier when the set has a large number of data points. For example, if
you want the median of 51 data points, it would be too time-consuming to list all
of them. Instead, you should recognize that the median will be the 26th data point.
330 PART 4 ■ MATH REVIEW
04-GRE-Test-2018_313-462.indd 330 12/05/17 12:03 pm