This page is part of my unofficial solutions manual to the GRE Paper Practice Book (2e), a free resource available on the ETS website. They publish the questions; I explain the answers. If you haven’t worked through the Practice Book, give Section 5 a shot before reading this!
5.8: Standard Deviation (SD) with Charts
Let’s just get this out of the way:
The GRE will seldom, if ever, ask you to calculate standard deviation directly. If it looks like you have to calculate SD to solve a problem, there’s probably a faster way.
Calculating the SD for even a modestly-sized data set is a tedious and time-consuming task, even by GRE standards. For a pair of data sets with 18 elements each (which is what problem 5.8 gives us to work with), the process could easily take five minutes or more. Instead, SD problems on the GRE usually test your general, qualitative sense of how SD works — and, in particular, your ability to compare the SDs of different data sets. For a detailed guide to this, see my post “Standard Deviation for the GRE.”
When comparing the SDs of two data sets, the rule is always the same:
The data set whose elements are more widely dispersed about the mean will have the larger SD.
So which data set is “more widely dispersed” in problem 5.8? First, let’s find the mean of each distribution. Because the distributions are symmetrical, it’s fairly straightforward to show that the mean in each case is 30:
- Pair up the 10s and 50s, which average out to 30.
- Pair up the 20s and 40s, which also average out to 30.
- Pair up the 30s, which (of course) average out to 30 as well.
Now, how are the values in each distribution dispersed relative to that mean? In distribution A, a third of the values (6 out of 18) are exactly equal to the mean, another third are relatively close to the mean (the 20s and the 40s), and the remaining third are far from the mean (the 10s and the 50s). Informally, we could say that A has a fairly narrow dispersion pattern.
B, in contrast, has very few values (2 out of 18) equal to the mean and lots of values (10 out of 18) at the far edges of the graph. When compared with A, the dispersion pattern of B is wide. It must therefore have the larger SD. Because Quantity B (the SD of distribution B) is larger than Quantity A (the SD of distribution A), the correct answer is (B).
Students are sometimes uncomfortable with this informal reasoning — in some cases, so much so that they would rather calculate the SD of each distribution directly, just to be sure. This is a costly tradeoff that robs minutes from the rest of the math section. Worse, the dozens of small calculations leave plenty of room for an arithmetic error; one such mistake, and the “direct” solution is no more accurate than an intuitive guess. For problems like 5.8, a broad comparison of the two data sets really is all that’s required. If that feels too much like guessing, here’s what I recommend: do it anyway. Then flag the problem for later review, and come back to it only if you finish the remainder of the section.
Math Review Reference
For more on this topic, see the following section of the GRE Math Review:
- 4.2: Numerical Methods for Describing Data (pp. 71-73)