twallack Wrote:I'm afraid both Ben and the MGMAT strategy book are incorrect. You may indeed be asked how to calculate the standard deviation on the GMAT.
Unfortunately, the GMAT strategy guide not only omits instructions on how to perform the calculation but says "you do not need to know -- it is very unlikely that the GMAT will ask you to calculate an exact SD." That advice led me astray.
I hope the MGMAT will correct the guide soon.
Hi twallack,
Thanks for your comment. I hope that my remarks were not taken to say that SD is not tested on the GMAT; it is on the test. I was suggesting that the calculation of SD is likely to be tested.
In terms of what I know about how SD is tested:
(1) We haven't really seen any official questions (in the OG and GMAT Prep) that requires specific calculation of the SD of a set of data (for example, what is the SD of {-2, 5, 6, 6, 9, 10}). We don't expect to see this because it requires very tedious calculations, including multi-step computations (finding the average, then taking the difference of each data point with the average, squaring, adding these squares up, divide by number in the set, and then taking the square root ... whew!).
(2) It's usually testing the ideas behind SD rather than the SD itself.
This is how I've seen it tested:
(1) You should know that the SD is a measure of spread: the more spread out a set is, the greater the standard deviation. (See OG12 Diagnostic #31) So,
(a) The set {4, 4, 4, 4, 4} has a SD of 0 because it has no spread.
(b) If you multiply each value of a set by 2, the SD increases because the spread of value increases.
(c) If you add 5 to each value of a set of 2, there is no change in the SD because there is no change in the spread.
(2) You should know that the SD = sqrt(variance). (See OG12 DS-134)
(3) You should know that the SD represents a fixed interval in a set of data. See
mean-deviation-t4558.htmlI'm curious to know what aspect of SD you think we've left out. We would really appreciate feedback by emailing
studentservices@manhattangmat.com.
Thanks!