Hey guys! Today we’re going to take a look at a DS problem that is a skills problem, focused on GMAT standard deviation.

#### Standard Deviation & Variance

What they’re asking here is do we have enough information to compute a standard deviation? It’s useful to think of standard deviation as clustering. If we have a whole series of points we can define how clustered or un-clustered the group of points is. That’s all that’s standard deviation, that’s all that variance is. So if we have all the points that works. What we should be on the lookout here for are parametric measurements. Especially things like the average number is, because while the average can be used to compute standard deviation, we need to know how each of the points differs from the average. But if we have each of the points we always get the average. That is, we can compute the average. So the average is a nice looking piece of information that actually has little to no value here. So let’s jump into the introduced information.

#### Statement 1

Number 1 BOOM – tells us that the average number of eggs is 4 and that’s great except that it doesn’t tell us about the clustering. If we run some scenarios here we could have every nest have 4 eggs or we could have 5 nests have 0, 5 nests have 8, or 9 nests have 0, 1 nest has 40. These are all different clusterings and we could end up with anything in between those extremes as well. So number 1 is insufficient.

#### Statement 2

Number 2: tells us that each of the 10 bird’s nests has exactly 4 eggs. What does this mean? We have all 10 points. They happen to all be on the average, which means the standard deviation is 0. that is there’s no clustering whatsoever. But 2 gives us all the information we need so B – 2 alone is sufficient is the answer here.

Hope this was useful guys, check out the links below for a video about how to compute standard deviation as a refresher, as well as other problems related to this one. Thanks for watching we’ll see you again real soon

If you enjoyed this GMAT problem, try another one next: Normative Distribution