Below are the essentials to statistics on the GRE math section. These concepts are essential, so make sure you master them before walking into the test!

## Mean

The mean is another way of saying average. To find the average add up the number of elements in the set and divide by the number of elements. For instance, let’s say I am counting the average number of push-ups my friends can do:

John: 10

Ray: 15

Cyrus: 22

Sue: 33

I find the total: . I then divide by the number of elements. In this case, I have four friends.

Another concept is range. The range of my friends’ push-ups above would be the difference between the greatest and the smallest: 33–10=23.

## Median

The median is the middle number in a list of numbers arranged in ascending order.

For instance, in the list 3, 4, 5, 8, 12, the number 5 is in the middle. Hence it is the median.

A little confusion arises when you have an even set of numbers. 3, 4, 4, 7, 9, 10. In this case 4 and 7, the innermost numbers, cannot both be medians. So we end up taking their average: [pmath](4+7)/2 = 5.5.

## Mode

Set A: {2, 2, 2, 4, 5.5, 7, 7}

The number that shows up most in a list of numbers is the mode. Because the number two shows up the most frequently it is known as the mode. A list can have more than one mode, so if we were to add another 7 to the list above, 2 and 7 would both be the mode. However, if each number shows up only once, then we say that a set either has no mode or that every number in the set is a mode (for the GRE don’t really worry about this last part—I just want to make sure I dot my mathematical “i”s).

## Standard Deviation

This one frightens many people, and understandably so. The good news is the GRE will rarely test you on finding the exact standard deviation on a list of numbers. (That fun task is reserved for stats 101 courses). Instead, the GRE wants you to have a sense of standard deviation.

Set A: 3, 4, 5, 6, 7

Set B: 2, 3, 5, 7, 8

Both sets have the same average. Notice how the spread in Set B is greater than that in Set A. Or, put another way, the numbers in Set A are clumped more closely together. The closer the spread (or the “clumpier”) the numbers, the lower the standard deviation.

The standard deviation for Set A is approximately 1.4 and that for Set B is approximately 2.3.

## Distribution curve

This post is about normal distribution.

The above are intended as primer for statistics you’ll be seeing in GRE questions. In all likelihood they will probably get you to about the 50% level. However, if you are looking to score higher, esp. in the top 10%, then you will definitely want to practice with difficult questions. But don’t fret so much, as there aren’t so many statistics questions on the GRE. 🙂

### Most Popular Resources

Hello Chris,

Thanks for giving us an idea about what to expect in GRE on Statistics. I was wondering whether Standard Deviation for Grouped Data(i.e for discrete and continuous frequency distribution) is also included in the syllabus?

Thanks,

Abir

Hi Abir,

You are very welcome! Glad to hear you are enjoying our blog 🙂

The GRE expects you to have a basic working understanding of standard deviation, but you won’t need to make any calculations for grouped data. Knowledge of discrete and continuous frequency distributions is far above what you’ll be expected to know for the GRE. Chris covers all of the most important points here, and you won’t need to go above and beyond the information in this blog post 🙂

Hello Chris,

Can you elaborate the derivation of 2.3 from Set B?

Hi Muhammad,

Chris talks a little about this in other comments to this post, but basically to calculate the standard deviation for Set B, we first need to see how much each term deviates from the mean, so Set B: {-3, -2, 0, 2, 3}.

Next, we have to find the sum of the squares of each of these terms, so:

(-3)^2 + (-2)^2 + 0^2 + 2^2 + 3^2 = 9 + 4 + 0 + 4 + 9 = 26

Now we find the average of this: 26/5 = 5.2

Finally, we divide by the square root of this 5.2, which results in approximately 2.3! 🙂

great work bro

Hey Chris,

Thanks for the reply. I actually did my Magoosh drills and watched the videos for this topic and i understood the concept. I am not a patron of SD formula, because its length gives me a heart attack. So i use the concepts taught in videos. Well, here is what i learnt, if i take the average of the data values away from the mean for set A i get an answer of 1.2

Set A: 3, 4, 5, 6, 7 (Average is 5 for this set) so the distance from the mean for the values

is 2+1+0+1+2= 6/5 = 1.2 ( Now my question is why are you getting 1.4 and I am getting 1.2 using this method) Is my method not precise?

For set B; If i use the same method, i get SD of 2, but your’s is 2.3 🙁

Thanks,

Amit

Hi Amit,

Yeah, the formula is pretty unwieldy :). These numbers aren’t too bad though. The thing you forgot to do is square the difference between each number and the mean. Therefore, it should be 2^2 + 1^2 + 1^1 + 2^2. 10/5 = 2. Finally, we take the square root of 2 (the 2 is known as the variance–I don’t think you have to know this for the GRE). That gives us 1.4ish.

Hope that makes sense!

Hi Chris,

This is about the standard deviation question above. Isn’t the SD in set A one, because the difference between the numbers is 1. I think standard deviation tells us about the deviation within a set.

Thanks,

Amit

Hi Amit,

SD is about how much numbers deviate from the mean, not how much they deviate from one another. The mean of Set A is 5. Notice that 3 and 7 are two away from 5 on the number line. When we use the SD formula, we will have to square that difference to get 2^2 + 2^2 = 8. When we do the same thing with 4 and 6, we get 1^2 + 1^2 = 2. That is a total of 8 + 2 = 10/5 = 2. To get SD we take square root of 2, which is 1.4 roughly.

Hope that makes sense!