SAT statistics questions usually involve finding the mean, median, and/or mode(s) of a set of numbers. You have probably dealt with with these concepts in your high school math classes but, as always, the SAT likes to put their own special twist on simple concepts such as these.

Whether or not you are familiar with these terms and the techniques needed to find a mean, median, or mode, this guide is for you. SAT questions are always tricky and knowing how to handle their version of these types of questions will serve you well as you go through your test.

**This will be your complete guide to SAT means, medians, and modes**--what they mean, how you’ll see them on the test, and how to solve even the most complicated of SAT statistics questions.

## What Are Means, Medians, and Modes?

Before we look at how to solve these kinds of problems, let us define our terms:

**A mean** is the statistical average of a group of numbers, found by adding up the sum of the numbers and then dividing by the amount of numbers in the group.

What is the average test score for the class if five students received scores of: 92, 81, 45, 95, and 68?

We must find the sum of all the numbers and then divide that number by the total amount, which in this case is 5.

$(92 + 81 + 45 + 95 + 68)/5$

$381/5$

$76.2$

**The mean (average) test score is 72.6.**

**The median** in a set is the number directly in the middle of the set of numbers after they have been arranged in order. (Note: the number will be halfway into the set, but is NOT necessarily the mid-value.)

For instance, in a set of numbers {2, 4, 5, 47, 99}, **the median would be 5** as it is in the middle of the set, despite the fact that 5 is NOT halfway between 2 and 99.

If you are given an even number of terms in the set, then you must take the mean (average) of both middle numbers.

Find the median value of the set of numbers {4, 12, 15, 3, 7, 10}

First, arrange the numbers in order from least to greatest.

3, 4, 7, 10, 12, 15

We have an even number of terms in our set, so we must take the average of the two middle terms.

$(7 + 10)/2$

$17/2$

$8.5$

**Our median is 8.5**

**The mode** of a set of numbers is the number or numbers that repeat the most frequently.

In the set of numbers {3, 4, 3, 4, 4, 5, 12}, **our mode is 4****.** Even though the number 3 occurred twice, the number 4 occurred three times and is thus our most frequently appearing number.

If each number in your set occurs only once, there is no mode.

In the set of numbers {1, 13, 8, 42, 11}, **there is no mode**, since no number repeats.

If multiple numbers in a set repeat the same number of times, your set will have more than one mode.

In the set {1, 2, 2, 2, 5, 5, 5, 7, 8, 8, 8}, **we have three modes--2, 5, and 8.** All three numbers occur exactly three times and no other numbers occur more frequently. Thus we have multiple modes.

*Ba-dum tss!*

## Typical Mean, Median, and Mode Questions

Because the statistical concepts of mean, median, and mode are fundamentally simple (and likely quite familiar to most of you), the SAT will try to complicate mean, median, and mode questions as much as they are able.

Unfortunately, these kinds of twists on simple concepts can come in a variety of different forms.

For **mean questions**, they may ask you for the average of a set with variables, or they may ask you to find the value to which the sum of a set of numbers must be raised or lowered in order to find a particular average.

Just keep in mind, that no matter how odd the question appears to be, the process for finding the mean is unchanging.

Now we can use two different strategies to solve this kind of problem--using algebra or using PIN. Let us look at both methods:

**Method 1--algebra**

If we approach this problem purely algebraically, then we simply use the values we are given and solve for the mean as usual. So we must find the sum of our set of values and divide that sum by the amount of numbers in our set (in this case 3). So:

${(12 - n) + (12) + (12 + n)}/3$

$36/3$

$12$

The $n$’s cancel each other out, so we are left with variables only.

**Our final answer is B**, 12.

**Method 2--plugging in numbers**

If you are someone who is uncomfortable using algebra for whatever reason, then you can solve this problem just as easily by plugging in your own number for $n$ and solving for the mean as usual.

Let us say that $n = 3$. (Why 3? Why not! Remember from your PIN lesson that, when you see a problem with a variable, you can replace that variable with any number you like. So long as it does not contradict any given information, literally any value for $n$ will work here. It’s completely up to you!)

Now, let us find the values in our set.

Our first value will be:

$12 - n$

$12 - 3$

$9$

Our second value will remain unchanged:

$12$

And our third value will be:

$12 + n$

$12 + 3$

$15$

Now let us find the mean of these three numbers:

$(9 + 12 + 15)/3$

$36/3$

$12$

**Our final answer is B,** 12.

You can see we got the same answer no matter our technique, so if algebra bothers you for any reason, you can still find your solution so long as you remember how to find your mean.

As for **questions on medians**, the SAT will often try to present you with a set of very large numbers or a set of numbers with some amount that are missing.

This question is asking about the median, so let us first start by arranging our numbers in ascending order.

3, 4, 10, 15, 18, 21

We are told that $x$ is our median, so it must lie in the middle of our set. This puts it between 10 and 15.

3, 4, 10, x, 15, 18, 21

This means that the only answer choice $x$ could possibly be is D, 14. All the other answer choices are too large or too small.

**Our final answer is D,** 14.

And lastly, **mode questions are actually quite rare on the SAT**. You should know what a “mode” means on the off chance that you will see a mode question on the test, but chances are you will only be asked about means and/or medians.

*Though the SAT may try to vary their questions, the principles behind them remain the same.*

## How to Solve Mean, Median, and Mode Questions

Because these questions often seem straightforward, it can be easy to find yourself rushing through them. But as you go through your test, remember to keep these SAT math tips in mind:

**1) Always (always!) make sure you are answering the right question**

Because the SAT will ask you to find means more than medians or modes, it is incredibly common for students who are rushing through the SAT to read “mean” when the question is actually asking for a “median.” If you’re trying to rush, it can become second nature to glance at an m-word and start in immediately on solving the problem.

Unfortunately, the test makers know that people will make errors like this and they will provide bait answers for anyone who makes this kind of mistake. As always when taking the SAT math sections, double-check that you are answering precisely the right question before you start in on solving the question (or at least before bubbling in your answer!).

**2) Write it out**

Make sure you take the time to rearrange your set of numbers in order when dealing with medians and modes, and make sure you write out your equations when dealing with means. It can be tempting to solve problems like these in your head, but a single misplaced digit is the difference between a correct answer and a wrong answer penalty.

In order to avoid making careless mistakes, always take a moment to write out your problem. It will not take as long as you think to reorganize your values and it will almost always highlight the path towards perfection.

**3) Use PIA/PIN when necessary**

If you find yourself stuck on a problem and have some extra time to spare, don’t hesitate to use the strategies of plugging in answers or plugging in numbers where applicable. Always keep in mind that it will often take you a little longer to solve a problem using these techniques, but doing so will almost always lead you to the right answer.

For example:

Let us say this was a problem you found yourself stuck on, but you luckily had some spare time before the section was over. Well, now is the time to use PIA!

We will be replacing our answer options with the value of $x$ and solving the problem as normal. As usual with PIA, let us start with the middle answer choice and go up or down from there.

This means that we will plug in answer choice C, 31, in place of our $x$ value.

We are trying to find the mean of 3 numbers, 6, 19, and 31, and seeing if we can get 19. So let us find the mean:

$(6 + 19 + 31)/3$

$56/3$

$18.67$

Our mean is just slightly too small. This means that we can eliminate answer choice C as well as answer choices A and B (as they are even smaller and will thus produce an even smaller mean.)

Our answer is likely going to be D, as E is much larger and will give us a far greater mean than 19. So let us test answer choice D, 32, to be sure.

$(6 + 19 + 32$/3$

$57/3$

$19$

Success! Answer choice D is correct.

**Our final answer is D,** 32.

*There are a variety of escape hatches when solving mean/median/mode problems, so never fear!*

## Test Your Knowledge

Now to test your statistics know-how on real SAT math problems!

**1)**

**2)**

**3)**

**4)**

**Answers:** E, C, 109, E

**Answer Explanations:**

**1)** This problem requires that we visualize a list of seven numbers in our set. So let’s break it into pieces. For now, we’ll give our set of seven numbers each a variable name:

$a, b, c, d, e, f, g$

We are told that the least and greatest numbers are 2 and 20, respectively. So now our list looks like this:

$2, b, c, d, e, f, 20$

We also know that 6 is our median, meaning it is exactly in the middle.

$2, b, c, 6, e, f, 20$

And lastly, we are told that the number 3 repeats the most often. Well, according to our list, we know that 3 can only show up twice maximum, since our median is 6. So let us fit in our 3’s.

$2, 3, 3, 6, e, f, 20$

Now, we can find our potential averages. Let us set up our problem with option I, mean of 7.

$(2 + 3 + 3 + 6 + e + f + 20)/7 = 7$

$(36 + e + f)/7 = 7$

$34 + e + f = 49$

$e + f = 15$

Now we know that $e$ and $f$ must be two different numbers (since only 3 repeats) and they must be real numbers between 6 and 20 that add up to 15.

Well $7 + 8 = 15$, so these could be our values for $e$ and $f$. This means that option I is possible and so we can eliminate answer choice D.

Now let us test option II using the same methods.

$(2 + 3 + 3 + 6 + e + f + 20)/7 = 8.5$

$(36 + e + f)/7 = 8.5$

$34 + e + f = 59.5$

$e + f = 25.5$

Again, we know that $e$ and $f$ must be two different, real numbers between 6 and 20 that add up to 25.5. Well, $12 + 13.5 = 25.5$, so $e$ and $f$ could easily be those numbers and fit our criteria.

This means option II also works and so we can eliminate answer choices A and C.

Finally, let us test option III.

$(2 + 3 + 3 + 6 + e + f + 20)/7 = 10$

$(36 + e + f)/7 = 10$

$34 + e + f = 70$

$e + f = 36$

$e$ and $f$ must be two different real numbers between 6 and 20 that add up to 36 and we can get this result with $17 + 19 = 36$. This means that option III also works, which means that all of three of our options work.

Our final answer is E, I, II, and III.

**2)** Because we are working with variables, we have two ways we can solve this problem--with algebra or using plugging in numbers. Let us go through both methods.

**Method 1--algebra**

We are told that the mean of $t$ and $t + 2$ equals $x$, so let us find the value of $x$ by manipulating the mean of $t$ and $t + 2$.

${t + (t + 2)}/2 = x$

${2t + 2}/2 = x$

We can reduce the value of the fraction here, since $2t + 2$ can be divided by 2. Which means we are left with:

$t + 1 = x$

Now let us use the same process for $y$, which we are told is the mean of $t$ and $t - 2$

${t + (t - 2)}/2 = y$

${2t - 2}/2 = y$

Again, we can reduce our fraction, since $2t - 2$ can be divided by 2. This gives us:

$t - 1 = y$

Now, we have values for both $x$ and $y$, so let us put them together to find their mean.

${(t + 1) + (t - 1)}/2$

${2t}/2$

$t$

The mean of $x$ and $y$ is $t$.

**Our final answer is C,** $t$.

**Method 2--plugging in numbers**

Alternatively, we can skip having to use algebra and plug in our own number for $t$ in order to solve the problem. Let us say that $t = 10$ and solve the problem from there. (Why 10? Why not!)

This means that the mean of $t$ and $t + 2$ still equals $x$. So:

${t + (t + 2)}/2 = x$

${10 + (10 + 2)}/2 = x$

$22/2 = x$

$11 = x$

And we will use the same value for $t$ and the same process to find $y$:

${t + (t - 2)}/2 = y$

${10 + (10 - 2)}/2 = y$

$18/2 = y$

$9 = y$

Now, we can find the mean of $x$ and $y$.

$(11 + 9)/2$

$20/2$

$10$

And, since we said that $t = 10$, the mean of $x$ and $y$ also equals $t$.

**Our final answer is, again, C**, $t$.

**3)** As you’ve seen with smaller sets of numbers, the median of an odd list of numbers is exactly in the middle, which means there is no need to average out two middle values. For example:

In the set of three numbers, $x, y, z$, the number $y$ is our median.

Whereas, in the set of four numbers, $w, x, y, z$, we will have to take the average of $x$ and $y$ to find our median.

For instance, let's imagine that this problem had a much smaller scale. If they had said that the median of *3* consecutive integers was 60, we would have:

$x, 60, z$

There is one number on either side of 60 and we know they are all consecutive. That means we can find $z$ (the largest number) by removing one number (60) from the number of integers in the set (3), and then dividing that result in half.

$3 - 1 = 2$

$2/2 = 1$

Now, we add this number to 60 in order to find the highest integer in our set.

$60 + 1 = 61$

That may have seemed like a waste of time, so let's look at it on a slightly larger scale. Let us say we have a set of 7 numbers, with 60 as our median.

$l, m, n, 60, p, q, r$

Now, we can follow the same steps to once again find $r$ (the largest integer value).

First, remove 1 integer from the set (60).

$7 - 1 = 6$

Now, divide that result in half.

$6/2 = 3$

And finally, add that number back to 60.

$60 + 3 = 63$

And we can once again do the same for our actual problem. Here we have 99 numbers, which means that our median is the whole number 60, directly in the middle. Just like with our examples sets of 3 and 5, we do not need to find an average.

We are also told that the set of numbers are consecutive integers, which means our job is simple in terms of finding our values to the right and left of 60.

All we have to do is simply subtract 1 integer (60) from 99 and divide the remainder in half. Then, we will once again use this number as the amount of our set on either side of our median. Let us put this into practice.

$99 - 1 = 98$

$98/2 = 49$

Now, 60 will be the number directly in the middle, and we will have 49 numbers before 60 and 49 numbers after. You can see why this makes sense, because:

$49 + 1 + 49 = 99$

(The 1 in the middle represents our median.)

Now, we must find the largest number in our set, so let us add 49 to 60, since our numbers are in consecutive order.

$60 + 49 = 109$

**So our final answer is 109. **

**4)** In order to find the new mean of our set of numbers, we must first find the sum of the original (erroneous) number of stamps. So let us set up the equation using the false set of stamps:

$x/10 = 88$

$x = 880$

The original sum of our stamps is 880. Now let us find out how many stamps we actually have to work with.

If 55 was entered in as 75, that means that there are actually 20 fewer stamps in the collection. And if 78 was entered as 88, there are actually 10 fewer stamps in the collection. This means that we actually have:

$20 + 10 = 30$ fewer stamps in the collection than we had previously thought.

So instead of a sum of 880, our sum is actually:

$880 - 30 = 850$

Now, we still have the same number of stamp collections (10), so let us find the new mean of the collection.

$850/10$

$85$

The actual mean of the collection is 85.

**Our final answer is E,** 85.

*You finished your statistics problems, whoo! The puppy is happy. *

## The Take Aways

Though you will see mean, median, and mode questions about 2 to 3 times per test, the questions themselves are often quite simple once you know your way around the techniques needed to solve them.

Never hesitate to use PIA or PIN if you have the time to spare and if you feel uncomfortable with algebra alone. Otherwise, make absolutely sure you are answering the proper question and don’t take for granted that these questions are simple (a careless error will still lose you precious points!). Just remember to keep your mind sharp and your work organized, and you’ll be able to tackle SAT statistics problems in no time.

## What’s Next?

Now that you've done your paces on SAT statistics questions, it's time to make sure you've got the rest of the SAT math topics well in hand. Are you all caught up with your probabilities? Your formulas? Your lines and angles? We've got guides that cover all the SAT math you'll need come test day.

**Don't know what score to aim for?** Check out how to gauge your current score level and how that stacks up for your schools.

**Looking to get a 600 on SAT math?** Look to our guide on how to improve your SAT math score.

**Trying for a perfect score?** If you're already at a 600 or above, check out our guide to getting a perfect 800 on the SAT math, written by a perfect-scorer.

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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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