Find and interpret center

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Difficulty: Beginner
Estimated time: 26 minutes
Domains: Problem Solving and Data Analysis
Techniques: Measures-of-centerMeanMedianCenter-comparisons

What you’ll learn

Find the mean of a list and explain it as a fair share.
Put data in order and find the median, whether the count is odd or even.
Read a median from a frequency table or a box plot.
Compare what the mean and the median tell you about the same data.
Pick the measure that fits what a question asks for.

Prerequisites

Read distributions and data displays0%
Read exact values, frequencies, and marked positions from one-variable displays.

Why this matters on the SAT

Know which center the question wants

An SAT center question might ask for the mean, the median, a comparison of the two, or the one that best describes the data. So your first job is to see which one it asks for. The mean uses every value. The median uses the middle position once the values are in order.

SAT example

The numbers of minutes a student practiced piano on five days were

12,\ 18,\ 15,\ 21,\ 14.

What was the mean number of minutes the student practiced per day?

A
$14$
B
$15$
C
$16$
D
$18$

Solution to the example

The values are out of order, so let Desmos do the work. Type the data in once, then ask for the mean:

Enter L=[12,18,15,21,14].
Enter mean(L).
Check that the list has all five values, each one exactly once.

Desmos returns $16$ , so the answer is C. The order of the values doesn’t change the mean.

Calculator loads as you approach

Type the data in once as a list, then use mean(L).

Understand the two centers

The mean is the sum of the values divided by how many values there are:

\text{mean}=\frac{\text{sum of all values}}{\text{number of values}}.

For a messy list like the one above, Desmos is the safe choice. But when a list is short, already in order, and easy to add, doing it by hand is often faster. Take

6,\ 9,\ 12,\ 17,\ 21.

The sum is $65$ , so the mean is

\frac{65}{5}=13.

What does $13$ actually mean here? Think of it as a fair share. If you pooled the total of $65$ and split it evenly among the five values, each would get $13$ . That doesn’t mean any value equals $13$ . In this list, none of them does.

The median is the middle of the data once it’s in order. At least half the values are at or below it, and at least half are at or above it. So you always put the data in order first:

Write the values from least to greatest.
If there’s an odd number of values, take the single middle value.
If there’s an even number, average the two middle values.

With an odd count of $n$ values, the middle is position

\frac{n+1}{2}.

With an even count, the two middle positions are $\frac n2$ and $\frac n2+1$ .

Check your understanding:

The mean of five repair times is $13$ minutes. What does that tell you for sure, and what doesn’t it tell you?

Common mistake:

The middle entry as the list is written may not be the median. In $13,7,11,9,9,15,20$ , the middle entry is $9$ , but the median turns out to be $11$ . Rewrite the values from least to greatest, count the positions, and then find the middle one or two. Before you go on, check that your ordered list has as many values as the question gave you.

Find medians with odd and even counts

Here’s that unordered list again:

13,\ 7,\ 11,\ 9,\ 9,\ 15,\ 20.

Enter L=[13,7,11,9,9,15,20], then median(L). Desmos returns $11$ .

To see where $11$ comes from, put the data in order:

7,\ 9,\ 9,\ \boxed{11},\ 13,\ 15,\ 20.

There are $7$ values, so the median is in position $\frac{7+1}{2}=4$ . The fourth value is $11$ , which matches Desmos.

Now try an even count:

4,\ 8,\ 10,\ 15,\ 17,\ 22.

With $6$ values, there’s no single middle. Positions $3$ and $4$ share it, and they hold $10$ and $15$ . The median is halfway between them:

\frac{10+15}{2}=12.5.

Notice that $12.5$ isn’t in the data. With an even count, the median doesn’t have to be one of the values.

Check your understanding:

The ordered data are $3,5,8,8,14,19,21,30$ . What is the median?

Example: Compare mean and median

When a question compares centers, work out each measure it names, one at a time. A single value can move the mean and leave the median right where it was.

Worked example

The values in data sets P and Q are listed in order from least to greatest.

Data set P: $5,7,9,10,14$

Data set Q: $5,7,9,10,29$

Which choice correctly compares the means and medians of the two data sets?

A
The means are equal, and the median of data set P is less than the median of data set Q.
B
The mean of data set P is less than the mean of data set Q, and the median of data set P is less than the median of data set Q.
C
The mean of data set P is less than the mean of data set Q, and the medians are equal.
D
The mean of data set P is greater than the mean of data set Q, and the medians are equal.

Step 1

Find both means

Data set P adds up to $45$ , so

\operatorname{mean}(P)=\frac{45}{5}=9.

Data set Q adds up to $60$ , so

\operatorname{mean}(Q)=\frac{60}{5}=12.

So P has the smaller mean.

Step 2

Find both middles

Both lists are already in order and have $5$ values, so each median is the third value. In both lists, that’s $9$ :

\operatorname{median}(P)=\operatorname{median}(Q)=9.

The medians are equal.

Step 3

Match both comparisons

The lists differ only in their last value: $14$ in P, $29$ in Q. That bigger value raises Q’s total, and with it Q’s mean. But it doesn’t change which value sits in the middle:

\operatorname{mean}(P)<\operatorname{mean}(Q) \quad\text{and}\quad \operatorname{median}(P)=\operatorname{median}(Q).

Choice C says both.

Check your understanding:

For data set Q, what do the mean of $12$ and the median of $9$ each tell you?

Common mistake:

Choice B gets the means right but says the medians differ. Each choice makes two claims, so find both means and both medians, and pick a choice only when both of its claims match your work.

Read the median from a display

In a frequency table, each frequency tells you how many ordered positions a value fills. So don’t just pick the middle row of the table. Count positions instead.

Say a table shows $20$ values:

The ordered positions each value fills

Value	Frequency	Ordered positions
$1$	$4$	$1$ through $4$
$2$	$6$	$5$ through $10$
$3$	$7$	$11$ through $17$
$4$	$3$	$18$ through $20$

With $20$ values, the two middle positions are $10$ and $11$ . Position $10$ holds a $2$ , and position $11$ holds a $3$ . So

\text{median}=\frac{2+3}{2}=2.5.

A box plot is even more direct. The line inside the box marks the median.

The line inside the box sits right on the labeled 30-minute tick.

That line lands on the labeled $30$ -minute tick, so the median wait time is $30$ minutes. One caution: a standard box plot doesn’t show the mean, so don’t try to rebuild or estimate one from the box.

Check your understanding:

A question asks for the median wait time shown in this box plot. What calculation do you need?

Compare mean and median, then choose a method

The two centers answer different questions:

The mean spreads the total evenly across all the values, so every value affects it.
The median marks the middle position, so values far from the center may not move it.

One way to remember it: every value pulls on the mean, but the median only cares who’s in the middle. Take

10,\ 11,\ 12,\ 13,\ 34.

The mean is

\frac{80}{5}=16,

but the median is $12$ . The $34$ adds a lot to the total, so it drags the mean above four of the five values. The median stays with the middle value. So when one value sits far from the rest, the median can give a better picture of a typical value. If the question asks for the mean, though, give the mean.

Check your understanding:

For the data $10,11,12,13,34$ , a report wants one value that describes a typical middle observation. Should it use the mean of $16$ or the median of $12$ ? Why?

Questions about adding, removing, correcting, or replacing values come later, in Analyze changed data and outliers.

Pick the quickest method you can trust:

If a raw list is out of order or has repeats, like $9,4,12,4,7$ , type it once as L=[...] and use mean(L) or median(L).
If a short list is already in order, like $6,9,12,17,21$ , read the median straight from the middle position ( $12$ here), and find the mean by hand when the total comes quickly.
If the data come in a frequency table, add up the frequencies to find which value fills the middle position or positions.
If a box plot marks the median, read the line and don’t rebuild the plot.

In Desmos, it’s the same three moves every time: enter the list once, ask for mean(L) or median(L), and compare your list with the question before you trust the result.

The calculator here holds $11$ values, out of order, with a mean of $14$ and a median of $14$ . Try changing the $19$ to $40$ . Both results update from the same list: the mean jumps to about $15.9$ , but the median stays at $14$ .

Check your understanding:

Why does the Desmos list method help with an unordered or repeated list, but not with the box plot above?

Calculator loads as you approach

One long raw list, entered once, gives both the mean and the median.

Common mistake:

Mean and median aren’t two names for the same average. Before you calculate, circle the measure the question names. Then check that your work matches it: every value for a mean, or the ordered middle for a median.

Practice problems

Each problem is a step harder: a raw list, then a frequency table, then a comparison where you find the one data set whose mean equals its median.

Find both centers

Practice problem

The distances, in kilometers, of five bicycle rides were

14,\ 8,\ 18,\ 14,\ 11.

Use a Desmos list with mean(L) and median(L).

Which choice correctly gives the mean and median distance?

Calculator loads as you approach

Enter L=[14,8,18,14,11], then use mean(L) and median(L).

Find the middle positions

Practice problem

The table summarizes the numbers of volunteer hours completed by $24$ students.

Volunteer hours	Frequency
$4$	$5$
$6$	$7$
$7$	$8$
$9$	$4$

What is the median number of volunteer hours?

Your answer

Calculator loads as you approach

Use the calculator if it helps, but counting positions by hand is quick here.

Spot equal centers

Practice problem

Each choice lists a data set in order from least to greatest. For which data set is the mean equal to the median?

Calculator loads as you approach

Check the totals here if it helps, then compare each mean with the middle value.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

The mean is the total divided by the number of values. It’s the fair-share value.
The median is the middle of the ordered data, so put the values in order first.
With an odd count, take the single middle value. With an even count, average the two middle values.
To compare means, use totals and counts. To compare medians, use the ordered middle positions.
In a frequency table, count positions. In a box plot, read the median line.
Every value pulls on the mean. A value far from the rest may barely move the median, because the median depends on position.
Use mean(L) or median(L) for unordered or repeated raw lists. Work by hand when that’s truly faster: easy totals, short ordered lists, frequency tables, and marked displays.

Next lesson

Recover totals and weighted means

Rebuild totals when a mean, missing value, frequency, or subgroup size gives only part of the data.

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Practice

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297 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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