Mean, Median, Mode, and Range

Learning goals

  • Compute the mean, median, mode, and range of a list, sorting first for the median
  • Explain why the mean is the balance point where deviations cancel
  • Choose median over mean when an outlier drags the data
  • Read the range as spread, not as a center
  • Name the mode as the summary that works for data with no numbers or order

The mean: a fair share

The mean, also called the arithmetic average, answers a simple question. If you pooled all the values together and split the total evenly, how much would each one get? You find it by adding every value and dividing by how many values there are.

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

Say five friends are carrying 44, 88, 66, 22, and 55 dollars. Together they hold 4+8+6+2+5=254 + 8 + 6 + 2 + 5 = 25 dollars. If they pour it all into one pile and redistribute it equally, each friend walks away with

255=5 dollars.\frac{25}{5} = 5 \text{ dollars}.

So the mean is 55 dollars. Notice what “fair share” really means here: replacing every value by the mean leaves the total untouched, because five fives still add to 2525. That is the defining property of the mean, and it is just the formula read backwards. That reading works because the sum equals the number of values times the mean.

Why the mean sits at the balance point

Calling the mean a “center” deserves a reason, not just a name. Look at the mean of 2,3,4,72, 3, 4, 7, which is 44. How far is each value from that mean? The value 22 sits 22 below it, 33 sits 11 below it, 44 sits right on it, and 77 sits 33 above it. Call this signed distance the deviation: negative when a value is below the mean, zero when a value equals the mean, and positive when it is above. Add the four deviations together:

(−2)+(−1)+0+(+3)=0.(-2) + (-1) + 0 + (+3) = 0.

The two below-mean deviations add to −3-3, exactly canceling the one above-mean deviation of +3+3. That is not a coincidence of this particular list. For any list of numbers, the deviations from the mean always add up to zero.

The mean as the balance point of 2, 3, 4, 7A ruler from 0 to 8 with weights at 2, 3, 4, and 7 and a fulcrum at the mean 4; the deviations plus 3 and minus 3 cancel.-2-10+3234701235678mean = 4
The data 2, 3, 4, 7 balanced on a ruler. The fulcrum sits at the mean, 4. The value at 7 is 3 units above the mean, which exactly balances the values at 2 and 3, together 3 units below it, so the deviations cancel.

Picture the values as weights sitting on a ruler, as in the figure. The mean is the point where you would put your finger to hold the ruler level: every value below the mean tips it one way, every value above tips it the other way, and because the deviations sum to zero, the two sides exactly balance. That is what makes the mean a true center of the data, not just its formula.

Check your understanding

A data set has mean 2020. Three of its four deviations from the mean are +5+5, +2+2, and −3-3. What must the fourth deviation be?

Answer choices

Worked example 1 Find the mean of 66, 99, 99, 44, and 77

Add the five values:

6+9+9+4+7=35.6 + 9 + 9 + 4 + 7 = 35.

There are 55 values, so divide the sum by 55:

mean=355=7.\text{mean} = \frac{35}{5} = 7.

The mean is 77. As a check, the deviations are (+2)+(+2)+(−1)+(−3)+0=0(+2) + (+2) + (-1) + (-3) + 0 = 0, exactly as the balance argument promised.

Check your understanding

Over four days a shop sold 1212, 1515, 99, and 88 sandwiches. What was the mean number sold per day?

Answer choices

The median: the middle value

The median is the value that lands in the middle once the data is sorted from least to greatest. It splits the ordered list into two equal halves. At least half the values sit at or below the median, and at least half sit at or above it. To find it, always sort first, then locate the middle.

When there is an odd number of values, one value sits in the dead center, and that value is the median. Sort 3,8,5,9,43, 8, 5, 9, 4 into 3,4,5,8,93, 4, 5, 8, 9; with five values the third one is the middle, so the median is 55. Two values fall below it and two fall above.

When there is an even number of values, no single value is in the center, and two values share the middle. The median is then the mean of those two middle values. Sort 7,2,10,47, 2, 10, 4 into 2,4,7,102, 4, 7, 10; the two middle values are 44 and 77, so the median is

4+72=5.5.\frac{4 + 7}{2} = 5.5.

Sorting is not optional. “Middle” means middle in order, not the middle of the list as it happened to be handed to you.

Finding the median for an odd and an even countTop row of five sorted values with the single middle value highlighted; bottom row of four sorted values with the two middle values highlighted and averaged.Odd count: one middle value34589median = 5Even count: average the middle two24710median = 5.5
The median splits the sorted data in half. With five values (top) the single middle value, 5, is the median. With four values (bottom) the two middle values 4 and 7 are averaged, giving 5.5.

Worked example 2 Find the median of 1414, 33, 99, 1414, and 66

Sort the five values from least to greatest:

3,  6,  9,  14,  14.3, \; 6, \; 9, \; 14, \; 14.

There are 55 values, an odd count, so the median is the single middle value. Counting in, five values means the third value is the middle:

median=9.\text{median} = 9.

Two values (33 and 66) lie below it and two (1414 and 1414) lie above, so 99 splits the data evenly. Notice that the repeated 1414s caused no trouble: as long as a value stays above the median, its own size does not change what the median is. The two 1414s could have been 1414 and 140140 and the median would still be 99.

Check your understanding

Find the median of 88, 22, 1111, and 55.

Answer choices

The mode: the most common value

The mode is the value that appears most often. Where the mean and median do arithmetic to find a center, the mode just counts. Tally how many times each value occurs, and the value with the highest tally is the mode.

In the list 4,7,4,2,9,44, 7, 4, 2, 9, 4 the value 44 appears three times while every other value appears once, so the mode is 44. A data set can have more than one mode: this happens when two or more values are tied for the highest tally, while at least one other value in the list has a lower tally. The list 1,1,5,8,81, 1, 5, 8, 8 has two modes, 11 and 88, each appearing twice, more than the once-appearing 55. But when every value in the list has the same tally, no value stands out as more frequent than the rest, so there is no mode at all. That covers both the case where nothing repeats, as in 3,6,93, 6, 9, and the case where everything repeats the same number of times, as in 1,1,2,21, 1, 2, 2.

The mode is the only one of these four summaries that works on data with no numbers and no order. The mean and range need numbers to add and subtract. The median needs an order to sort into. The mode just needs you to count, so it works even on colors or names. Suppose six classmates name their favorite color: blue, red, blue, green, red, blue. You cannot add “blue” and “red” together, and you cannot sort colors from least to greatest, but you can still count. Blue is named three times, red twice, and green once, so blue is the mode.

Check your understanding

Five classmates name their favorite snack: chips, pretzels, chips, popcorn, chips. What is the mode?

Answer choices

The range: how spread out the data is

The first three numbers all try to pin down the center of the data. The range answers a different question: how spread out is it? The range is the distance from the smallest value to the largest, found by subtracting the minimum from the maximum.

range=maximum−minimum.\text{range} = \text{maximum} - \text{minimum}.

For the test scores 72,88,95,60,8172, 88, 95, 60, 81, the largest is 9595 and the smallest is 6060, so the range is 95−60=3595 - 60 = 35. A small range means the smallest and largest values are close together; a large range means they are far apart. Range looks only at those two extreme values, so it says nothing about how the values in between are arranged.

Two classes can share the very same mean and still feel completely different. Suppose one class scores 78,79,80,81,8278, 79, 80, 81, 82: the mean is 8080, and the range is only 82−78=482 - 78 = 4, so every score sits close to 8080. A second class scores 50,75,80,95,10050, 75, 80, 95, 100: the mean is also 8080, but the range is 100−50=50100 - 50 = 50, so the scores are spread far apart even though the mean looks identical. The range is a first, rough measure of that spread.

Worked example 3 Find the mean, median, mode, and range of 5,8,3,8,65, 8, 3, 8, 6

Start with the mean. Add the five values and divide by 55:

mean=5+8+3+8+65=305=6.\text{mean} = \frac{5 + 8 + 3 + 8 + 6}{5} = \frac{30}{5} = 6.

For the median, sort the data into 3,5,6,8,83, 5, 6, 8, 8. With five values the middle one is the third, so

median=6.\text{median} = 6.

For the mode, count repeats. The value 88 appears twice and everything else once, so the mode is 88.

For the range, subtract the smallest value from the largest:

range=8−3=5.\text{range} = 8 - 3 = 5.

So this data set has mean 66, median 66, mode 88, and range 55. The mean and median happen to agree here; the next section shows a data set where they disagree instead.

Check your understanding

For the data 10,4,7,4,10,410, 4, 7, 4, 10, 4, which statement is true?

Answer choices

Mean versus median: which center to trust

When the data is lopsided, the mean and the median can tell very different stories, and knowing why lets you pick the more useful one. The mean uses the actual size of every value, so a single far-off value, an outlier, drags the mean toward that extreme. The median uses only position, so one outlier barely moves it.

Picture five households with yearly incomes, in thousands of dollars, of 3030, 3535, 4040, 4545, and 10001000. The mean income is

30+35+40+45+10005=11505=230,\frac{30 + 35 + 40 + 45 + 1000}{5} = \frac{1150}{5} = 230,

which is larger than four of the five incomes. Reporting “the average income is 230230 thousand” would mislead anyone, because nobody here lives like that except the single wealthy household. The median tells a more typical story. The data is already sorted, so the middle value is

median=40,\text{median} = 40,

which sits right among the typical households. Replace the outlier with a typical income, say 5050 thousand, and the mean becomes 30+35+40+45+505=40\frac{30 + 35 + 40 + 45 + 50}{5} = 40, matching the median exactly. It is the single value of 10001000 that drags the mean all the way up to 230230, while the median barely moves.

This is the practical rule of thumb. When the data is roughly even, the mean and median land close together and either one describes the center well. When the data is skewed by a few extreme values, the median usually describes a typical value better. The mean is still the right tool when you genuinely care about the total being shared out. Splitting a bill or finding a true per-person amount is one of those cases.

Check your understanding

A small company has salaries, in thousands of dollars, of 4040, 4242, 4545, 4848, and 200200. Which measure best describes a typical salary here, and why?

Answer choices

Common mistakes

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Why the deviations from the mean always add to zero

The deviations from the mean always add to zero#

Call the values x1,x2,…,xnx_1, x_2, \ldots, x_n, so there are nn of them, and let mm stand for their mean. By the definition of the mean, the sum of the values divided by nn equals mm. Multiplying both sides by nn gives a fact we will use in a moment: the values add up to n×mn \times m.

Now form the deviation of each value, xi−mx_i - m, and add all the deviations together:

(x1−m)+(x2−m)+⋯+(xn−m).(x_1 - m) + (x_2 - m) + \cdots + (x_n - m).

Separate the two kinds of term. All the values collect into their sum x1+x2+⋯+xnx_1 + x_2 + \cdots + x_n. The mean mm is then subtracted once for each of the nn values, so those subtractions remove n×mn \times m in total:

(x1+x2+⋯+xn)−n×m.(x_1 + x_2 + \cdots + x_n) - n \times m.

But the values add up to exactly n×mn \times m, so this becomes n×m−n×m=0n \times m - n \times m = 0. The deviations cancel completely, for any data set at all.

A bit of history (optional)

A cargo ship is caught in a storm. It is riding far too low. To save it, the crew heaves part of the cargo over the side. The ship limps into port. Whose loss is the cargo now at the bottom of the sea?

The old answer is that everyone pays, but not equally. Traders around the Mediterranean, the sea south of Europe, refused to leave one unlucky owner carrying the whole loss. Instead, the cost was totaled and then divided among every merchant whose goods came home safe, each paying a share in proportion to what they had at risk. Roman law had that rule fifteen hundred years ago, and shipping still uses it under the name general average.

Our word likely grew out of that rule. Mediterranean sailors used avaria for a loss or a charge tied to a shared sea voyage, and average is what English made of it over time. So the word did not begin as a summary of a list. It began as a shared loss, divided among the merchants who benefited.

That shared-loss idea is a cousin of the mean, not the mean itself, since a real general average payment depends on how much each merchant had at stake, not an equal split. But the core idea, pooling a cost rather than leaving one person to bear it all, is the same spirit that later gave average its modern meaning: add the values, split the sum evenly, and hand every value the same share. The total does not budge, just as five fives still add to twenty-five.