# Mean, median, and mode

Mean, median, mode. These are very important tools in the toolbox of statisticians.

These important “measures of center” are all using data points for the approximation and understanding of “average” or “middle value” of any given set of data.

Two other measures of importance are the midrange and range, which are using the least and greatest values of a data set for helping to describe the data spread.

Now, why would we need or want to find out about the middle of any data set? Also, why would we need three different measures rather than just one?

Well, let’s look a little closer at these three measures of center to understand in what way they may help you understand sets of numbers and data.

Mean, Median, Mode

So “Mean” is the mathematical term for, as you may know, “average.” We also refer to this as “arithmetic mean,” and we can find it is we add all the numbers or data values in the data set together and divide that number by the total number of items in the data set.

Often, we can find the average (“Mean”) of $$2$$ familiar numbers (like $$10$$ & $$16$$) in our head without a lot of calculation. So which number is lying halfway between $$10$$ and $$16$$? Right, $$13$$. If we want to solve this mathematically, we have to add the values $$10$$ & $$16$$ (giving us $$26$$) and divide that value by two (because we have two values of numbers in this data set). So $$26 \div 2$$ is $$13$$.

Understanding and knowing this process may be very helpful when we have to discover the mean when we have more than $$2$$ numbers.

To give you an example, when we are asked to come up with the mean of these numbers: $$2, \,3, \,5,\, 4, \,5,\,$$ and $$5$$, we first need to come up with the sum. So $$2 + 3 + 5 + 4 + 5 + 5$$ makes $$24$$. After that, we divide this sum $$(24)$$ by the total number of values in our set, being 6. So our mean of this data set is $$24$$ divided by $$6$$, which makes $$4$$.

Notice that, in this previous data set, our mean is $$4$$ and that the data set contains a value of $$4$$ as well. This something that is not always occurring. Take a look at the following example. Here, our mean, as you will see, is $$18$$, though $$18$$ doesn’t show at all in that data set.

An example:

Problem: What is the mean of this data set: $$4, \,7,\, 28,\,$$ and $$33$$?

$$4 + 7 + 28 + 33 = 72$$

We need to add all four values

$$72$$ divided by four $$(7\,\frac{2}{4})$$ is $$18$$. So divide the total of the values by the number of values.

The Answer is: $$18$$

Now, let’s take a look at the idea of “median.” Well, the median is that middle value when we order the data. In case we have $$2$$ middle values, our median is the two middle values’ average.

If we want to calculate the median, we first have to put our data into the correct numerical order, starting with the least to the greatest. After that, we need to identify the middle value or middle values.

As an example, let’s take a look at these values: $$4,\,1, \,5, \,3, \,2,\, 6,\,$$ and $$7$$. If we want to find this set’s median, we need to put the values in order, from the least to the greatest.

### $$1\, 2 \,3 \,{\color{Red} 4} \,5 \,6 \,7$$

Now we need to identify the set’s middle value. Well, we have $$3$$ values to the left of the value “four” and $$3$$ values to the right of “four”. Our middle value is the $$“4”$$, so our median is $$4$$.

If, however, our data set includes an even number of values or data items, our median is going to be the two center values’ mean.

An example:

Problem: What is the median of this set? $$2, \,3, \,5, \,4, \,5, \,5$$

First, arrange these values (least to greatest).

$$2, \,3, \,{\color{Red} 4,} \,{\color{Red} 5,} \,5, \,5$$

This data set has two middle values. Take the average (mean) of these $$2$$ values.

So, $$9$$ divided by $$2\,(\frac{9}{2})$$ which is $$4.5$$

The answer:  This data set’s median is $$4.5$$.

And now, let’s take a look at the “mode.” Well. the mode can be found if we look at that data value that is appearing most often. When you have a 2-way tie, most often that data set is bimodal. In that case, you may use both values as the set’s modes.

There are also sets that don’t have a mode. That’s the case when a data set has no value that is occurring most often. Look at our example data set, $$2,\, 3, \,4, \,5, \,5, \,5\,$$. As you see, the number five $$(“5”)$$ is appearing three times while all other values are appearing only once. So our mode is the value $$5$$.

An example

Problem: What is the mode of this data set: $$12,\, 5, \,12, \,4, \,5, \,8,\, 0, \,12,\, 1,\,$$ and $$12\,?$$ $$0, \,1,\, 4,\, 5, \,5,\, 8, \,12, \,12, \,12, \,12$$

First, arrange all these values (from least to greatest). This, however, is not a necessary step, but it sometimes is helpful to discover the mode if the values or numbers will be put in ascending order.

$$0,\, 1,\, 4, \,5, \,5, \,8, {\color{Red} \,12, \,12,\, 12,\, 12}$$

So, find the number or value that most often occurs.

The answer is: Our mode here is: $$12$$

Now, let’s take a look at some examples that include a few relevant data.

An example:

Problem: Carlos attained these scores on his math tests: $$84, \,92, \,74, \,98, \,82$$.
Now, come up with the mean, median, and mode of Carlos’ scores.

$$\frac{84+92+74+98+82}{5}$$ For finding the mean, we need to add together all the scores and divide that value by the number of the tests.

So: $$430$$ divided by $$5$$ is $$86 (\frac{430}{5} = 86)$$. The mean we needed to find is $$86$$.

$$74, \,82, \,84, \,92, \,98$$ For finding the median, first we need to put Carlos’ test scores in order (from least to greatest).

The median is $$84$$. We have five scores in total. In our ordered list, the middle score is the third. This our median.

$$74, \,82, \,84, \,92, \,98$$ Because all numbers appear exactly just one time, in the data set is no mode.

The mean in our data set is: $$86$$

The median in our data set is: $$84$$

Our data set contains no mode.

What exactly can we learn from Carlos’ test scores mean, median, and mode? Note that the values of these measures are not the same.

Both the median and the mean are giving us a picture and an idea of Carlos’ performance. When we look at these measures, we can notice that this data set’s middle is in the $$mid-80s$$ range. The set’s mean is $$86$$, while the set’s median is $$84$$.

Now, that’s all we’re actually after when we use the mean and median: to find the middle, or center, of our data set. You have noticed as well that our set contains no mode, as Carlos didn’t attain the same scores on any of the tests.

When dealing with test-taking, modes are generally meaningless unless we can see many $$0s$$ (zeros), which often means that a student didn’t do any homework at all, or that he really doesn’t have a clue of the subject!

An example:
Problem: What are the mean, median, and mode values of the following number set:
$$12, \,11,\, 13, \,11, \,12, \,10, \,10, \,11, \,13, \,14$$.

$$\frac{12+11+13+11+12+10+10+11+13+14}{10} =$$ For finding the mean, we have to add all the numbers together and divide the sum by the total number of values.

$$10, \,10, \,11, \,11,\, \textbf{11}, \,\textbf{12}, \,12, \,13,\, 13,\, 14$$ For finding the median, we first have to order the values (from least to greatest).

$$\frac{11+12}{2} = \frac{23}{2}=11.5$$ As we have ten $$(10)$$ numbers (ten is an even number), the median value is the middle two values’ mean. These are the $$5th$$ and the $$6th$$, so our median is halfway between the values $$11$$ and $$12$$.

$$10, \,10, \,\textbf{11}, \,\textbf{11},\, \textbf{11}, \,12, \,12, \,13,\, 13,\, 14$$ For finding the data set’s mode, we need to look for that number that is showing most often: 11.

The value $$11$$ is appearing most often $$(3\, times)$$.

The mean in this data set is: $$11.7$$

The median in this data set is $$11.5$$

The mode in this data set is: $$11$$

In our example, the mean, median, and mode values are very close. This indicates a certain consistency in our data set that has an average (middle) value of around $$11$$ .

If the above-listed data set would, for example, represent the ages of members on a water polo team, we would have a pretty good idea that about everyone on that team would be around $$11$$ years of age, with some younger and older members in the team.

During a $$7-day$$ period in the month of July, meteorologists recorded that over that period, the daily median high temperature was $$91^{\circ}$$.

Now, which of these statements is/are true?

1) On each of these $$7$$ days, the high temperature measured was exactly $$91^{\circ}$$.
2) The high temperature was in no case lower than $$92^{\circ}$$.
3) Half of the high temperatures in the $$7$$ days were above $$91^{\circ}$$, and half of the high temperatures were below $$91^{\circ}$$.

A) $$1$$ only
B) $$2$$ only
C) $$3$$ only
D) $$1, \,2,$$ and $$3$$

A) $$1$$ only
This answer is incorrect. Just the fact that the median high in that period was $$91^{\circ}$$ doesn’t mean that over that time, the temperature had reached $$91^{\circ}$$ on every single day. Statement $$3$$ is the only correct answer.

B) $$2$$ only
This answer is incorrect. We know that the median is $$91^{\circ}$$. Thus, $$91^{\circ}$$ is part of a data set. This means that the measured temperature must have been under $$92^{\circ}$$ at least once during that week. Statement $$3$$ is the only correct answer.

C) $$3$$ only
This the correct answer. Half of the measured temperatures were actually above $$91^{\circ}$$, and half of the measured temperatures were below $$91^{\circ}$$ because the median is always representing the value at a point that half of the values in a data set is higher and half of the data set values is lower.

D) $$1, \,2,$$ and $$3$$ This answer is incorrect. Statement number $$1$$ is incorrect as a median high value of $$91^{\circ}$$ doesn’t necessarily indicate that the measured temperature reached $$91^{\circ}$$ on all days, and statement number $$2$$ is not correct because we know that the median is $$91^{\circ}$$, meaning that $$91^{\circ}$$ is a part of a data set which means that the measured temperature must have been, at least $$1$$ time during that week, lower than $$92^{\circ}$$. Statement $$3$$ is the only correct answer.

1. Calculate the mean for the following data points:
$$2,\; 5,\; 3,\; 2$$
A.
B.
C.

Question 1 of 3

2. Calculate the mean for the following data points:
$$8,\; 6,\; 1$$
A.
B.
C.

Question 2 of 3

3. Find the median for this set of numbers.
$$12,\; 6,\; 18,\; 29\; and\; 42$$
A.
B.
C.
D.

Question 3 of 3

Next lesson: Range and Midrange

Last Updated on April 12, 2021.