# Statistics/Summary/Averages/mean

## Contents

### Mean, Median and Mode[edit]

#### Mean[edit]

The mean, or more precisely the arithmetic mean, is simply the arithmetic average of a group of numbers (or **data set**) and is shown using -bar symbol . So the mean of the variable * is , pronounced "**x*-bar". It is calculated by adding up all of the values in a data set and dividing by the number of values in that data set
**:.**For example, take the following set of data: {1,2,3,4,5}. The mean of this data would be:

Here is a more complicated data set: {10,14,86,2,68,99,1}. The mean would be calculated like this:

#### Median[edit]

The median is the "middle value" in a set. That is, the median is the number in the center of a data set that has been ordered sequentially.

For example, let's look at the data in our second data set from above: {10,14,86,2,68,99,1}. What is its median?

- First, we sort our data set sequentially: {1,2,10,14,68,85,99}
- Next, we determine the total number of points in our data set (in this case, 7.)
- Finally, we determine the central position of or data set (in this case, the 4th position), and the number in the central position is our median - {1,2,10,
**14**,68,85,99}, making 14 our median.

Helpful Hint! | |
---|---|

An easy way to determine the central position or positions for any ordered set is to take the total number of points, add 1, and then divide by 2. If the number you get is a whole number, then that is the central position. If the number you get is a fraction, take the two whole numbers on either side. |

Because our data set had an odd number of points, determining the central position was easy - it will have the same number of points before it as after it. But what if our data set has an even number of points?

Let's take the same data set, but add a new number to it: {1,2,10,14,68,85,99,*100*} What is the median of this set?

When you have an even number of points, you must determine the *two* central positions of the data set. (See side box for instructions.) So for a set of 8 numbers, we get (8 + 1) / 2 = 9 / 2 = 4 1/2, which has 4 and 5 on either side.

Looking at our dataset, we see that the 4th and 5th numbers are 14 and 68. From there, we return to our trusty friend the mean to determine the median. (14 + 68) / 2 = 82 / 2 = **41**.
find the median of 2, 4, 6, 8 => firstly we must count the numbers to determine its odd or even
as we see it is even so we can write: M=(4+6)/2=10/2=5
5 is the median of above sequential numbers.

#### Mode[edit]

The mode is the most common or "most frequent" value in a data set. Example: the mode of the following data set (1, 2, **5**, **5**, 6, 3) is 5 since it **appears** **twice**. This is the most common value of the data set.
Data sets having one mode are said to be **unimodal**, with two are said to be **bimodal** and with more than two are said to be **multimodal** . An example of a unimodal dataset is {1, 2, 3, **4**, **4**, **4**, 5, 6, 7, 8, 8, 9}. The mode for this data set is 4. An example of a bimodal data set is {1, **2**, **2**, **3**, **3**}. This is because both 2 and 3 are modes.
**Please** **note**: If all points in a data set occur with equal frequency, it is equally accurate to describe the data set as having many modes or no mode.

#### Midrange[edit]

The midrange is the arithmetic mean strictly between the minimum and the maximum value in a data set.

#### Relationship of the Mean, Median, and Mode[edit]

The relationship of the mean, median, and mode to each other can provide some information about the relative shape of the data distribution. If the mean, median, and mode are approximately equal to each other, the distribution can be assumed to be approximately symmetrical. If the mean > median > mode, the distribution will be skewed to the right. If the mean < median < mode, the distribution will be skewed to the left.

### Questions[edit]

1. There is an old joke that states: "Using median size as a reference it's perfectly possible to fit four ping-pong balls and two blue whales in a rowboat." Explain why this statement is true.