Section 5.1: Measures of Central Tendency
SECTION 5.1 – MEASURES OF CENTRAL TENDENCY
The “Center” is an important aspect of a distribution. When we refer to the “Center” we are trying to get an idea of the location of the middle of the data set. There are three common measures of central tendency. They are the mean, median, and mode.
Let’s begin by trying to find the most “typical” value of a data set.
Note that we just used the word “typical” although in many cases you might think of using the word “average.” We need to be careful with the word “average” as it means different things to different people in different contexts. One of the most common uses of the word “average” is what mathematicians and statisticians call the arithmetic mean, or just plain old mean for short. “Arithmetic mean” sounds rather fancy, but you have likely calculated a mean many times without realizing it; the mean is what most people think of when they use the word “average”.
Mean
The mean of a set of data is the sum of the data values divided by the number of values.
Example 1
Marci’s exam scores for her last math class were: 79, 86, 82, 94. Find the mean.
Solution: The mean of these values would be: [latex]\frac{79~+~86~+~82~+~94}{4}[/latex] = 85.25
Typically we round means to one more decimal place than the original data had. In this case, we would round 85.25 to 85.3.
Example 2
The one hundred families in a particular neighborhood are asked their annual household income, to the nearest $5 thousand dollars. The results are summarized in a frequency table below. Find the mean.
| Income ($1,000’s) | Frequency |
| 15 | 6 |
| 20 | 8 |
| 25 | 11 |
| 30 | 17 |
| 35 | 19 |
| 40 | 20 |
| 45 | 12 |
| 50 | 7 |
Solution: Calculating the mean by hand could get tricky if we try to type in all 100 values:
[latex]\frac{15~+~\cdot\cdot\cdot~+~15~+~20~+~\cdot\cdot\cdot~+~20~+~25~+~\cdot\cdot\cdot~+~25~+~\cdot\cdot\cdot}{100}[/latex]
We could calculate this more easily by noticing that adding 15 to itself six times is the same as 15 × 6 = 90. Using this simplification, we get:
[latex]\frac{15~\cdot~6~+~20~\cdot~8~+~25~\cdot~11~+~30~\cdot~17~+~35~\cdot~19~+~40~\cdot~20~+~45~\cdot~12~+~50~\cdot~7}{100}[/latex] = [latex]\frac{3390}{100}[/latex] = 33.9
The mean household income of our sample is 33.9 thousand dollars ($33,900).
Example 3
Extending off the last example, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand). Adding this to our sample, find the mean.
Solution: Adding $5 million to our sample, our mean is now:
[latex]\frac{15~\cdot~6~+~20~\cdot~8~+~25~\cdot~11~+~30~\cdot~17~+~35~\cdot~19~+~40~\cdot~20~+~45~\cdot~12~+~50~\cdot~7~+~5000~\cdot~1}{101}[/latex] = [latex]\frac{8390}{101}[/latex] = 83.069
While 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a “typical” value.
Imagine the data values on a see-saw or balance scale. The mean is the value that keeps the data in balance, like in the picture below.
If we graph our household data, the $5 million data value is so far out to the right that the mean has to adjust up to keep things in balance.
For this reason, when working with data that have outliers – as described in the previous section – it is common to use a different measure of center, the median.
Median
The median of a set of data is the value in the middle when the data is in order To find the median, begin by listing the data in order from smallest to largest.
If the number of data values, n, is odd, then the median is the middle data value. The position in the ordered list of this value can be found using the formula: [latex]\frac{n~+~1}{2}[/latex]
Note: this formula gives you the POSITION of the median, not the value of the median itself.
If the number of data values is even, there is no one middle value, so we find the mean of the two middle values. The position of the media is still [latex]\frac{n~+~1}{2}[/latex], but this will not be a whole number.
The median will be found by taking the mean of the numbers in the positions immediately below and immediately above the position found using the formula.
Example 4
Returning to the football touchdown data, we would start by listing the data in order from smallest to largest.
6 9 12 12 14 14 14 15 16 18 18 18 18 19 19 20
20 21 21 21 22 22 22 23 28 28 29 32 33 33 37
Solution: Since there are 31 data values, an odd number, the median will be the middle number, in the [latex]\frac{(31~+~1)}{2}[/latex] = [latex]\frac{32}{2}[/latex] = 16th position (leaving 15 values below and 15 above). The 16th data value is 20, so the median number of touchdown passes in the 2000 season was 20 passes. Notice that for this data, the median is fairly close to the mean we calculated earlier, 20.5.
Example 5
Find the median of these quiz scores: 15 20 18 16 14 18 12 15 17 17
Solution: We start by listing the data in order: 12 14 15 15 16 17 17 18 18 20
Since there are 10 data values, an even number, there is no one middle number. To find the median, use the formula to find the position of the median with n = 10, [latex]\frac{(10~+~1}{2}[/latex] = [latex]\frac{11}{2}[/latex] = 5.5, so we need the mean of the 5th and 6th numbers in the ordered list. The 5th number in the ordered list is 16, and the 6th number in the ordered list is 17. So we find the mean of the two middle numbers, 16 and 17, and get (16+17)/2 = 16.5.
The median quiz score was 16.5.
Example 6
Let us return now to our original household income data. Find the median.
| Income ($1000’s) | Frequency |
| 15 | 6 |
| 20 | 8 |
| 25 | 11 |
| 30 | 17 |
| 35 | 19 |
| 40 | 20 |
| 45 | 12 |
| 50 | 7 |
There are 6 data values of $15, so → Values 1 to 6 are $15 thousand
The next 8 data values are $20, so → Values 7 to (6+8) = 14 are $20 thousand
The next 11 data values are $25, so → Values 15 to (14+11) = 25 are $25 thousand
The next 17 data values are $30, so → Values 26 to (25+17) = 42 are $30 thousand
The next 19 data values are $35, so → Values 43 to (42+19) = 61 are $35 thousand
Solution: Here we have 100 data values. If we didn’t already know that, we could find it by adding the frequencies. Since 100 is an even number, using the position formula we get [latex]\frac{100~+~1}{2}[/latex] = [latex]\frac{101}{2}[/latex] = 50.5. We need to find the mean of the middle two data values – the 50th and 51st data values. To find these, we start counting up from the bottom (see above). From this we can tell that values 50 and 51 will be $35 thousand, and the mean of these two values is $35 thousand.
The median income in this neighborhood is $35 thousand.
In addition to the mean and the median, there is one other common measurement of the “typical” value of a data set: the mode.
Mode
The mode is the element of the data set that occurs most frequently.
The mode is fairly useless with data like weights or heights where there are a large number of possible values. The mode is most commonly used for categorical data, for which median and mean cannot be computed.
Example 7
In our vehicle color survey, we collected the data below. Find the mode.
| Color | Blue | Green | Red | White | Black | Grey |
| Frequency | 25 | 52 | 41 | 36 | 39 | 23 |
Solution: For this data, Green is the mode, since it is the data value that occurred the most.
It is possible for a data set to have more than one mode if several categories have the same frequency, or no modes if each every category occurs only once.