Measures of Central Tendency and Dispersion
Standard
S.ID.2 Use statistics appropriate to the shape of the data distribution to compare center and spread of two or more different data sets.
Goals for this section:
Students should be able to find mean, median, mode, and the range of a set of data.
Outliers
Outliers are values that are much greater than or less than the other values in a set. For this class we will use common sense to identify outliers but generally in statistics there are other formulas.
- Example: 112, 113, 118, 23, 112, 114, 115, 112, 567, 114
- 23 and 567 are the outliers. They're really "out" there!
Measures of Central Tendency
We use measures of central tendency to describe the value that is in the center.
Mean
Sometimes called "average". Take the sum of all values and divide by total number of data values.
When do we use mean? When we have no outliers!
Median
Think "middle". The median is your middle value. If there's two values in the middle, add them and divide by 2.
When do we use median? When there is an outlier!
Mode
Think "more of" for mode. This is the data value that appears the most. It is possible to have one mode, no mode, or several modes.
When do we use mode? When the data is nonnumeric or when choosing the popular item!
Example: Bowling Scores
Jason has been keeping track of his final bowling scores:
170, 171, 175, 178, 177, 171, 171, 142, 173, 189, 177, 170
Find the mean, median, and mode.
Good practices: Identify how many values there are, which we will call n. Order the set from least to greatest.
There are 12 bowling scores, so n = 12.
142, 170, 170, 171, 171, 171, 173, 174, 177, 177, 178, 189
Mean
- Add all the values.
- 142 + 170 + 170 + 171 + 171 + 171 + 173 + 174 + 177 + 177 + 178 +189 = 2063
- Divide by n.
- 2063 / n = 2063 / 12 = 171.917
Median
142, 170, 170, 171, 171, 171, 173, 174, 177, 177, 178, 189
There's two values in the middle!
- Add the two values.
- 171 + 173 = 344
- Divide by 2.
- 172
Mode
There are three 171 data values. 171 is the mode.
Working backwards
Suppose your test scores in a class are 78, 78, and 76. Your average right now is 77.3. You've got another test coming up and you really want to raise your test score average to a B- (80). What score would you need to get in order to raise your average?
Step 1
Identify what the question is asking for.
We need to find the score (which we will call x since it is unknown) that will raise your test average to a B-.
Step 2
Create your equation.
This is asking for an average so we're using mean. So we have to add all the values up and then divide by n (the total number of values).
- x is the score we're looking for which is added to the current scores you have.
- With the addition of x, we now have a total of 4 scores (n=4).
- Our average must equal 80.
Our equation:
( 78 + 78 + 76 + x ) / 4 = 80
Step 3
Solve the equation.
( 78 + 78 + 76 + x ) / 4 = 80
- Multiply 4 on both sides.
- 78 + 78 + 76 + x = 320
- Add 78 + 78 + 76 to simplify the left side of the equation.
- 232 + x = 320
- Subtract 232 on both sides.
- x = 88
Step 4
Interpret result.
You need to score an 88 on the next test in order to get a B- average!
Measures of Dispersion
We use measures of dispersion to describe how spread out the values in a data set are. In this lesson we will only focus on range.
Range
The difference between the greatest and smallest data values.
Example
Given: 4, 5, 3, 7, 9, 12, 15, 2, 6
What is the range?
- Which value is the greatest? 15
- Which value is the smallest? 2
- Now subtract. Your range is 13.