Not all roles available for this page.
Sign in to view assessments and invite other educators
Sign in using your existing Kendall Hunt account. If you don’t have one, create an educator account.
Elena, Jada, and Lin enjoy playing basketball during recess. Lately, they have been practicing free throws. They record the number of baskets they make out of 10 attempts. Here are their data sets for 12 school days.
Elena
2
2
2
2
4
5
5
6
8
9
9
9
Jada
2
4
5
4
6
6
4
7
3
4
8
7
Lin
3
6
6
4
5
5
3
5
4
6
6
7
Here are the dot plots showing the number of baskets that Elena, Jada, and Lin each made over 12 school days.
On each dot plot, mark the location of the mean with a triangle. Then, contrast the dot plot distributions. Write 2–3 sentences to describe the shape and spread of each distribution.
Discuss these questions with your group. Explain your reasoning.
The tables show the number of baskets made by Jada and Lin in several games. Recall that the mean of Jada and Lin’s data is 5.
Record the distance between the number of baskets Jada made in each game and the mean.
| Jada | 2 | 4 | 5 | 4 | 6 | 6 | 4 | 7 | 3 | 4 | 8 | 7 |
| distance from 5 |
Now find the average of the distances in the table. Show your reasoning, and round your answer to the nearest tenth.
This value is the mean absolute deviation (MAD) of Jada’s data. Jada’s MAD: _________
Find the mean absolute deviation of Lin’s data. Round it to the nearest tenth.
| Lin | 3 | 6 | 6 | 4 | 5 | 5 | 3 | 5 | 4 | 6 | 6 | 7 |
| distance from 5 |
Lin’s MAD: _________
Elena’s distribution has a MAD of about 2.5. Compare the MADs and dot plots of the three students’ data. Do you see a relationship between each student’s MAD and the distribution on her dot plot? Explain your reasoning.
Your teacher will give your group a deck of cards. Shuffle the cards, and put the deck face down on the playing surface.
| player A | round 1 | round 2 | round 3 | round 4 | round 5 |
|---|---|---|---|---|---|
| sum of cards | |||||
| distance from 22 |
Average distance from 22: ____________
| player B | round 1 | round 2 | round 3 | round 4 | round 5 |
|---|---|---|---|---|---|
| sum of cards | |||||
| distance from 22 |
Average distance from 22: ____________
| player C | round 1 | round 2 | round 3 | round 4 | round 5 |
|---|---|---|---|---|---|
| sum of cards | |||||
| distance from 22 |
Average distance from 22: ____________
Whose average distance from 22 is the smallest? Who won the game?
We use the mean of a data set as a "measure of center" of its distribution, but two data sets with the same mean could have very different distributions.
This dot plot shows the number of stickers on each page of a 22-page sticker book.
The mean number of stickers is 21. All the pages have within 3 stickers of the mean, and most of them are even closer. These pages are all fairly close in the number of stickers on them.
This dot plot shows the number of stickers on each page of another sticker book that has 30 pages.
In this sticker book, the mean number of stickers on each page is also 21, but some pages have less than half that number of stickers and others have more than one-and-a-half times as many. There is a lot more variability in the number of stickers.
There is a number that we can use to describe how far away, or how spread out, data points generally are from the mean. This measure of spread is called the mean absolute deviation (MAD).
To find the MAD, we find the distance between each data value and the mean, and then calculate the mean of those distances. For instance, the point that represents 18 stickers is 3 units away from the mean of 21 stickers.
We can find the distance between each point and the mean of 21 stickers and then organize the distances into a table, as shown.
The values in the first row of the table are the number of stickers on each page in the first book. Their mean, 21, is the mean number of stickers on a page.
The values in the second row of the table are the distances, or absolute deviation, between the values in the first row and 21. The mean of these distances is the MAD of the number of stickers on a page, about 1.2 stickers.
What can we learn from the averages of these distances once they are calculated?
In the first book, the distances are all between 0 and 3. The MAD is 1.2 stickers, which tells us that the number of stickers are typically within 1.2 of the mean number, 21. We could say that a typical page has between 19.8 and 22.2 stickers.
In the second book, the distances are all between 0 and 13. The MAD is 5.6 stickers, which tells us that the number of stickers are typically within 5.6 of the mean number, 21. We could say that a typical page has between 15.4 and 26.6 stickers.
The MAD is also called a measure of the variability of the distribution. In these examples, it is easy to see that a higher MAD suggests a distribution that is more spread out, showing more variability.
The mean absolute deviation (MAD) is one way to measure how spread out a data set is. To find the MAD, find the distance between each data value and the mean. Add all the distances. Then divide by how many distances there are.
The MAD is 2.4. So, these travel times are typically 2.4 minutes away from the mean of 11 minutes.