Mean Absolute Deviation Worksheets - Free Printable

Problem Overview:

The task involves calculating the Mean Absolute Deviation (MAD) for two sets of data. The Mean Absolute Deviation is a measure of variability that shows how far, on average, each data point is from the mean of the dataset.

#### Step 1: Understand the Formula for MAD
The formula for Mean Absolute Deviation is:
\[
\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n}
\]
Where:
- \( x_i \) is each data point in the dataset.
- \( \bar{x} \) is the mean of the dataset.
- \( n \) is the number of data points.
- \( |x_i - \bar{x}| \) represents the absolute difference between each data point and the mean.

#### Step 2: Solve Part 1
The first part of the problem asks us to find the Mean Absolute Deviation for the data sets labeled as Set A and Set B.

##### Set A:
Data: \( 5, 7, 9, 11, 13 \)

1. Calculate the Mean (\( \bar{x} \)) of Set A:
\[
\bar{x} = \frac{5 + 7 + 9 + 11 + 13}{5} = \frac{45}{5} = 9
\]

2. Calculate the Absolute Deviations:
- For \( x_1 = 5 \): \( |5 - 9| = 4 \)
- For \( x_2 = 7 \): \( |7 - 9| = 2 \)
- For \( x_3 = 9 \): \( |9 - 9| = 0 \)
- For \( x_4 = 11 \): \( |11 - 9| = 2 \)
- For \( x_5 = 13 \): \( |13 - 9| = 4 \)

3. Sum the Absolute Deviations:
\[
\sum |x_i - \bar{x}| = 4 + 2 + 0 + 2 + 4 = 12
\]

4. Calculate the MAD:
\[
\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n} = \frac{12}{5} = 2.4
\]

So, the Mean Absolute Deviation for Set A is:
\[
\boxed{2.4}
\]

##### Set B:
Data: \( 6, 8, 10, 12, 14 \)

1. Calculate the Mean (\( \bar{x} \)) of Set B:
\[
\bar{x} = \frac{6 + 8 + 10 + 12 + 14}{5} = \frac{50}{5} = 10
\]

2. Calculate the Absolute Deviations:
- For \( x_1 = 6 \): \( |6 - 10| = 4 \)
- For \( x_2 = 8 \): \( |8 - 10| = 2 \)
- For \( x_3 = 10 \): \( |10 - 10| = 0 \)
- For \( x_4 = 12 \): \( |12 - 10| = 2 \)
- For \( x_5 = 14 \): \( |14 - 10| = 4 \)

3. Sum the Absolute Deviations:
\[
\sum |x_i - \bar{x}| = 4 + 2 + 0 + 2 + 4 = 12
\]

4. Calculate the MAD:
\[
\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n} = \frac{12}{5} = 2.4
\]

So, the Mean Absolute Deviation for Set B is:
\[
\boxed{2.4}
\]

#### Step 3: Solve Part 2
The second part of the problem asks us to find the Mean Absolute Deviation for the scores of students in three schools (Teams A, B, and C).

##### Team A:
Scores: \( 75, 80, 85, 90, 95 \)

1. Calculate the Mean (\( \bar{x} \)) of Team A:
\[
\bar{x} = \frac{75 + 80 + 85 + 90 + 95}{5} = \frac{425}{5} = 85
\]

2. Calculate the Absolute Deviations:
- For \( x_1 = 75 \): \( |75 - 85| = 10 \)
- For \( x_2 = 80 \): \( |80 - 85| = 5 \)
- For \( x_3 = 85 \): \( |85 - 85| = 0 \)
- For \( x_4 = 90 \): \( |90 - 85| = 5 \)
- For \( x_5 = 95 \): \( |95 - 85| = 10 \)

3. Sum the Absolute Deviations:
\[
\sum |x_i - \bar{x}| = 10 + 5 + 0 + 5 + 10 = 30
\]

4. Calculate the MAD:
\[
\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n} = \frac{30}{5} = 6
\]

So, the Mean Absolute Deviation for Team A is:
\[
\boxed{6}
\]

##### Team B:
Scores: \( 70, 75, 80, 85, 90 \)

1. Calculate the Mean (\( \bar{x} \)) of Team B:
\[
\bar{x} = \frac{70 + 75 + 80 + 85 + 90}{5} = \frac{400}{5} = 80
\]

2. Calculate the Absolute Deviations:
- For \( x_1 = 70 \): \( |70 - 80| = 10 \)
- For \( x_2 = 75 \): \( |75 - 80| = 5 \)
- For \( x_3 = 80 \): \( |80 - 80| = 0 \)
- For \( x_4 = 85 \): \( |85 - 80| = 5 \)
- For \( x_5 = 90 \): \( |90 - 80| = 10 \)

3. Sum the Absolute Deviations:
\[
\sum |x_i - \bar{x}| = 10 + 5 + 0 + 5 + 10 = 30
\]

4. Calculate the MAD:
\[
\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n} = \frac{30}{5} = 6
\]

So, the Mean Absolute Deviation for Team B is:
\[
\boxed{6}
\]

##### Team C:
Scores: \( 65, 70, 75, 80, 85 \)

1. Calculate the Mean (\( \bar{x} \)) of Team C:
\[
\bar{x} = \frac{65 + 70 + 75 + 80 + 85}{5} = \frac{375}{5} = 75
\]

2. Calculate the Absolute Deviations:
- For \( x_1 = 65 \): \( |65 - 75| = 10 \)
- For \( x_2 = 70 \): \( |70 - 75| = 5 \)
- For \( x_3 = 75 \): \( |75 - 75| = 0 \)
- For \( x_4 = 80 \): \( |80 - 75| = 5 \)
- For \( x_5 = 85 \): \( |85 - 75| = 10 \)

3. Sum the Absolute Deviations:
\[
\sum |x_i - \bar{x}| = 10 + 5 + 0 + 5 + 10 = 30
\]

4. Calculate the MAD:
\[
\text{MAD} = \frac{\sum |x_i - \bar{x}|}{n} = \frac{30}{5} = 6
\]

So, the Mean Absolute Deviation for Team C is:
\[
\boxed{6}
\]

#### Final Answers:
1. Set A: \( \boxed{2.4} \)
2. Set B: \( \boxed{2.4} \)
3. Team A: \( \boxed{6} \)
4. Team B: \( \boxed{6} \)
5. Team C: \( \boxed{6} \)

Thus, the final boxed answers are:
\[
\boxed{2.4, 2.4, 6, 6, 6}
\]