Question

By how much does the mean absolute deviation of Dataset A exceed that of Dataset B?

Dataset A: 9, 9, 10, 10, 10, 11, 12, 14, 14, 18
Dataset B: 7, 8, 8, 9, 10, 10, 11, 11, 12, 15

Answer 1

To find out how much the mean absolute deviation (MAD) of Dataset A exceeds that of Dataset B, we first need to calculate the MAD for each dataset.

**Step 1: Calculate the Mean for each dataset.**

**Dataset A:**
\[ A = [9, 9, 10, 10, 10, 11, 12, 14, 14, 18] \]

Mean of Dataset A:
\[
\text{Mean}_A = \frac{9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18}{10} = \frac{ 9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18 }{10} = \frac{ 117 }{10} = 11.7
\]

**Dataset B:**
\[ B = [7, 8, 8, 9, 10, 10, 11, 11, 12, 15] \]

Mean of Dataset B:
\[
\text{Mean}_B = \frac{7 + 8 + 8 + 9 + 10 + 10 + 11 + 11 + 12 + 15}{10} = \frac{ 8 + 8 + 9 + 10 + 10 + 11 + 11 + 12 + 15 }{10} = \frac{ 111 }{10} = 11.1
\]

**Step 2: Calculate the Mean Absolute Deviation (MAD) for each dataset.**

The formula for the mean absolute deviation is:
\[
\text{MAD} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \text{Mean}|
\]

**MAD for Dataset A:**
\[
|9 - 11.7| = 2.7\\
|9 - 11.7| = 2.7\\
|10 - 11.7| = 1.7\\
|10 - 11.7| = 1.7\\
|10 - 11.7| = 1.7\\
|11 - 11.7| = 0.7\\
|12 - 11.7| = 0.3\\
|14 - 11.7| = 2.3\\
|14 - 11.7| = 2.3\\
|18 - 11.7| = 6.3
\]

\[
\text{MAD}_A = \frac{2.7 + 2.7 + 1.7 + 1.7 + 1.7 + 0.7 + 0.3 + 2.3 + 2.3 + 6.3}{10} = \frac{ 21.1 }{10} = 2.11
\]

**MAD for Dataset B:**
\[
|7 - 11.1| = 4.1\\
|8 - 11.1| = 3.1\\
|8 - 11.1| = 3.1\\
|9 - 11.1| = 2.1\\
|10 - 11.1| = 1.1\\
|10 - 11.1| = 1.1\\
|11 - 11.1| = 0.1\\
|11 - 11.1| = 0.1\\
|12 - 11.1| = 0.9\\
|15 - 11.1| = 3.9
\]

\[
\text{MAD}_B = \frac{4.1 + 3.1 + 3.1 + 2.1 + 1.1 + 1.1 + 0.1 + 0.1 + 0.9 + 3.9}{10} = \frac{ 19.5 }{10} = 1.95
\]

**Step 3: Calculate the difference in MADs.**
\[
\text{Difference} = \text{MAD}_A - \text{MAD}_B\\
= 2.11 - 1.95 = 0.16
\]

Therefore, the mean absolute deviation of Dataset A exceeds that of Dataset B by \( 0.16 \).