Question

By how much does the mean absolute deviation of Dataset A exceed that of Dataset B? Round your answer to two decimal places.

Dataset A: {9,9,10,10,10,11,12,14,14,18}


Dataset B: {7,8,8,9,10,10,11,11,12,15}

1 answer

To find the mean absolute deviation (MAD) of both datasets, we first need to calculate the mean of each dataset, then the absolute deviations from the mean, and finally the mean of those absolute deviations.

Step 1: Calculate the mean of each dataset

Dataset A: {9, 9, 10, 10, 10, 11, 12, 14, 14, 18}

Mean of Dataset A: \[ \text{Mean} = \frac{9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18}{10} = \frac{ 9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18}{10} = \frac{ 9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18}{10} = \frac{ 9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18}{10} = \frac{ 9 + 9 + 10 + 10 + 10 + 11 + 12 + 14 + 14 + 18}{10} \] Calculate the sum: \[ = \frac{ 121}{10} = 12.1 \]

Dataset B: {7, 8, 8, 9, 10, 10, 11, 11, 12, 15}

Mean of Dataset B: \[ \text{Mean} = \frac{7 + 8 + 8 + 9 + 10 + 10 + 11 + 11 + 12 + 15}{10} = \frac{ 7 + 8 + 8 + 9 + 10 + 10 + 11 + 11 + 12 + 15}{10} \] Calculate the sum: \[ = \frac{ 91 }{10} = 9.1 \]

Step 2: Calculate the absolute deviations for each dataset

Absolute deviations for Dataset A: \[ |x - \text{mean}| \] Where \(x\) are the elements of Dataset A.

\[ |9 - 12.1| = 3.1, \] \[ |9 - 12.1| = 3.1, \] \[ |10 - 12.1| = 2.1, \] \[ |10 - 12.1| = 2.1, \] \[ |10 - 12.1| = 2.1, \] \[ |11 - 12.1| = 1.1, \] \[ |12 - 12.1| = 0.1, \] \[ |14 - 12.1| = 1.9, \] \[ |14 - 12.1| = 1.9, \] \[ |18 - 12.1| = 5.9. \]

The absolute deviations are: \[3.1, 3.1, 2.1, 2.1, 2.1, 1.1, 0.1, 1.9, 1.9, 5.9\]

Sum of absolute deviations (Dataset A): \[ 3.1 + 3.1 + 2.1 + 2.1 + 2.1 + 1.1 + 0.1 + 1.9 + 1.9 + 5.9 = 23.4 \]

Mean Absolute Deviation for Dataset A: \[ \text{MAD}_A = \frac{23.4}{10} = 2.34 \]

Absolute deviations for Dataset B: \[ |x - \text{mean}| \] Where \(x\) are the elements of Dataset B.

\[ |7 - 9.1| = 2.1, \] \[ |8 - 9.1| = 1.1, \] \[ |8 - 9.1| = 1.1, \] \[ |9 - 9.1| = 0.1, \] \[ |10 - 9.1| = 0.9, \] \[ |10 - 9.1| = 0.9, \] \[ |11 - 9.1| = 1.9, \] \[ |11 - 9.1| = 1.9, \] \[ |12 - 9.1| = 2.9, \] \[ |15 - 9.1| = 5.9. \]

The absolute deviations are: \[2.1, 1.1, 1.1, 0.1, 0.9, 0.9, 1.9, 1.9, 2.9, 5.9\]

Sum of absolute deviations (Dataset B): \[ 2.1 + 1.1 + 1.1 + 0.1 + 0.9 + 0.9 + 1.9 + 1.9 + 2.9 + 5.9 = 18.0 \]

Mean Absolute Deviation for Dataset B: \[ \text{MAD}_B = \frac{18.0}{10} = 1.80 \]

Step 3: Find the difference between the mean absolute deviations

Difference: \[ \text{MAD}_A - \text{MAD}_B = 2.34 - 1.80 = 0.54 \]

Final Answer

The mean absolute deviation of Dataset A exceeds that of Dataset B by 0.54 (rounded to two decimal places).