Allina list the ages of all of the family members who live in her house below what is the mean absolutely deviation of their ages of 913.43 and 55 a 28 b38 c76 d19

Answer 1

To calculate the mean absolute deviation (MAD) of the ages provided, we will follow these steps:

1. List the ages:

   • 13
   • 43
   • 55
   • 28
   • 38
   • 76
   • 19

2. Calculate the mean (average) of the ages. \[ \text{Mean} = \frac{13 + 43 + 55 + 28 + 38 + 76 + 19}{7} = \frac{272}{7} \approx 38.857 \]

3. Calculate the absolute deviations from the mean for each age:

   • |13 - 38.857| = | -25.857 | = 25.857
   • |43 - 38.857| = | 4.143 | = 4.143
   • |55 - 38.857| = | 16.143 | = 16.143
   • |28 - 38.857| = | -10.857 | = 10.857
   • |38 - 38.857| = | -0.857 | = 0.857
   • |76 - 38.857| = | 37.143 | = 37.143
   • |19 - 38.857| = | -19.857 | = 19.857

4. List the absolute deviations:

   • 25.857
   • 4.143
   • 16.143
   • 10.857
   • 0.857
   • 37.143
   • 19.857

5. Calculate the mean of the absolute deviations: \[ \text{Mean Absolute Deviation} = \frac{25.857 + 4.143 + 16.143 + 10.857 + 0.857 + 37.143 + 19.857}{7} = \frac{114.0}{7} \approx 16.286 \]

So, the mean absolute deviation of the ages is approximately 16.286.