Question

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**.