Question

The mean absolute deviation of Dataset #1 is 48.2. Dataset #2 follows. Which dataset is more spread out?
Dataset #2: 401, 467, 489, 503, 568
(1 point)
Dataset #1 is more spread out because it has a mean absolute deviation of 48.2 whereas Dataset #2 has a mean absolute deviation of 0.
Dataset #2 is more spread out because it has a mean absolute deviation of 41.28, whereas Dataset #1 has a mean absolute deviation of 48.2.
Dataset #1 is more spread out because it has a mean absolute deviation of 48.2 whereas Dataset #2 has a mean absolute deviation of 41.28.
Dataset #2 is more spread out because it has a mean absolute deviation of 485.6 whereas Dataset #1 has a mean absolute deviation of only 48.2.

Answer 1

To determine which dataset is more spread out, we can compare the mean absolute deviations (MAD) of both datasets.

You provided that the mean absolute deviation of Dataset #1 is 48.2.

Now, let's calculate the mean absolute deviation for Dataset #2: 401, 467, 489, 503, 568.

1. First, calculate the mean of Dataset #2:

\[
\text{Mean} = \frac{401 + 467 + 489 + 503 + 568}{5} = \frac{2428}{5} = 485.6
\]

2. Next, calculate the absolute deviations from the mean:

- |401 - 485.6| = 84.6
- |467 - 485.6| = 18.6
- |489 - 485.6| = 3.4
- |503 - 485.6| = 17.4
- |568 - 485.6| = 82.4

3. Now, find the mean absolute deviation:

\[
\text{MAD} = \frac{84.6 + 18.6 + 3.4 + 17.4 + 82.4}{5} = \frac{206.4}{5} = 41.28
\]

Now, we can compare the mean absolute deviations:
- MAD of Dataset #1 = 48.2
- MAD of Dataset #2 = 41.28

Since 48.2 > 41.28, Dataset #1 is more spread out.

So the correct choice is:
**Dataset #1 is more spread out because it has a mean absolute deviation of 48.2 whereas Dataset #2 has a mean absolute deviation of 41.28.**