the values for three different sets of data are shown below.

| data set | values |
|----------|--------------|
| 1 | 52, 55, 59, 53, 50 |
| 2 | 40, 49, 43, 42, 90, 38 |
| 3 | 24, 30, 26, 31, 40, 26 |

without calculating any statistics, jadyn knows that data set 1 would have the least mean absolute deviation among the three sets. which statement explains how she knows?

- sets 2 and 3 have an even number of values.
- set 1 has the least number of values.
- sets 2 and 3 contain outliers.
- set 1 contains an outlier.

Question

the values for three different sets of data are shown below.

| data set | values       |
|----------|--------------|
| 1        | 52, 55, 59, 53, 50 |
| 2        | 40, 49, 43, 42, 90, 38 |
| 3        | 24, 30, 26, 31, 40, 26 |

without calculating any statistics, jadyn knows that data set 1 would have the least mean absolute deviation among the three sets. which statement explains how she knows?

- sets 2 and 3 have an even number of values.
- set 1 has the least number of values.
- sets 2 and 3 contain outliers.
- set 1 contains an outlier.

Accepted Answer

# Brief Explanations:
To determine which statement explains why Set 1 has the least mean absolute deviation (MAD), we analyze the data:
- **Set 1**: Values are 52, 55, 59, 53, 50. These are all close to each other (range ~9), so less spread.
- **Set 2**: Values are 40, 49, 43, 42, 90, 38. The value 90 is an outlier, increasing spread.
- **Set 3**: Values are 24, 30, 26, 31, 40, 26. Values are more spread out than Set 1.

Mean absolute deviation depends on the spread of data (how far values are from the mean). A smaller spread (less variation) means a smaller MAD. Set 1 has values clustered closely, while Sets 2 (has 90 as an outlier) and 3 (more spread) have greater variation. Wait, no—wait, the question is about why Set 1 has the least MAD. Wait, the options:

Wait, the options are:
- Sets 2 and 3 have an even number of values. (Not relevant to MAD; MAD depends on spread, not count parity.)
- Set 1 has the least number of values. (Set 1 has 5 values, Set 2 has 6, Set 3 has 6. But MAD isn’t determined by number of values alone; a small set with spread can have high MAD. So this is incorrect.)
- Sets 2 and 3 contain outliers. (Set 2 has 90 (outlier), Set 3: let's check. Set 3 values: 24,26,26,30,31,40. 40 is higher than others, but maybe not an outlier? Wait, Set 2 has 90 (clear outlier), Set 3: the values are 24,26,26,30,31,40. The range is 40-24=16. Set 1 range is 59-50=9. So Set 2 has an outlier (90), Set 3 has more spread than Set 1. So if Sets 2 and 3 have outliers (or more spread), Set 1 has less spread. Wait, the correct option: Wait, the options are:

Wait, the options given:

1. Sets 2 and 3 have an even number of values. → Irrelevant to MAD.
2. Set 1 has the least number of values. → Number of values doesn’t determine MAD (e.g., a small set with big spread can have high MAD).
3. Sets 2 and 3 contain outliers. → Set 2 has 90 (outlier), Set 3: let's see, 40 is higher than the rest (24,26,26,30,31), so maybe 40 is an outlier? Or Set 3’s values are more spread. So if Sets 2 and 3 have outliers (or greater spread), Set 1 has values closer together (less spread), so lower MAD. But wait, the question is “which statement explains how she knows [Set 1 has least MAD]”. Wait, maybe I misread. Wait, the options:

Wait, the original options (from the image):

- Sets 2 and 3 have an even number of values.
- Set 1 has the least number of values.
- Sets 2 and 3 contain outliers.
- Set 1 contains an outlier.

Wait, no—wait, the user’s image:

Options:

- Sets 2 and 3 have an even number of values.
- Set 1 has the least number of values.
- Sets 2 and 3 contain outliers.
- Set 1 contains an outlier.

Wait, let's re-express:

Data:

Set 1: 52,55,59,53,50 (5 values, all in 50-59 range, no outlier)

Set 2: 40,49,43,42,90,38 (6 values; 90 is way higher than others → outlier)

Set 3: 24,30,26,31,40,26 (6 values; 40 is higher than 24,26,26,30,31 → maybe an outlier, or just spread)

Mean absolute deviation (MAD) measures the average distance from the mean. A data set with values closer together (less spread) has a smaller MAD.

Set 1’s values are clustered (50-59), so mean is around 54, and distances from mean are small (e.g., 52-54=2, 55-54=1, etc.).

Set 2 has 90, which is far from the mean (mean of Set 2: (40+49+43+42+90+38)/6 = (40+49=89; 89+43=132; 132+42=174; 174+90=264; 264+38=302; 302/6 ≈50.33). 90-50.33≈39.67, a large deviation, so MAD will be high.

Set 3: mean is (24+30+26+31+40+26)/6 = (24+30=54; +26=80; +31=111; +40=151; +26=177; 177/6=29.5). 40-29.5=10.5, 24-29.5=5.5, etc. So spread is more than Set 1.

Now, the options:

- “Sets 2 and 3 have an even number of values.” → Even number of values doesn’t affect MAD. Eliminate.
- “Set 1 has the least number of values.” → Set 1 has 5, Sets 2 and 3 have 6. But number of values doesn’t determine MAD (e.g., a set with 2 values far apart has high MAD). Eliminate.
- “Sets 2 and 3 contain outliers.” → Set 2 has 90 (outlier), Set 3: 40 is higher than the rest (24,26,26,30,31), so 40 is an outlier? Or just a high value. Either way, Sets 2 and 3 have values that are more spread out (Set 2 has 90, Set 3 has 40 far from the lower values), while Set 1 has no outliers and values clustered. So if Sets 2 and 3 have outliers (causing more spread), Set 1 has less spread, hence lower MAD. This makes sense.
- “Set 1 contains an outlier.” → Set 1’s values are 50,52,53,55,59—all close, no outlier. Eliminate.

Thus, the correct statement is “Sets 2 and 3 contain outliers.” Wait, no—wait, the option is “Sets 2 and 3 contain outliers.” So that explains why Set 1 has the least MAD: because Sets 2 and 3 have outliers (increasing their spread, hence increasing their MAD), while Set 1 has no outliers and values clustered (lower MAD).
# Answer:
Sets 2 and 3 contain outliers.

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