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Question
the values for three sets of data are shown below.
| data set | data values |
|---|---|
| 2 | 63, 29, 35, 28, 30 |
| 3 | 2, 5, 3, 8 |
without calculating any statistics, anna knows that data set 3 would have the least mean absolute deviation among the three sets. which statement explains how she knows?
- sets 1 and 2 contain outliers.
- set 3 has the least mean.
- set 3 contains an outlier.
- sets 1 and 2 have an odd number of values.
To determine the mean absolute deviation (MAD), we consider the spread of data. MAD measures the average distance of data points from the mean. A smaller MAD means data points are closer to the mean (less spread out).
- For Set 1: Values are 42, 48, 50, 88, 49. The value 88 is much larger than others, so it's an outlier, causing more spread.
- For Set 2: Values are 63, 29, 35, 28, 30. The value 63 is much larger than others, so it's an outlier, causing more spread.
- For Set 3: Values are 2, 5, 3, 8. These values are relatively close to each other (no extreme outliers), so the data is more clustered around the mean, leading to a smaller MAD.
Now, let's analyze the options:
- Option 1: Sets 1 and 2 contain outliers. This is true (88 in Set 1, 63 in Set 2) and explains why their MAD would be larger (more spread) compared to Set 3 (no extreme outliers, less spread).
- Option 2: Set 3 has the least mean. MAD depends on spread, not the mean's value. We can't conclude this without calculating the mean.
- Option 3: Set 3 contains an outlier. Looking at Set 3 values (2, 5, 3, 8), there's no extreme outlier (all values are within a small range).
- Option 4: Sets 1 and 2 have an odd number of values. The number of values (odd/even) doesn't directly relate to MAD; MAD depends on spread, not the count's parity.
So the correct statement is "Sets 1 and 2 contain outliers" as it explains why Set 3 has the least MAD (since outliers in 1 and 2 increase their spread, while Set 3 has no such outliers, leading to less spread and thus smaller MAD).
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A. Sets 1 and 2 contain outliers.