To determine the skewness of each data set, we compare the mean and the median of each set.
### Set A: [tex]\(\{32, 12, 24, 46, 18, 22, 14\}\)[/tex]
- Mean: [tex]\(\frac{32 + 12 + 24 + 46 + 18 + 22 + 14}{7} = \frac{168}{7} = 24\)[/tex]
- Median: Ordering the data [tex]\(\{12, 14, 18, 22, 24, 32, 46\}\)[/tex], the median is 22.
Since the mean (24) is greater than the median (22), Set A shows a positive skew.
### Set B: [tex]\(\{4, 12, 11, 14, 11, 5, 12, 13, 18, 14\}\)[/tex]
- Mean: [tex]\(\frac{4 + 12 + 11 + 14 + 11 + 5 + 12 + 13 + 18 + 14}{10} = \frac{114}{10} = 11.4\)[/tex]
- Median: Ordering the data [tex]\(\{4, 5, 11, 11, 12, 12, 13, 14, 14, 18\}\)[/tex], the median is the average of the 5th and 6th numbers, [tex]\(\frac{12 + 12}{2} = 12\)[/tex].
Since the mean (11.4) is less than the median (12), Set B shows a negative skew.
### Set C: [tex]\(\{5, 4, 9, 12, 14, 26, 22, 18\}\)[/tex]
- Mean: [tex]\(\frac{5 + 4 + 9 + 12 + 14 + 26 + 22 + 18}{8} = \frac{110}{8} = 13.75\)[/tex]
- Median: Ordering the data [tex]\(\{4, 5, 9, 12, 14, 18, 22, 26\}\)[/tex], the median is the average of the 4th and 5th numbers, [tex]\(\frac{12 + 14}{2} = 13\)[/tex].
Since the mean (13.75) is greater than the median (13), Set C shows a positive skew.
### Set D: [tex]\(\{1, 1, 1, 2, 2, 3, 3, 4, 5, 6\}\)[/tex]
- Mean: [tex]\(\frac{1 + 1 + 1 + 2 + 2 + 3 + 3 + 4 + 5 + 6}{10} = \frac{28}{10} = 2.8\)[/tex]
- Median: Ordering the data, [tex]\(\{1, 1, 1, 2, 2, 3, 3, 4, 5, 6\}\)[/tex], the median is the average of the 5th and 6th numbers, [tex]\(\frac{2 + 3}{2} = 2.5\)[/tex].
Since the mean (2.8) is greater than the median (2.5), Set D shows a positive skew.
### Summary
- Positive Skew: Sets A, C, D
- Negative Skew: Set B
Thus, the sets that show a positive skew are sets A, C, and D, and the set that shows a negative skew is set B.
