A fund posted the following annual returns over six years: 9%, −3%, 12%, 9%, 3% and 6%. Which statement correctly compares the mean, median and mode of these returns?
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Answer: D
- Mean = (9 − 3 + 12 + 9 + 3 + 6) ÷ 6 = 36 ÷ 6 = 6%.
- Sort: −3, 3, 6, 9, 9, 12. Even count, so median = (6 + 9) ÷ 2 = 7.5%.
- Mode = most frequent value = 9%.
- Mean (6%) < median (7.5%) < mode (9%); the −3% year pulls the mean down (negative skew).
The mean is greater than the median, which is greater than the mode: Reversed: the mean (6%) is the smallest of the three and the mode (9%) the largest.
The mean equals the median, and both are below the mode: Takes the median as 6%; with an even number of values the median is the average of the two middle values (6% and 9%), or 7.5%.
The median is below the mean, which is below the mode: The median (7.5%) is above the mean (6%), not below it.
The mean is below the median, which is below the mode: Correct — mean = 36% ÷ 6 = 6%; sorted (−3, 3, 6, 9, 9, 12), the median is (6 + 9) ÷ 2 = 7.5%; the mode is 9%, the only repeated value.
For an even number of observations, the median is the average of the two middle values; a low outlier drags the mean below the median.