Suppose I have a set of numbers, for example - exam scores of 20 students.
Can there be more than half of students scoring above the mean ? similarly, can more than half the class score above the median and mode ?
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1$\begingroup$ Yes and no. You can't have more than half above the median; at most in the presence of ties more than half can be equal to or above the median. (Trivial example: all students get the same mark.) Otherwise imagine 9 at 0, 6 at 50 and 5 at 100. Then the mode is 0 and 11 are above. The mean is 40 and 11 are above. Many such examples. $\endgroup$ – Nick Cox Feb 1 '18 at 17:30
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$\begingroup$ @NickCox You should post that as an answer rather than a comment. $\endgroup$ – Kodiologist Feb 1 '18 at 17:35
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$\begingroup$ thanks @NickCox, can you share some more insights into this ? any study material ? similarly, what will be the case with mode and mean ? $\endgroup$ – RicoRicochet Feb 1 '18 at 17:36
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$\begingroup$ Any good introductory statistics text should discuss the possibilities. @A.G. has now provided a very good analysis. $\endgroup$ – Nick Cox Feb 1 '18 at 17:41
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$\begingroup$ Note that "above" can be replaced by "below" simply by negating all values. Thus, if there is any statistic whatsoever where the answer is always "no," then by definition that statistic is a median. $\endgroup$ – whuber♦ Feb 1 '18 at 18:17
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It is not possible for over half of the students to score above the median:
- If there is an even number of students $n$, the median is the mean of the 2 scores in the "middle" of the sorted set of scores, meaning that at most half ($\frac{n}{2}$) of the students will score above or below the median.
- If there is an odd number of students $n$, the median is the number in the "middle" of the sorted set of scores. In this case, at most less than half ($\frac{n-1}{2}$, to be precise) students may score above the median.
Notice that I said "at most" above: when multiple students score at the median, the number of students scoring over the median must be even less than the $\frac{n}{2}$ or $\frac{n-1}{2}$ figures I gave above.
But is is possible for more than half the students to score above the mean: this may happen when the distribution is negatively-skewed, like in image (a) below, and the mean is below the median.
And it is possible for more than half the students to score above the mode: this may happen when the distribution is positively-skewed, like in image (c) below, and the mode is below the median.
(For the avoidance of doubt, the perfectly symmetrical distribution in image (b) above is actually a bit of a special case, where the mean exists (unlike the Cauchy distribution) and there is a single mode. If those conditions do not hold, we cannot state that the mean, median & mode coincide, so being perfectly symmetrical is neither necessary nor sufficient to do so. One famous example of such a symmetric distribution where those conditions do hold is the Normal/Gaussian ("bell curve") distribution.)
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2$\begingroup$ Good (+1). Some readers might appreciate notes that the normal is just an example of a perfectly symmetrical distribution; and that distributions being symmetrical is neither necessary nor sufficient for mode to coincide with median and mean. $\endgroup$ – Nick Cox Feb 1 '18 at 17:43
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$\begingroup$ Thanks for the feedback. I've edited it to add that info. $\endgroup$ – A. G. Feb 1 '18 at 17:54
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1$\begingroup$ @NickCox - In other words, "normal" is completely abnormal. $\endgroup$ – Daniel R Hicks Feb 1 '18 at 21:51
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1$\begingroup$ Edwin Jaynes suggested that we call it the central distribution, but I've never seen that quoted. There is always Gaussian.... $\endgroup$ – Nick Cox Feb 1 '18 at 21:54
