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I came across the following GRE question. I had no problem finding the mean. However, the answer for the median is given to be 1. I don't understand how they arrive at this.

Find the mean and median of the values of the random variable $X$, whose relative frequency distribution is given in the table below. $$\begin{array}{c|c} \,\,\,X\,\,\,& \,\,\text{Relative Frequency}\,\,\\ \hline \\ 0 & 0.18 \\ 1 & 0.33 \\ 2 &0.10 \\ 3 &0.06 \\ 4 &0.33 \\ \hline \end{array}$$

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  • $\begingroup$ I don't either. It looks like the answer should be 0 to me. Perhaps the book has a mistake. $\endgroup$ – Wintermute Mar 31 '13 at 12:57
  • $\begingroup$ I'm curious -- why do you think it should be zero? I don't see that. $\endgroup$ – Joebevo Mar 31 '13 at 13:02
  • $\begingroup$ Put the values of $X$ in order or relative frequency. Zero is the value in the middle. $\endgroup$ – Wintermute Mar 31 '13 at 13:20
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Imagine that there were $100$ observations. We got the result $0$ a total of $18$ times, and the result $1$ a total of $33$ times, and so on. Thus $51$ of the observations are $\le 1$. It follows that the median is $\le 1$. But only $18$ of the observations are $\le 0$. It follows that the median is $1$.

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  • $\begingroup$ First paragraph above is great, but I don't think the followup "remark" makes sense. Clearly the data are discrete, so I'm not sure what a "small change" could possibly mean for them. $\endgroup$ – Daniel R. Collins Sep 12 '15 at 6:35
  • $\begingroup$ To the extent I can remember that far back, I was thinking of the data as already processed into artificially chosen bins, as in a coarse-grained histogram. Or perhaps as representing degree of "liking." $\endgroup$ – André Nicolas Sep 12 '15 at 12:17
  • $\begingroup$ That would have to be shown in the frequency table, with bins 0 to 1, 1 to 2, etc. As-is the table is communicating that we're dealing with discrete data. $\endgroup$ – Daniel R. Collins Sep 12 '15 at 20:07
  • $\begingroup$ Since you think the remark could cause confusion, I will delete it. $\endgroup$ – André Nicolas Sep 12 '15 at 20:14
  • $\begingroup$ I think that's an improvement, thanks for attending to that. $\endgroup$ – Daniel R. Collins Sep 12 '15 at 20:20
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Mean $$0\times0.18+1\times0.33+2\times0.1+3\times0.06+4\times0.33= 2.03$$ median:

there are 18 of 0's, 33of 1's, 10 of 2's, 6 of 3's, and 33 of 4's. the middle value is $(50^{th}+51^{st})/2$=(1+1)/2=1.

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  • $\begingroup$ That makes sense thank you $\endgroup$ – fullMoon Nov 13 '15 at 15:59
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As you know, the relative frequency is the associated frequency divided by the total number of data, but in this question we don't know the total number of data. If we assume it to be 100, the frequency corresponding to each X value can be manipulated by multiplying each X value by 100. Now for each X value we have 18, 33, 10, 6, and 33 frequencies respectively. In order to calculate the median, we should first order the numbers from smallest to highest, as the middle value is the median. In this question we have 100 numbers (an even number), so the position of the median is located at the Y values corresponding to X values 50 and 51 and is found by averaging these two values. The corresponding Y value for both X=50 and X=52 is 1, and (1 + 1)/2 = 1, which is the answer.

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