# Find the median given a table of relative frequencies

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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I don't either. It looks like the answer should be 0 to me. Perhaps the book has a mistake. –  Wintermute Mar 31 '13 at 12:57
I'm curious -- why do you think it should be zero? I don't see that. –  Joebevo Mar 31 '13 at 13:02
Put the values of $X$ in order or relative frequency. Zero is the value in the middle. –  Wintermute Mar 31 '13 at 13:20

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 $1$.

Remark: Our data are coarse-grained, and a small change in them could produce a large change in median. So although the answer of $1$ is technically correct, it feels uncomfortable.

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As you know the relative frequency is the associated frequency divided by the total number of data. In this question we don't know the total number of data. But if we assume it as 100, the frequency for each X can be manipulated by multiplying each X with 100. Now for each X we have, 18,33,10,6,33 frequencies respectively. In order to calculate the mean, we should first order the numbers from the smallest to highest and then the middle value is the mean. In this question we have 100 numbers (an even number), so the position of the mean is located at the number 50 and 51 divided by 2. In the location of 50 and 51 we have 1. And 1 divided by two gives us one. Which is the answer.

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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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