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. $$\begin{array}{c|c}
\,\,\,X\,\,\,& \,\,\text{Relative Frequency}\,\,\\ \hline \\ 0 & 0.18 \\ 1 & 0.33 \\ 2 &0.10 \\ 3 &0.06 \\ 4 &0.33 \\ \hline
\end{array}$$

 A: 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.
A: 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 $1$.
A: 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. 
