Calculate the median for a continuous distribution given below.

Class interval	Frequency
10-20	11
20-30	13
30-40	13
40-50	9
50-60	4

Hint: We first explain the terms class marks, cumulative frequencies. We use the formula of ${{x}_{me}}={{x}_{l}}+\dfrac{{{f}_{m}}-{{f}_{m-1}}}{2{{f}_{m}}-{{f}_{m-1}}-{{f}_{m+1}}}\times c$ where ${{x}_{me}}$ denotes the median value. We place the values to find the solution.

Complete step-by-step solution:
To find the median for a continuous distribution, we need to find the total frequency along with cumulative frequency and the class mark.
The frequencies are given which in total gives $n=11+13+13+9+4=50$. The value of $\dfrac{n}{2}=\dfrac{50}{2}=25$.
The class intervals and their upper and lower class is given which gives the class marks as the middle point of those upper and lower classes.
We also find the less than type cumulative frequencies.

The median class will be the class where the value of $\dfrac{n}{2}=25$ lies in the Cumulative Frequency column. So, the median class is 20-30 as 25 crosses that value of 24 in cumulative frequencies’ column.
We follow the formula of ${{x}_{me}}={{x}_{l}}+\dfrac{{{f}_{m}}-{{f}_{m-1}}}{2{{f}_{m}}-{{f}_{m-1}}-{{f}_{m+1}}}\times c$.
Here ${{x}_{me}}$ denotes the median value. ${{f}_{m}},{{f}_{m-1}},{{f}_{m+1}}$ are the frequency values of the with respect the median class frequency being ${{f}_{m}}$. $c$ is the class width.
So, ${{x}_{me}}=20+\dfrac{13-11}{2\times 13-11-13}\times 10=30$.
The median value for the given distribution is 30.

Note: If the cumulative frequency for an interval is exactly 50%, then the median value would be the endpoint of this interval. A cumulative frequency diagram is also a good way to find the interquartile range, which is the difference between the upper quartile and lower quartile.