Find the median from the following table. \[\]

Marks Scored	Below 20	Below 40	Below 60	Below 80	Below 100
Number of students	6	10	20	36	50

A.56 marks \[\]
B.60 marks\[\]
C.66 marks\[\]
D.70 marks\[\]

Hint: We divide the data of marks into 5 class intervals ${{I}_{1}}=0-20,{{I}_{2}}=20-40,{{I}_{3}}=40-60$ , ${{I}_{4}}=60-80,{{I}_{5}}=80-100$. The number of students below a particular mark is cumulative frequency. We find the frequency of any interval ${{I}_{k}}$ as ${{f}_{k}}={{\left( cf \right)}_{k}}-{{\left( cf \right)}_{k-1}}$. We then use the formula for median $m=l+\dfrac{\dfrac{n}{2}-{{\left( cf \right)}_{m}}}{{{f}_{m}}}\times w$ where $l,w,{{\left( cf \right)}_{m}},{{f}_{m}}$ are respectively the lower limit, width, cumulative frequency, frequency corresponding to median interval ${{I}_{m}}$ and $n$ is the sum of frequencies. \[\]

Complete step-by-step solution:
If the data is divided into class intervals say ${{I}_{1}},{{I}_{2}},...,{{I}_{n}}$ and corresponding frequencies ${{f}_{1}},{{f}_{2}},...,{{f}_{n}}$ then we calculate the cumulative frequencies. The cumulative frequency $cf$ of the class interval ${{I}_{k}}$ is given by
\[{{\left( cf \right)}_{k}}=\sum\limits_{i=1}^{c}{{{f}_{i}}}\]
The frequency of the interval ${{I}_{k}}$ if the cumulative frequencies are given is
\[{{f}_{k}}={{\left( cf \right)}_{k}}-{{\left( cf \right)}_{k-1}}\]
The median class interval ${{I}_{m}}$ is the interval that corresponds to cumulative frequency which is just greater than half of the sum of frequencies. We estimate the mean using the formula
\[m=l+\dfrac{\dfrac{n}{2}-{{\left( cf \right)}_{m}}}{{{f}_{m}}}\times w\]
Here $l$ is the lower limit, $w$ is the width, ${{\left( cf \right)}_{m}}$ is the cumulative frequency, ${{f}_{m}}$ is the frequency of the median interval ${{I}_{m}}$ and $n$ is the sum of frequencies.
We are given the question of the number of students below a certain mark which is our cumulative frequency. So we need to find the interval and the frequencies. If we divide the data into 5 intervals for marks as ${{I}_{1}}=0-20,{{I}_{2}}=20-40,{{I}_{3}}=40-60,{{I}_{4}}=60-80,{{I}_{5}}=80-100$ then cumulative frequencies are ${{\left( cf \right)}_{1}}=0-10,{{\left( cf \right)}_{2}}=20-40,{{\left( cf \right)}_{3}}=40-60,{{\left( cf \right)}_{4}}=60-80,{{\left( cf \right)}_{5}}=80-100$ and we find the frequencies as
\[\begin{align}
  & {{f}_{1}}={{\left( cf \right)}_{1}}=6 \\
 & {{f}_{2}}={{\left( cf \right)}_{2}}-{{\left( cf \right)}_{1}}=10-6=4 \\
 & {{f}_{3}}={{\left( cf \right)}_{3}}-{{\left( cf \right)}_{2}}=20-10=10 \\
 & {{f}_{4}}={{\left( cf \right)}_{4}}-{{\left( cf \right)}_{3}}=36-20=16 \\
 & {{f}_{5}}={{\left( cf \right)}_{5}}-{{\left( cf \right)}_{14}}=50-36=14 \\
\end{align}\]
Let us draw the cumulative frequency table.

The sum of the frequencies is
\[n={{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}=50\]
Here half of the frequency is 25 and the cumulative frequency just greater than 25 is 36 which corresponds to the interval 60-80. So the median interval is ${{I}_{m}}=60-80$. The median interval has width $w=80-60=20$, lower limit $l=60$, frequency ${{f}_{m}}=16$ and cumulative frequency${{\left( cf \right)}_{m}}=36$. We use the median formula and have
\[m=60+\dfrac{\dfrac{50}{2}-20}{16}\times 20=60+\dfrac{5}{16}\times 20=66.25\]
We round after the decimal point and find the median as 66 marks. So the correct option is C. \[\]

Note: We note that 50% of the data lie below median. This is why the median is primarily used as a robust statistic where quality cannot fall below a breakdown point of 50%. The mean is different from the median as mean as the sum of values divided by the number of values.