Question

Find the median for the following frequency distribution

Height (in cm)	$160-162$	$163-165$	$166-168$	$169-171$	$172-174$
Frequency	$15$	$117$	$136$	$110$	$14$

Answer 1

Hint: We will first find the total number$\left( \text{N} \right)$ of students by summing the frequencies of different classes in the given distribution and convert the given exclusive frequency into inclusive frequency by adding $0.5$ to the upper limit of each class and subtracting the same $0.5$from lower limit of each class. Now we will identify the class in which the value $\dfrac{\text{N}}{2}$ lies and define that class as Median Class. Now the value of Median is obtained from the below formula
$\text{Median}=l+\left[ \dfrac{\dfrac{\text{N}}{2}-cf}{f} \right]h$
Where
     $l$ is the lower limit of the median class
     $\text{N}$ is the total number of students
     $cf$ is the Cumulative frequency of class preceding the median class
     $f$ is the frequency of the median class
     $h$ is the height of the class

Complete step-by-step answer:
Given that,
Frequency distribution table is

Height (in cm)	$160-162$	$163-165$	$166-168$	$169-171$	$172-174$
Frequency	$15$	$117$	$136$	$110$	$14$

The above frequency table is exclusive, so we are going to convert it into inclusive by adding $0.5$ to the upper limit of each class and subtracting the same $0.5$from lower limit of each class, then

Class Interval (Exclusive)	Class Interval (Inclusive)	Class Interval Frequency	Cumulative Frequency
$160-162$	$159.5-162.5$	$15$	$15$
$163-165$	$162.5-165.5$	$117$	$132\left( 15+117 \right)$
$166-168$	$165.5-168.5$	$136$	$268\left( 132+136 \right)$
$169-171$	$168.5-171.5$	$110$	$378\left( 268+110 \right)$
$172-174$	$171.5-174.5$	$14$	$392\left( 378+14 \right)$
			$\text{N}=392$

Here the sum of the frequencies is $\text{N}=392$
Now the value of $\dfrac{\text{N}}{2}$ is $\dfrac{\text{N}}{2}=\dfrac{392}{2}=196$
Hence the value $\dfrac{\text{N}}{2}$ lies in $132$,$268$. Hence the median class is $165.5-168.5$
Now

lower limit of the median class is $l=165.5$

Cumulative frequency of class preceding the median class is $cf=132$

Frequency of the median class is $f=136$

Height of the class is $h=168.5-165.5=3$

Then the value of the median is given by
$\begin{align}
  & \text{Median}=l+\left[ \dfrac{\dfrac{\text{N}}{2}-cf}{f} \right]h \\
 & =165.5+\left[ \dfrac{196-132}{136} \right]\times 3 \\
 & =165.5+\left[ \dfrac{64}{136} \right]\times 3 \\
 & =165.5+1.41 \\
 & =166.91
\end{align}$
Hence the value of Median is $166.91$.

Note: Here you need to know what is the median. Median is the Middle Most value of the data and it separates the higher half of the data set from the lower half of the data set. To find the median you need to arrange the data in an ascending order or descending order.