Question

A survey regarding the height (in cm) of 51 girls of Class X of a school was conducted and following data were obtained. Find the median height.

Height in cm	Number of girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

Answer 1

Hint: We will be using the formula $\text{Median}=l+\left( \dfrac{\dfrac{n}{2}-cf}{f} \right)\times h$ , where l is the lower limit of median class, cf is the cumulative frequency of the class preceding the median class, n is the no. of observations, f is frequency of median class and h is the class size. By finding these terms and substituting the values in the formula, the median height can be obtained.

Complete step-by-step answer:
We need to find the median height of girls from the given data. In the given data, we can see that each data is written less than some number. So let us convert this into proper class. Hence, the data can be written as

Height in cm	Number of girls
0- 140	4
140- 145	\[11-4=7\]
145- 150	\[29-11=18\]
150- 155	\[40-29=11\]
155- 160	\[46-40=6\]
160- 165	\[51-46=5\]

Now, let us calculate the median. We will be using the formula
$\text{Median}=l+\left( \dfrac{\dfrac{n}{2}-cf}{f} \right)\times h...(i)$
where l = lower limit of median class
cf = cumulative frequency of the class preceding the median class
n = no. of observations
f = frequency of median class
h = class size
We must draw a table in the following manner to find f and cf . cf can be found by adding the frequency in each step. We should also find the sum of frequency, that is, $\sum{{{f}_{i}}}$ and this will be the same as n.

Height in cm	Number of girls$\left( {{f}_{i}} \right)$	Cumulative frequency (cf)
0- 140	4	4
140- 145	7	\[4+7=11\]
145- 150	18	\[11+18=29\]
150- 155	11	\[29+11=40\]
155- 160	6	\[40+6=46\]
160- 165	5	\[46+5=51\]
	$\sum{{{f}_{i}}}=51$

We know that $n=\sum{{{f}_{i}}}=51$ .
$\Rightarrow \dfrac{n}{2}=\dfrac{51}{2}=25.5$
Now, we must select the median class. For this, we must check the cf from the table that is close to $\dfrac{n}{2}$ , that is, the cf which is nearest to 25.5.
From the above table, we can see that the cf close to 25.5 is 29. Hence, we will be taking the class 145-150 as the median class.
Hence, the lower limit of this class, $l=145$
And frequency, $f=18$
Let us find the class interval.
$h=150-145=5$
We know that cf is the cumulative frequency of the class preceding the median class. Hence,
$cf=11$
Now, let us substitute these values in the formula (i). Hence,
$\text{Median}=145+\left( \dfrac{25.5-11}{18} \right)\times 5$
Let us simplify this. We will get
$\begin{align}
  & \text{Median}=145+\left( \dfrac{14.5}{18} \right)\times 5 \\
 & =145+4.02 \\
 & =149.02 \\
\end{align}$
Hence, the median height is 149.02.

Note: When we write the given data in the form of class, we do not take the found value of previous class to subtract. Instead we take the original given value. For example, in the class 145- 150 , we did 40-29=11 not 40-18. But this is not the case when we find cf for each class. We subtracted value from the previous cf that we found. Be careful with the formula $\text{Median}=l+\left( \dfrac{\dfrac{n}{2}-cf}{f} \right)\times h$ . You may write it as $\text{Median}=l+\left( \dfrac{\dfrac{n}{2}-f}{cf} \right)\times h$ leading to the wrong solution.