Question

For the following frequency distribution, find:

(i) The median (ii) lower quartile (iii) upper quartile

Variate	$25$	$31$	$34$	$40$	$45$	$48$	$50$	$60$
Frequency	$3$	$8$	$10$	$15$	$10$	$9$	$6$	$2$

Answer 1

Step 1: Make the required cumulative frequency distribution table.

First, we have to prepare a table to find the value of cumulative frequency.

Let, the frequency be $\left(f\right)$

The cumulative frequency is obtained by adding the frequency to the cumulative frequency of the predecessors.

Variate	Frequency $\left(f\right)$	Cumulative frequency
$25$	$3$	$3$
$31$	$8$	$3+8=11$
$34$	$10$	$11+10=21$
$40$	$15$	$21+15=36$
$45$	$10$	$36+10=46$
$48$	$9$	$45+9=55$
$50$	$6$	$55+6=61$
$60$	$2$	$61+2=63$

Step 2: Calculate the Median.

From the table,

Here, the number of observation $n=63$ which is odd.

Then, Median $={\left(\frac{n+1}{2}\right)}^{th}term$

$$\begin{gathered} \Rightarrow {\left(\frac{63+1}{2}\right)}^{th}term \\ \Rightarrow {\left(\frac{64}{2}\right)}^{th}term \\ \Rightarrow {32}^{th}term \\ \Rightarrow 40 \end{gathered}$$

Hence the median is $40$

Step 3: Calculate the Lower quartile (part (ii)).

Lower quartile $Q_1={\left(\frac{n+1}{4}\right)}^{th}term$

$$\begin{gathered} Q_1={\left(\frac{63+1}{4}\right)}^{th}term \\ {Q}_1={\left(\frac{64}{4}\right)}^{th}term \\ {Q}_1={\left(16\right)}^{th}term \end{gathered}$$

Therefore, from the table $Q_1=34$.

Step 4: Calculate the Upper quartile (part (iii)).

Upper Quartile $Q_3=3{\left(\frac{n+1}{4}\right)}^{th}term$

$$\begin{gathered} Q_3=3{\left(\frac{63+1}{4}\right)}^{th}term \\ {Q}_3=3{\left(\frac{64}{4}\right)}^{th}term \\ {Q}_3=3{\left(\frac{16}{4}\right)}^{th}term \\ {Q}_3={(3\times 16)}^{th}term \\ {Q}_3={48}^{th}term \end{gathered}$$

Therefore, from the table $Q_3=48$.

Hence the Arithmetic Mean of the given distribution is $40$, the lower quartile is $34$ and the upper quartile is $48$.