# How to find mode of the following frequency distribution?

## Class $\textcolor{w h i t e}{X X X X X X X}$Frequency$\textcolor{w h i t e}{X X X}$ $0 - 10 \textcolor{w h i t e}{X X X X X X X X} 22$ $10 - 20 \textcolor{w h i t e}{X X X X X X \times} 24$ $20 - 30 \textcolor{w h i t e}{X X X X X X \times} 36$ $30 - 40 \textcolor{w h i t e}{X X X X X X \times} 14$ $40 - 50 \textcolor{w h i t e}{X X X X X X \times} 12$

##### 1 Answer
Nov 24, 2016

Mode is $23.53$

#### Explanation:

Let us first work out the cumulative frequency, which is as given below:

Class $\textcolor{w h i t e}{X X X X X X X}$Frequency$\textcolor{w h i t e}{X X X}$Cum. Frequency

$0 - 10 \textcolor{w h i t e}{X X X X X X X X} 22 \textcolor{w h i t e}{X X X X X X X X X X} 22$
$10 - 20 \textcolor{w h i t e}{X X X X X X \times} 24 \textcolor{w h i t e}{X X X X X X X X X X} 46$
$20 - 30 \textcolor{w h i t e}{X X X X X X \times} 36 \textcolor{w h i t e}{X X X X X X X X X X} 82$
$30 - 40 \textcolor{w h i t e}{X X X X X X \times} 14 \textcolor{w h i t e}{X X X X X X X X X X} 96$
$40 - 50 \textcolor{w h i t e}{X X X X X X \times} 12 \textcolor{w h i t e}{X X X X X X X X X X} 108$

As the highest frequency is ${f}_{m} = 36$, whose modal class is $20 - 30$ and class interval is $10$; the frequency just before is ${f}_{m - 1} = 24$ and frequency just after is ${f}_{m + 1} = 14$.

Formula for Mode is $M o \mathrm{de} = {L}_{1} + \frac{{f}_{m} - {f}_{m - 1}}{2 {f}_{m} - {f}_{m - 1} - {f}_{m - 2}} \times i$

$. : M o \mathrm{de} = 20 + \frac{36 - 24}{2 \times 36 - 24 - 14} \times 10$

= $20 + \frac{12}{34} \times 10 = 20 + 3.53 = 23.53$