Question

Find the mode of the following frequency distribution.

Class	Frequency
0-10	8
10-20	10
20-30	10
30-40	16
40-50	12
50-60	6
60-70	7

Answer 1

Hint: As we know mode is the one which has the highest frequency and we will find the class which has the highest frequency which is called modal class and this modal class will be having a range. Later on, we get the lower boundary of the modal class and highest of each class. Now, modal frequency is the frequency of the class having the highest frequency. After that we need the frequency of class proceeding to modal class and frequency of class succeeding to modal class.
After that we will use a formula to find mode that is \[l + h\left( {\dfrac{{{f_m} - {f_1}}}{{2{f_m} - {f_1} - {f_2}}}} \right)\] and we can now substitute all the value in the given formula to obtain the answer.

Complete step-by-step answer:
From the table given in the question,
The class 30-40 has the highest frequency 16
Here, the modal class is 30-40
Lower boundary of modal class is $l$= 30
Height of each class is $h$ = 10-0 = 10
Modal frequency is ${f_m}$ = 16
Now, Frequency of the class proceeding to modal class is ${f_1}$ = 10
Frequency of class succeeding to modal class if ${f_2}$ = 12
So, mode is calculated by
  \[m = l + h\left( {\dfrac{{{f_m} - {f_1}}}{{2{f_m} - {f_1} - {f_2}}}} \right)\]
Now, substituting $l$= 30, $h$=10, ${f_m}$ = 16, ${f_1}$ = 10 and ${f_2}$ = 12 in the above equation,
We get,
$
 \Rightarrow m = 30 + 10 \times \dfrac{{16 - 10}}{{32 - 10 - 12}} \\
  \Rightarrow m = 30 + 10 \times \dfrac{6}{{10}} \\
 \Rightarrow m = 30 + 6 \\
 \Rightarrow m = 36 \\
 $
Hence, the mode is 36.
Hence, the answer is 36.

Note: As we know, high attention is needed to do these types of questions because we have to take a lot of values from the table and we will not mismatch the values. It should also be noted that there is a difference between modal frequency and mode.