Question

The age wise participation of students in the Annual Function of a school is shown in the following distribution.
$\begin{array}{cccccccc}\text{Age (in years)}& 5-7& 7-9& 9-11& 11-13& 13-15& 15-17& 17-19\\ \text{Number of}& & & & & & & \\ \text{students}& x& 15& 18& 30& 50& 48& x\end{array}$

Find the missing frequencies when the sum of frequencies is $181$. Also, find the mode of the data.

Answer 1

Given that the sum of frequencies is $181$.
$x+15+18+30+50+48+x=181$
$\Rightarrow 2x+161=181$
$\Rightarrow 2x=20$
$\therefore x=10$
So, the missing frequency is $10$
Here the maximum class frequency is 50 and the class corresponding to this frequency is $13-15$.
So the model class is $13-15$
lower limit (l) of modal class=13
class size(h)=2
Frequency $f_1$of the modal class$=50$
Frequency $f_0$ of class preceding the modal class=30
Frequency $f_2$ of class succeeding the modal class=48
Now, let us substitute these values in the formula
Mode$=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$
$=13+\left[\frac{50-30}{(2\times 50)-30-48}\right]\times 2$
$=13+\frac{20}{100-78}\times 2$
$=13+\frac{20}{22}\times 2$
$=13+\frac{20}{11}$
$=13+1.82$
Therefore , the mode of the data is 14.82
Modal class is the class with the maximum frequency
The mode is a value inside the modal class.
So, it cannot be less than the lower class limit and can't be greater than upper class.
Hence, mode of the data $14.82\in$ [13, 15] is the correct value.