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.