Question

Find the mode of the following data



C.I :-- 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 ,60-70 ,70-80 ,80-90.


f :-- 5, 3, 4 ,9 ,3, 4 ,7, 3, 7,​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}\cline{1-10}\multicolumn{10}{|c|}{Frequency\:Table}\\ \cline{1-10}Class-Interval & 0-10 & 10-20 & 20-30& 30-40& 40-50& 50-60 & 60-70 & 70-80 & 80-90\\ \cline{1-10}Frequency & 5 & 3 & 4& 9&3 & 4& 7& 3&7\\ \cline{1-10} \end{tabular}$

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The Mode of the following data.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of the mode:

$\boxed{\bf{Mode=L+\bigg\lgroup\dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} }\bigg\rgroup h}}}}}}$

Therefore;

• Lower class limit of the modal class,[L]= 30
• Class-Interval of the modal class,,[h]= 10
• Frequency of the modal class,[f1]= 9
• Frequency of the class preceding the modal class,[f0]= 4
• Frequency of the class succeeding the modal class,[f2]= 3

$\dag$ Substituting the value of formula of the mode:

$\implies\tt{Mode=30+\bigg\lgroup\dfrac{9-4}{2(9)-4-3} \bigg\rgroup \times 10}}\\\\\\\implies\tt{Mode=30+\bigg\lgroup\dfrac{5}{18-7} \bigg\rgroup\times 10}}\\\\\\\implies\tt{Mode=30+\dfrac{5}{11} \times 10}}\\\\\\\implies\tt{Mode=30+\cancel{\dfrac{50}{11}} }\\\\\\\implies\tt{Mode=30+4.54}\\\\\\\implies\tt{\pink{Mode=34.54}}$

Thus;

The mode is 34.54 .

Answer 2

Given:

C.I :-- 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 ,60-70 ,70-80 ,80-90.

f :-- 5, 3, 4 ,9 ,3, 4 ,7, 3, 7,

Solution:

Mode:

✿Mode is defined as the value which occurred more frequently in a series.

Formula:

✿Mode = l+{(f1-f2)/(2f1-f0-f2)}×h

Where,

→ l= lower limit of the modal class

→ h= size of the class interval(assuming all sizes to be equal)

→ f1= frequency of the modal

→f0= frequency of the class preceding the modal class

→f2= frequency of the class succeeding the modal class

✿ here the maximum class frequency is 9 and the class corresponding to this frequency is 30-40.So,the modal class is 30-40.

Now,

→modal class = 30-40, lower limit =30

→class size(h)=10

→ frequency(f1)=9

→f1 = 4

→f2 = 3

Now put the value of this in given formula,

→Mode = 30+[(9-4)/(2(9)-4-3)]×10

→Mode = 30 + [(5)/18-7)]×10

→Mode = 30 + (5/11)×10

→Mode = 30 + 4.54

→Mode = 34.545

Therefore the mode of the above data is 34.545