Question

Grade 10, Statistics.

Find the mode of the following data:

Class
20 - 40
40 - 60
60 - 80
80 - 100
100 - 120
120 - 140

Frequency
8
14
22
16
15
13

Answer 1

Answer :-

Mode of the given data is 71.42

Explanation :-

$\begin{tabular}{| c | l |} \cline{1-2} Class & Frequency \\ \cline{1-2} 20-40 & 8 \\ \cline{1-2} 40-60 & 14 \\ \cline{1-2} 60-80 & 22 \\ \cline{1-2} 80-100 & 16 \\ \cline{1-2} 100-120 & 15 \\ \cline{1-2} 120-140 & 13 \\ \cline{1-2} \end{tabular}$

Class 60 - 80 has the highest frequency among all.

So Modal class = is 60 - 80

From table,

Frequency of the modal class (f1) =22

Frequency of the class preceeding modal class (f0) = 14

Frequency of the class succeeding modal class (f2) = 16

Lower boundary of the modal class (l) = 60

Class size (h) = Difference of lower boundaries = 40 - 20 = 20

$\sf Mode = l + \bigg( \dfrac{f_1 - f_0 }{2f_1 - f_0 - f_2 } \bigg) \times h \\ \\ \\ \sf = 60 + \bigg( \dfrac{22 - 14}{2(22) - 14 - 16} \bigg) \times 20 \\ \\ \\ \sf = 60 + \bigg( \dfrac{8}{44 - 30} \bigg) \times 20 \\ \\ \\ \sf = 60 + \dfrac{160}{14} \\ \\ \\ \sf = 60 + 11.42 \\ \\ \\ \sf = 71.42$

Mode = 71.42

Answer 2

Answer :

The Mode of the given data is 71.42.

Step-by-step explanation :

Given :

$\begin{tabular}{|c|c|}\cline{1-2}Class & Frequency \\\cline{1-2}\ 20\;-\;40 &8 \\\cline{1-2}\ 40\;-\;60 & 14 \\\cline{1-2}\ 60\;-\;80 & 22 \\\cline{1-2}\ 80\;-\;100 & 16 \\\cline{1-2}\ 100\;-\;120 &15 \\\cline{1-2}\ 120\;-\;140 &13 \\\cline{1-2}\end{tabular}$

To Find :

The mode.

Solution :

First of all Check that which is highest frequency among all.

If you see the table, We will found that :

The Class 60 - 80 has the highest frequency.

So, It implies - The Modal class - 60 - 80

Now,

Frequency of the class succeeding modal class (f₂) = 16

Frequency of the class preceeding modal class (f₀) = 14

Lower boundary of the modal class (l) = 60

Difference of lower boundaries = 40 - 20 = 20

____________{ Given in Table! }

We know that :

$\bigstar\; \boxed{\sf Mode = l+\bigg( \dfrac{f_1 - f_0 }{2f_1 - f_0 - f_2} \bigg) \times h}}$

Put the values,

$\implies \sf 60 +\bigg(\dfrac{22-14}{2(22) - 14 - 16}\bigg) \times 20 \\\\\implies \sf 60 + \bigg( \dfrac{8}{44 - 30}\bigg) \times 20 \\ \\\implies \sf 60+\dfrac{160}{14}\\ \\\implies \sf 60 + 11.42 \\ \\\implies\sf 71.42$