# Ex.14.2 Q6 Statistics Solution - NCERT Maths Class 10

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## Question

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find mode of the data.

 Number of cars $$0 -10$$ $$10 - 20$$ $$20 - 30$$ $$30 - 40$$ $$40 - 50$$ $$50 - 60$$ $$60 - 70$$ $$70 - 80$$ Frequency $$7$$ $$14$$ $$13$$ $$12$$ $$20$$ $$11$$ $$15$$ $$8$$

## Text Solution

What is known?

The number of cars passing through a spot on a road for $$100$$ periods each of $$3$$ minutes

What is unknown?

The Mode of the data.

Reasoning:

Modal Class is the class with highest frequency.

Mode $$= l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$$

Where,

Class size,$$h$$

Lower limit of modal class,$$l$$

Frequency of modal class,$$f_1$$

Frequency of class preceding modal class,$$f_0$$

Frequency of class succeeding the modal class,$$f_2$$

Steps:

From the table, it can be observed that the maximum class frequency is $$20,$$ belonging to class interval $$40 − 50$$

Therefore, Modal class$$=40 − 50$$

Class size,$$h=10$$

Lower limit of modal class,$$l=40$$

Frequency of modal class,$$f_1=20$$

Frequency of class preceding modal class,$$f_0=12$$

Frequency of class succeeding the modal class,$$f_2=11$$

Mode $$= l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h$$

$\begin{array}{l} = 40 + \left( {\frac{{20 - 12}}{{2 \times 20 - 12 - 11}}} \right) \times 10\\ = 40 + \left( {\frac{8}{{40 - 23}}} \right) \times 10\\ = 40 + \frac{8}{{17}} \times 10\\ = 40 + 4.705\\ = 40.705\\ = 40.7 \end{array}$

Hence the mode is $$40.7$$

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