# Ex.14.2 Q5 Statistics Solution - NCERT Maths Class 10

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## Question

The given distribution shows the number of runs scored by some top batsman of the world in one- day international cricket matches.

 Runs scored Number of batsmen $$3000 – 4000$$ $$4000 – 5000$$ $$5000 – 6000$$ $$6000 – 7000$$ $$7000 – 8000$$ $$8000 – 9000$$ $$9000 – 10000$$ $$10000 –11000$$ $$4$$ $$18$$ $$9$$ $$7$$ $$6$$ $$3$$ $$1$$ $$1$$

Find the mode of the data.

## Text Solution

What is known?

The number of runs scored by some top batsman of the world in one- day international cricket matches.

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 $$18,$$ belonging to class interval $$4000 − 5000$$

Therefore, Modal class$$=4000 − 5000$$

Class size,$$h=1000$$

Lower limit of modal class,$$l=4000$$

Frequency of modal class,$$f_1=18$$

Frequency of class preceding modal class,$$f_0=4$$

Frequency of class succeeding the modal class,$$f_2=9$$

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

$\begin{array}{l} = 4000 + \left( {\frac{{18 - 4}}{{2 \times 18 - 4 - 9}}} \right) \times 1000\\ = 4000 + \left( {\frac{{14}}{{36 - 13}}} \right) \times 1000\\ = 4000 + \frac{{14}}{{23}} \times 1000\\ = 4000 + 608.695\\ = 4608.695\\ = 4608.7 \end{array}$

Hence the mode is $$4608.7$$

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