# Ex.14.2 Q2 Statistics Solution - NCERT Maths Class 10

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## Question

The following data gives information on the observed lifetimes (in hours) of $$225$$ electric components

 Lifetime (in hours) $$0 - 20$$ $$20 - 40$$ $$40 - 60$$ $$60 - 80$$ $$80 -100$$ $$100 -120$$ Frequency $$10$$ $$35$$ $$52$$ $$61$$ $$38$$ $$29$$

Determine the modal lifetimes of the components.

## Text Solution

What is known?

The observed lifetimes (in hours) of $$225$$ electric components.

What is unknown?

The modal lifetimes of the components.

Reasoning:

Modal Class is the class with highest frequency.

\begin{align} {\bf{Mode}} = l + \left( \frac{{f_1} - {f_0}}{2{f_1} - {f_0} - {f_2}} \right) \times h\end{align}

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:

 Lifetime (in hours) Frequency $$0 – 20$$ $$20 – 40$$ $$40 – 60$$ $$60 – 80$$ $$80 – 100$$ $$100 – 120$$ $$10$$ $$35$$ $$52$$ $$61$$ $$38$$ $$29$$

From the table, it can be observed that the maximum class frequency is $$61,$$ belonging to class interval $$60 − 80$$

Therefore, Modal class$$=60 − 80$$

Class size,$$h=20$$

Lower limit of modal class,$$l=60$$

Frequency of modal class,$$f_1=61$$

Frequency of class preceding modal class,$$f_0=52$$

Frequency of class succeeding the modal class,$$f_2=38$$

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

$\begin{array}{l} = 60 + \left( {\frac{{61 - 52}}{{2 \times 61 - 52 - 38}}} \right) \times 20\\ = 60 + \left( {\frac{9}{{122 - 90}}} \right) \times 20\\ = 60 + \frac{9}{{32}} \times 20\\ = 60 + 5.625\\ = 65.625 \end{array}$

Hence,the modal lifetimes of the components are $$65.62$$ hours.

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