Ex.14.2 Q2 Statistics Solution - NCERT Maths Class 10


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.

 Video Solution
Statistics
Ex 14.2 | Question 2

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

\[\begin{align}&= 60 + \left( {\frac{{61 - 52}}{{2 \times 61 - 52 - 38}}} \right) \times 20\\&= 60 + \left( {\frac{9}{{122 - 90}}} \right) \times 20\\&= 60 + \frac{9}{{32}} \times 20\\&= 60 + 5.625\\&= 65.625\end{align}\]

Hence,the modal lifetimes of the components are \(65.62\) hours.

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