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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