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.
Text Solution
What is known?
The observed lifetimes (in hours) of \(225\) electric components.
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.