Ex.14.2 Q6 Statistics Solution - NCERT Maths Class 10
Question
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find mode of the data.
Number of cars | \(0 -10\) | \(10 - 20\) | \(20 - 30\) | \(30 - 40\) | \(40 - 50\) | \(50 - 60\) | \(60 - 70\) | \(70 - 80\) |
Frequency | \(7\) | \(14\) | \(13\) | \(12\) | \(20\) | \(11\) | \(15\) | \(8\) |
Text Solution
What is known?
The number of cars passing through a spot on a road for \(100\) periods each of \(3\) minutes
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 \(20,\) belonging to class interval \(40 − 50\)
Therefore, Modal class\(=40 − 50\)
Class size,\(h=10\)
Lower limit of modal class,\(l=40\)
Frequency of modal class,\(f_1=20\)
Frequency of class preceding modal class,\(f_0=12\)
Frequency of class succeeding the modal class,\(f_2=11\)
Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{array}{l}
= 40 + \left( {\frac{{20 - 12}}{{2 \times 20 - 12 - 11}}} \right) \times 10\\
= 40 + \left( {\frac{8}{{40 - 23}}} \right) \times 10\\
= 40 + \frac{8}{{17}} \times 10\\
= 40 + 4.705\\
= 40.705\\
= 40.7
\end{array}\]
Hence the mode is \(40.7\)