Ex.14.2 Q5 Statistics Solution - NCERT Maths Class 10
Question
The given distribution shows the number of runs scored by some top batsman of the world in one- day international cricket matches.
Runs scored |
Number of batsmen |
\(3000 – 4000\) \(4000 – 5000\) \(5000 – 6000\) \(6000 – 7000\) \(7000 – 8000\) \(8000 – 9000\) \(9000 – 10000\) \(10000 –11000\) |
\(4\) \(18\) \(9\) \(7\) \(6\) \(3\) \(1\) \(1\) |
Find the mode of the data.
Text Solution
What is known?
The number of runs scored by some top batsman of the world in one- day international cricket matches.
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 \(18,\) belonging to class interval \(4000 − 5000\)
Therefore, Modal class\(=4000 − 5000\)
Class size,\(h=1000\)
Lower limit of modal class,\(l=4000\)
Frequency of modal class,\(f_1=18\)
Frequency of class preceding modal class,\(f_0=4\)
Frequency of class succeeding the modal class,\(f_2=9\)
Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{align} &= 4000\!+\!\!\left( {\frac{{18 - 4}}{{2 \times 18 - 4 - 9}}} \right)\!\!\times\!\!1000\\ &= 4000 + \left( {\frac{{14}}{{36 - 13}}} \right) \times 1000\\ &= 4000 + \frac{{14}}{{23}} \times 1000\\& = 4000 + 608.695\\ &= 4608.695\\& = 4608.7\end{align}\]
Hence the mode is \(4608.7\)