Ex.14.2 Q5 Statistics Solution - NCERT Maths Class 10

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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.

What is unknown?

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

\[\begin{array}{l}
 = 4000 + \left( {\frac{{18 - 4}}{{2 \times 18 - 4 - 9}}} \right) \times 1000\\
 = 4000 + \left( {\frac{{14}}{{36 - 13}}} \right) \times 1000\\
 = 4000 + \frac{{14}}{{23}} \times 1000\\
 = 4000 + 608.695\\
 = 4608.695\\
 = 4608.7
\end{array}\]

Hence the mode is \(4608.7\)

  
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