Question

The maximum bowling speed $\left( {{\text{km/hour}}} \right)$ of $33$ players at a cricket coaching centre is given below. Find the modal bowling speed of players.

Bowling speed(km/hr)	85-100	100-115	115-130	130-145
No. of players	9	11	8	5

${\text{A}}.$ ${\text{Rs}}{\text{.101}}$${\text{km/hour}}$
${\text{B}}.$ ${\text{Rs}}{\text{.106}}$${\text{km/hour}}$
${\text{C}}{\text{.}}$${\text{Rs}}{\text{.115}}$${\text{km/hour}}$
${\text{D}}{\text{.}}$${\text{Rs}}{\text{.118}}$${\text{km/hour}}$

Answer 1

Hint: From the question, we have to choose the correct option by using the mode formula. First, we are going to find the terms used in the formula using the given table. After substitution on the formula and further simplification, we get the required result.

Formula used: The class which has the highest frequency will be the modal class of the distribution. It can be calculated by using the following formula:
\[{\text{Mode}} = {\text{L}} + \left( {\dfrac{{{{\text{f}}_{\text{m}}} - {{\text{f}}_{{\text{m - 1}}}}}}{{2{{\text{f}}_{\text{m}}} - {{\text{f}}_{{\text{m - 1}}}} - {{\text{f}}_{{\text{m + 1}}}}}}} \right) \times {\text{h}}\]
Where, \[{\text{L}} = \] lower boundary of modal class
\[{{\text{f}}_{\text{m}}} = \] Frequency of modal class
 \[{{\text{f}}_{{\text{m - 1}}}} = \] Frequency of pre-modal class
\[{{\text{f}}_{{\text{m + 1}}}} = \] Frequency of post-modal class
${\text{h}} = $ Difference between class intervals

Complete step-by-step solution:
Now, we are going to find the terms used in the mode formula by using the given table and substitute the values on the mode formula. Then, we get the maximum or modal bowling speed, from the given table, the maximum frequency \[{{\text{f}}_{\text{m}}} = 11\].
Therefore, the modal class of maximum frequency $ = 100 - 115$.
Lower limit of the modal class, ${\text{L = 100}}$.
The pre-modal class of maximum frequency $ = 85 - 100$.
The frequency of the pre-modal class, \[{{\text{f}}_{{\text{m - 1}}}} = 9\] .
The post-modal class of maximum frequency $ = 115 - 130$.
The frequency of the post-modal class, \[{{\text{f}}_{{\text{m + 1}}}} = 8\].
Difference between class intervals, ${\text{h}} = 115 - 100 = 15$.
Now, we have to substitute the above values in the mode formula. Then, we get
$\Rightarrow$Mode \[ = {\text{L}} + \left( {\dfrac{{{{\text{f}}_{\text{m}}} - {{\text{f}}_{{\text{m - 1}}}}}}{{2{{\text{f}}_{\text{m}}} - {{\text{f}}_{{\text{m - 1}}}} - {{\text{f}}_{{\text{m + 1}}}}}}} \right) \times {\text{h}}\]
$\Rightarrow$Mode $ = 100 + \left( {\dfrac{{11 - 9}}{{2\left( {11} \right) - 9 - 8}}} \right) \times 15$
On simplifying, we get the required solution.
$\Rightarrow$Mode $ = 100 + \left( {\dfrac{2}{{22 - 17}}} \right) \times 15$
Subtracting the terms we get,
$\Rightarrow$Mode $ = 100 + \left( {\dfrac{2}{5}} \right) \times 15$
Dividing the required terms,
$\Rightarrow$Mode $ = 100 + \left( 2 \right) \times 3$
Multiplying we get,
$\Rightarrow$Mode \[ = 100 + 6\]
Adding the terms,
$\Rightarrow$Mode $ = 106$.
Therefore, the modal bowling speed is $106$${\text{km/hour}}$.

$\therefore $ The correct option is ${\text{B}}$.

Note: In statistics, the mode is the most commonly observed value in a set of data. For the normal distribution, the mode is also the same value as the mean and median. In many cases, the modal value will differ from the average value in the data. The mode can be computed in an open ended frequency table.
The mode is easy to understand and calculate. The mode is easy to identify in a data set and in a discrete frequency distribution. Mode is most useful as a measure of central tendency when examining categorical data, such as the model of cars or flavours of soda, for which a mathematical average median value based on ordering cannot be calculated.