Question

Find the mode and mean deviation w.r.t mode for the following table

Marks	\[0 - 10\]	\[10 - 20\]	\[20 - 30\]	\[30 - 40\]	\[40 - 50\]
No. of Students	\[5\]	\[12\]	\[30\]	\[10\]	\[3\]

Answer 1

Hint: We are required to measure mean deviation, which indicates how far the values are from the middle value.
Since we have grouped data, we must calculate the class midpoints to obtain the term\[{f_i}{d_i}\]
The mean deviation will be calculated using the mode formula.
 Formula used:
$Mode = l + \;\left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - \;\;{f_2}}}} \right)\; \times \;h$
$Mean\;Deviation\;about\;Mode\; = \;\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}$

 Complete answer:
We'll start by calculating mode.
The formula for determining Mode is:
$Mode = l + \;\left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - \;\;{f_2}}}} \right)\; \times \;h$
where \[l = \]the modal class's the lower boundary
$f = $Frequency of modal class
\[\;{f_1} = \]Frequency of class preceding the modal class
\[{f_2} = \]Frequency of class succeeding the modal class
$h = $size of the class interval
Let's tabulate the data and then go into calculation mode.
For the first-class interval,$0 - 10$ we will measure the following terms.
$Class\;midpoints\; = \;\dfrac{{10\; + \;0}}{2}\; = \;5$
We now know that the modal class is the one with the highest frequency.
The highest frequency is $30$, which corresponds to the$20 - 30$class.
As a result, the Modal class ranges from$20 - 30$.
We'll now get the values for \[l,f,{f_1},{f_2}\] and $h$ as follows:
$l = 20,h = 10,f = 30,{f_1} = 12,$ and ${f_2} = 10$
To calculate mode, we will substitute these values.
$Mode = l + \;\left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - \;\;{f_2}}}} \right)\; \times \;h$
$= 20 + \;\left( {\dfrac{{30\; - \;12}}{{2 \times 30\; - \;12\; - \;10}}} \right)\; \times \;10 $
$= 20 + \;\left( {\dfrac{{18}}{{38}}} \right)\; \times \;10 $
$= \;20 + \;4.7 $
$= \;24.7 $
We will measure $\left| {{x_i}\; - \;M} \right|$ in the table using this value of Mode ($M$).
Let us now measure the\[{f_i}{d_i}\]deviation for the first-class interval of $0 - 10$
We'll do the same thing with the other class intervals and tabulate the results as follows:

Class Interval (\[C.I.\])	Frequency ($f$)	Class midpoints (${x_i}$)	Deviation $\left| {{d_i}} \right|\; = \;\left| {{x_i}\; - \;M} \right|\; = \;\left| {{x_i}\; - \;24.7} \right|$	\[{f_i}{d_i}\]
\[0 - 10\]	\[5\]	\[5\]	\[19.7\]	\[98.5\]
\[10 - 20\]	\[12\]	\[15\]	\[9.7\]	\[116.4\]
\[20 - 30\]	\[30\]	\[25\]	\[0.3\]	\[9\]
\[0 - 40\]	\[10\]	\[35\]	\[10.3\]	\[103\]
\[40 - 50\]	\[3\]	\[45\]	\[20.3\]	\[60.9\]
	$\sum {{f_i}} \; = \;60$			$\sum {{f_i}{d_i}} \; = \;387.8$

Let's evaluate the mean deviation about mode now.
$Mean\;Deviation\;about\;Mode\; = \;\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}\; = \;\dfrac{{387.8}}{{60}}\; = \;6.46$
Therefore, we get the$Mode = 24.7$and the\[Mean{\text{ }}Deviation{\text{ }}about{\text{ }}Mode\]$ = 6.46$

Note:
We just make use of the mean, median, and mode; where mean is the average that will be used of the standard deviation also.
Median is all about the midpoint in the distribution.
And we calculated midpoints deviation and we will build three columns. It's important to keep in mind to multiply the proper columns and terms.