Question

Find the mode of the following distribution:

Answer 1

Hint: In this example, first we will convert the given class into a continuous class interval by subtracting $0.5$ from the lower limit and by adding $0.5$to the upper limit. Then, we will write modal class. The modal class is the class in which the frequency (number of workers) is highest. Then, we will use the formula
$Z = l + \left[ {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right] \times h$ to find mode $\left( Z \right)$ of the given distribution.

Complete step-by-step answer:
Let us convert a given class into continuous class intervals by subtracting $0.5$ from the lower limit and by adding $0.5$to the upper limit. Therefore, we have the following distribution.

From the table, we can say that $42.5 - 48.5$is modal class because the highest frequency is $20$given for this class.
Now we have
$l = $ Lower limit of modal class$ = 42.5$
${f_1} = $ Frequency of modal class$ = 20$
${f_0} = $ Frequency of the class just before the modal class $ = 12$
${f_2} = $ Frequency of the class just after the modal class $ = 15$
$h = $ Length of modal class $ = 48.5 - 42.5 = 6$
Now we are going to find the value of mode by using the formula $Z = l + \left[ {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right] \times h$.
Therefore, $Z = 42.5 + \left[ {\dfrac{{20 - 12}}{{2\left( {20} \right) - 12 - 15}}} \right] \times 6$
$ \Rightarrow Z = 42.5 + \left[ {\dfrac{{8 \times 6}}{{40 - 27}}} \right]$
$ \Rightarrow Z = 42.5 + \left[ {\dfrac{{48}}{{13}}} \right]$
$ \Rightarrow Z = 42.5 + 3.69$
$ \Rightarrow Z = 46.19 \approx 46.2$
Hence, mode is $46.2$ for the given distribution.


Note: If the highest frequency occurs in the first class of the given distribution then we cannot find ${f_0}$. If the highest frequency occurs in the last class of the given distribution then we cannot find ${f_2}$. In this case, first we will find mean and median of given distribution. Then, we will use the relation between mode, median and mean to find the value of mode. The relation is $Z = 3M - 2\bar X$ where $Z$ is the mode, $M$ is the median and $\bar X$ is the mean.