Hint: Ungrouped and grouped data are two terms that are frequently used to describe data. Data that is presented as individual data points is referred to as ungrouped data. Grouped data is data given in intervals whereas Ungrouped data without a frequency distribution. The most frequent value of a variable is known as its mode. It is the variable's value that corresponds to the distribution's highest frequency.Complete step-by-step answer:Let us write the formula of mode for grouped data.Mode for grouped data is given as , where is the lower limit of modal class,  is the size of class interval, is the frequency of the modal class,  is the frequency of the class preceding the modal class, and  is the frequency of the class succeeding the modal class.The modal class is the class with the highest frequency.Additional information:Example of finding Mode for grouped data: Marks obtained by 50 students of a class are formulated below. The highest mark is 30. We have to find the mode.Marks obtainedNumber of students0-101810-202420-308We know that mode is given byWe have to consider the maximum class frequency, that is, the highest number of students. Here, this is 24. Hence, modal class is 10-20.Now, we can find the lower limit, .We know that the size of class interval, Frequency of modal class, We know that is the frequency of the class preceding the modal class. Hence, Let us find . It is the frequency of the class succeeding the modal class. Hence,Now let us substitute these values in the formula (i). We will getWhen we solve this, we will getNote: > The advantage of Mode over Mean and Median is that it can be applied to any type of dataset, whereas Mean and Median cannot be used with nominal data.  It is also not affected by outliers. > Its drawback is that it can't be used for in-depth analysis.

Hint: To solve this question, we will first make 3 bar graphs of frequencies \\[{{f}_{1}},{{f}_{2}},{{f}_{3}}.\\] Mode is the value of the highest bar as that is of the maximum frequency. Finally, we will calculate the midpoint of the largest bar to get the value of the mode formula. Complete step-by-step answer:Let us first define the mode for grouped data. The mode of a list of data values is simply the most common values (or the values if any). When the data are grouped as in a histogram, we will normally talk only about the modal class (the class, or group with the greatest frequency) because we don’t know the individual values. The derivation of the mode formula is given by using the bar graph.\n \n \n \n \n Let the frequency of the modal class be \\[{{f}_{1}}.\\] The frequency of the class first after the modal class is \\[{{f}_{2}}.\\] From the above figure, we see that, triangle AEB is similar to triangle DEC. \\[\\Rightarrow \\Delta AEB\\sim \\Delta DEC\\]The relative side ratio is also equal. \\[\\Rightarrow \\Delta AEB\\sim \\Delta DEC\\]\\[\\Rightarrow \\dfrac{AB}{CD}=\\dfrac{BE}{DE}\\]And BE is nothing but \\[{{f}_{1}}-{{f}_{0}}\\] and \\[DE={{f}_{1}}-{{f}_{2}}.\\]\\[\\Rightarrow \\dfrac{AB}{CD}=\\dfrac{BE}{DE}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{{{f}_{1}}-{{f}_{2}}}\\]\\[\\Rightarrow \\dfrac{AB}{CD}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{{{f}_{1}}-{{f}_{2}}}\\]Again we have \\[\\Delta BEF\\sim \\Delta BDC\\] from the figure.\\[\\Rightarrow \\dfrac{FE}{BC}=\\dfrac{BE}{BD}\\]Clearly, \\[BE={{f}_{1}}-{{f}_{0}}\\] and \\[BD=BE+ED\\]\\[\\Rightarrow BD=\\left( {{f}_{1}}-{{f}_{0}} \\right)+\\left( {{f}_{1}}-{{f}_{2}} \\right)\\]\\[\\Rightarrow BD={{f}_{1}}-{{f}_{0}}+{{f}_{1}}-{{f}_{2}}\\]\\[\\Rightarrow BD=2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}\\]Therefore, we have, \\[\\dfrac{FE}{BC}=\\dfrac{BE}{BD}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\]\\[\\Rightarrow \\dfrac{FE}{BC}=\\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\]\\[\\Rightarrow FE=\\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\times BC\\]We know that \\[BC={{f}_{1}},\\] so we can write\\[\\Rightarrow FE=\\left( \\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \\right)\\times {{f}_{1}}\\]Let, FE be x.\\[\\Rightarrow x=\\left( \\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \\right)\\times {{f}_{1}}\\]Therefore, the mode can be obtained by adding this value of x to \\[{{I}_{0}}.\\]\\[\\Rightarrow \\text{Mode}={{I}_{0}}+x\\]Substituting the value of x as obtained from above, we get, \\[\\Rightarrow \\text{Mode}={{I}_{0}}+x\\]\\[\\Rightarrow \\text{Mode}={{I}_{0}}+\\left( \\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \\right)\\times {{f}_{1}}\\]\\[\\Rightarrow \\text{Mode}={{I}_{0}}+\\dfrac{\\left( {{f}_{1}}-{{f}_{0}} \\right)}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\times {{f}_{1}}\\]Hence, the mode formula is determined. \\[\\Rightarrow \\text{Mode}={{I}_{0}}+\\dfrac{\\left( {{f}_{1}}-{{f}_{0}} \\right)}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\\times {{f}_{1}}\\]Note: We have used the bar graph to determine the mode formula. So, \\[{{f}_{0}}\\] is considered a point after the first bar and the midpoint of the highest bar is the mode. The highest bar is in the middle. So, we have assumed x = midpoint of the largest bar and hence calculate \\[{{I}_{0}}+x\\] to get the mode value.