Question

If the mode is 34.5, then find the missing frequency?

0-10	4
10-20	8
20-30	10
30-40	x
40-50	8

Answer 1

Hint: To solve the question above, we will first find out the modal class or class with highest frequency. Then, we will find the lower limit of modal class. Then we will calculate the frequency of modal class. After calculating all these, we will apply the formula of mode which is shown below:
\[Mode = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\]
Where l= lower limit of modal class, \[{f_1}\] is the frequency of modal class, \[{f_2}\]is the frequency of class preceding modal class, \[{f_0}\]is the frequency of the class succeeding modal class and h is the class size.

Complete step-by-step answer:
In the question, we are given that the mode of a given data is 34.5. With the help of this, we will find out which is modal class. As 34.5 lies between 30 and 40. The class containing (30-40) will be the modal class. The lower limit of the modal class will be the first element of this class. The first element is = 30. So the lower limit of the modal class will be = 30. Here, we can see that the difference between the first element and last element of class = 10. So class size will be = 10. The class preceding modal class is the class containing (20-30) and the class succeeding the modal class is the class containing (40-50). Now, the mode of a grouped data is given by the formula:
\[Mode = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\]
Where, l is the lower limit of modal class and in our case, it is 30, h is the class size and is equal to 10, \[{f_0}\] is the frequency of class preceding modal class and it is equal to 10, \[{f_1}\] is the frequency of modal class and in our case, it is equal to x. The frequency of class succeeding modal class is denoted by \[{f_2}\] and it is = 8. Thus, we will get:
\[\begin{array}{l}34.5 = 30 + \left( {\dfrac{{x - 10}}{{2x - 10 - 8}}} \right) \times 10\\ \Rightarrow 34.5 - 30 = \dfrac{{x - 10}}{{2x - 18}} \times 10\\ \Rightarrow 4.5 = \dfrac{{x - 10}}{{2\left( {x - 9} \right)}} \times 10\\ \Rightarrow \dfrac{{4.5 \times 2}}{{10}} = \dfrac{{x - 10}}{{x - 9}}\\ \Rightarrow \dfrac{9}{{10}} = \dfrac{{x - 10}}{{x - 9}}\\ \Rightarrow 9\left( {x - 9} \right) = 10\left( {x - 10} \right)\\ \Rightarrow 9x - 81 = 10x - 100\\ \Rightarrow 10x - 9x = 100 - 81\\ \Rightarrow x = 19\end{array}\]
Thus, the missing frequency is 19.

Note: The method which we have used for solving the question is valid only when the unknown frequency is frequency \[{f_0},{f_1}{\text{ and }}{{\text{f}}_{\text{2}}}\]. If the frequency which is missing is of some other class than the modal class, preceding class or succeeding class, then we won't be able to calculate the missing frequency.