Question

The mode of the following data is $85.7$. Find the missing frequency in it.

Size	Frequency
45-55	7
55-65	12
65-75	17
75-85	$f$
85-95	32
95-105	6
105-115	10

A) 33
B) 31
C) 30
D) 32

Answer 1

Here we will use the concept and formula of the mode to find the value of the missing frequency. First, we will select the modal class from the given data. Then we will apply the formula of the mode and equate it to the given value. Then we will solve it to get the value of the missing frequency.

Formula used:
Mode $= L + h\left( {\dfrac{{{f_m} - {f_1}}}{{2{f_m} - {f_1} - {f_2}}}} \right)$, where $L$ is the lower limit of the modal class, $h$ is the size of the class interval, ${f_m}$ is the frequency of the model class, ${f_1}$ is the frequency of the class preceding the modal class and ${f_2}$ is the frequency of class succeeding the modal class.

Complete Step by Step Solution:
It is given that the mode of the data is $85.7$.
Now we will select the modal class from the given data in the question. Modal class is the class with the maximum frequency in the data. Therefore the modal class is 85-95.
Now we will use the formula of the mode to find its value. Therefore, we get
Lower limit of the modal class is 85. Therefore, $L = 85$
Size of the class interval in the given data is 10. Therefore, $h = 10$
Frequency of the modal class is 32. Therefore, ${f_m} = 32$
Frequency of the class preceding the modal class i.e. class 75-85 is $f$. Therefore, ${f_1} = f$
Frequency of the class succeeding the modal class i.e. class 95-105 is 6. Therefore, ${f_2} = 6$
Now we have to put all the values in the formula of mode to find its value, we get
Mode $= 85 + 10\left( {\dfrac{{32 - f}}{{2\left( {32} \right) - f - 6}}} \right)$
$\Rightarrow$ Mode $= 85 + \left( {\dfrac{{320 - 10f}}{{58 - f}}} \right)$
It is given that the value of the mode of the data given is equal to $85.7$. So we will put the value of mode in the above equation, we get
$\Rightarrow 85.7 = 85 + \left( {\dfrac{{320 - 10f}}{{58 - f}}} \right)$
Now we will solve this equation to get the value of the required missing frequency in the data. Therefore, we get
$\Rightarrow 85.7 - 85 = \left( {\dfrac{{320 - 10f}}{{58 - f}}} \right)$
$\Rightarrow 0.7\left( {58 - f} \right) = \left( {320 - 10f} \right)$
$\Rightarrow 40.6 - 0.7f = 320 - 10f$
Now we take the terms with $f$ on the LHS of the equation and constants on the RHS of the equation, we get
$\Rightarrow 10f - 0.7f = 320 - 40.6$
Subtracting the like terms, we get
$\Rightarrow 9.3f = 279.4$
Dividing both sides by $9.3$, we get
$\Rightarrow f = \dfrac{{279.4}}{{9.3}} = 30$
Hence, the missing frequency from the data is equal to 30.

So, option C is the correct option.

Note:
Here in this type of question we should remember the formula of the mode of the data with the frequencies. Here, we should not get confused between mean, median and mode. Median is the middle value of the given list of numbers or it is the value that is separating the data into two halves i.e. upper half and lower half. Mode is the number or value which occurs a maximum number of times in a given set of numbers. Mean is the average of the given data.