Question

Find the missing frequencies ${f_1}$ and ${f_2}$, if the mean of the frequency table is 62.8 and the sum of all frequencies is 50.

Class	Frequency
0-20	5
20-40	${f_1}$
40-60	10
60-80	${f_2}$
80-100	7
100-120	8
Total	50

A. ${f_1} = 12,{f_2} = 8$
B. ${f_1} = 18,{f_2} = 2$
C. ${f_1} = 6,{f_2} = 6$
D. ${f_1} = 8,{f_2} = 12$

Answer 1

Hint: Given the set of observations where it is a distribution, given the class intervals in the class column in the table and how often they are occurring are given in the frequency column of the table. Here, given the class intervals which have a particular range, it has an upper limit as well as a lower limit. The difference between the upper limit and the lower limit gives the class width. Here we have to find the class mark for each class interval.

Complete step-by-step solution:
We can observe in each class interval that the width of the class interval is 20.
As the class width is given by : upper limit – lower limit
Now the class mark is given by: $\dfrac{\text{upper limit + lower limit}}{{2}}$
The class marks of each class is denoted by ${x_i}$,
Hence calculating the class mark of each class:
Class mark of class 0-20 is given by :
$ \Rightarrow \dfrac{{0 + 20}}{2} = 10$
$ \Rightarrow $Class mark of 0-20 is 10.
$ \Rightarrow $Class mark of 20-40 is 30.
$ \Rightarrow $Class mark of 40-60 is 50.
$ \Rightarrow $Class mark of 60-80 is 70.
$ \Rightarrow $Class mark of 80-100 is 90.
$ \Rightarrow $Class mark of 100-120 is 110.
Now we have to find out the values of \[{f_1}\] and \[{f_2}\].
Now given that the sum of all the frequencies is 50, which is given below:
$ \Rightarrow \sum {{f_i} = 50} $
$ \Rightarrow 5 + {f_1} + 10 + {f_2} + 7 + 8 = 50$
\[ \Rightarrow {f_1} + {f_2} = 20\]
We got an equation which is the sum of \[{f_1}\] and \[{f_2}\]is 20.
Also given the mean of the frequency table is 62.8, which is given below:
$ \Rightarrow \dfrac{{{{\sum {{f_i}x} }_i}}}{{\sum {{f_i}} }} = 62.8$
$\because \sum {{f_i} = 50} $
$\therefore \dfrac{{\sum {{f_i}{x_i}} }}{{50}} = 62.8$
$ \Rightarrow \dfrac{{5(10) + {f_1}(30) + 10(50) + {f_2}(70) + 7(90) + 8(110)}}{{50}} = 62.8$
$ \Rightarrow 30{f_1} + 70{f_2} + 2060 = 50(62.8)$
$ \Rightarrow 30{f_1} + 70{f_2} = 1080$
Hence we have 2 variables and 2 equations which are :
\[ \Rightarrow {f_1} + {f_2} = 20\], which we will consider as equation 1.
$ \Rightarrow 30{f_1} + 70{f_2} = 1080$, which is equation 2.
To solve \[{f_1}\] and \[{f_2}\], multiplying the equation 1 with 30, and subtracting it from equation 2, as given below :
$ \Rightarrow 30({f_1} + {f_2} = 20)$, when multiplied with 30 gives:
$ \Rightarrow 30{f_1} + 30{f_2} = 600$, which is equation 1.
Now subtracting equation 1 from equation 2, as given below :
$30{f_1} + 70{f_2} = 1080$
$30{f_1} + 30{f_2} = 600$
$ \Rightarrow 40{f_2} = 480$
$\therefore {f_2} = 12$
So finding \[{f_1}\] from the expression \[{f_1} + {f_2} = 20\], which is equation 1, as given below:
$ \Rightarrow {f_1} + 12 = 20$
\[\therefore {f_1} = 8\]
Hence \[{f_1} = 8\] and ${f_2} = 12$.
The values of \[{f_1}\] and ${f_2}$ are 8 and 12 respectively.

Note: Remember that when given the class intervals in the class column of a given distribution then first we have to find the class mark of the class which is given the average of the lower and upper limit of the class.