Question

Find the missing frequency ${f_1}$ and ${f_2}$in the table given below, it is being given that the mean of the given frequency distribution is 50.

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	${f_1}$	32	${f_2}$	19	120

Answer 1

Hint: Here we have to find out the frequencies ${f_1}$ and ${f_2}$, which are in the frequencies of which a particular class occurs those many times, which is called as the frequencies. So given the mean of the frequency distribution, to find out the unknown frequencies which are ${f_1}$ and ${f_2}$. First of all we have to understand what mean is, mean is the average of the given data, which is calculated by the ratio of sum of all the observations to the total number of observations.

Complete step-by-step solution:
Here in order to find the mathematical expression of mean, we have to find out the expression for the sum of all the observations and also the expression for the total number of observations.
The total number of observations is 120 which is already given :
$ \Rightarrow \sum\limits_i {{f_i}} = 120$
The sum of all the observations is given by $\sum\limits_i {{f_i}{x_i}} $, as the observations have occurred for multiple number of times, hence multiplying the observation with how many number of times it has occurred which is the frequency.
$\because $Here each class is not of an individual number but rather it is an interval, in such a case consider the middle value of the interval for each class, which is given below:
$ \Rightarrow $For the class 0-20 , the ${x_1} = 10$
$ \Rightarrow $For the class 20-40 , the ${x_2} = 30$
$ \Rightarrow $For the class 40-60 , the ${x_3} = 50$
$ \Rightarrow $For the class 60-80 , the ${x_4} = 70$
$ \Rightarrow $For the class 80-100 , the ${x_5} = 90$
$\therefore $The expression of mean is given by:
$ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{17(10) + {f_1}(30) + 32(50) + {f_2}(70) + 19(90)}}{{17 + {f_1} + 32 + {f_2} + 19}}$
\[ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{3480 + 30{f_1} + 70{f_2}}}{{{f_1} + {f_2} + 68}}\]
Given that the mean of the frequency distribution is 50, is equated to the above obtained mean expression to find out the values of \[{f_1}\] and \[{f_2}\] is shown below:
\[ \Rightarrow \dfrac{{3480 + 30{f_1} + 70{f_2}}}{{{f_1} + {f_2} + 68}} = 50\]
Also given that $\sum\limits_i {{f_i}} = 120$, from here another equation can be obtained:
$ \Rightarrow {f_1} + {f_2} + 68 = 120$
Hence we have 2 variables and 2 equations to get the solutions of the two unknown variables ${f_1}$and${f_2}$.
Now from the frequency summation :
$ \Rightarrow {f_1} + {f_2} + 68 = 120$
$ \Rightarrow {f_1} + {f_2} = 52$
From this equation finding the expression of${f_2}$ :
$ \Rightarrow {f_2} = 52 - {f_1}$
Substituting the ${f_2}$ expression in the mean expression to find the value of ${f_1}$ :
\[ \Rightarrow \dfrac{{3480 + 30{f_1} + 70(52 - {f_1})}}{{{f_1} + (52 - {f_1}) + 68}} = 50\]
\[ \Rightarrow \dfrac{{3480 + 30{f_1} + 3640 - 70{f_1}}}{{{f_1} - {f_1} + 120}} = 50\]
\[ \Rightarrow \dfrac{{7120 - 40{f_1}}}{{120}} = 50\]
\[ \Rightarrow 7120 - 40{f_1} = 50(120)\]
\[ \Rightarrow 7120 - 40{f_1} = 6000\]
\[ \Rightarrow 7120 - 6000 = 40{f_1}\]
\[ \Rightarrow 1120 = 40{f_1}\]
$ \Rightarrow {f_1} = \dfrac{{1120}}{{40}}$
$ \Rightarrow {f_1} = 28$
$\therefore {f_1} = 28$
Now finding ${f_2}$ by substituting ${f_1}$ in :
$ \Rightarrow {f_2} = 52 - {f_1}$
$ \Rightarrow {f_2} = 52 - 28$
$ \Rightarrow {f_2} = 24$
$\therefore {f_2} = 24$

The values of ${f_1} = 28$ and ${f_2} = 24$

Note: Here while calculating the value of the mean, the sum of all the observations is the sum of the frequency times the observation, but here the observations of a class are given in the form of an interval, hence the observation value here is taken to be the middle value of the interval of the class.