Question

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequencies ${f_1}$ and ${f_2}$

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

Answer 1

According to the question we have to compute the missing frequencies ${f_1}$and ${f_2}$when the mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. So, first of all we have to understand about the mean as explained below:
Mean: The mean is basically a way to find the average of the given data but before that we have to arrange the all data or given number into ascending order means we have to arrange all the given data from the smaller number to the largest number and after that we have to add up all the given data or number and same as we have to determine the total number of given data or terms.

Formula used: $\Rightarrow \overline x = \dfrac{{\sum {fx} }}{{\sum f }}........................(a)$
Now, we have to find the value of ${x_i}$which can be obtained by dividing the sum of upper limit and lower limit by 2. Hence, we will obtain a linear expression in terms of the missing frequencies ${f_1}$ and ${f_2}$
After that we will obtain the another expression with the help of the formula (a) and on solving the obtained linear expressions we can obtain missing frequencies ${f_1}$ and ${f_2}$

Complete step-by-step solution:
Step 1: First of all we have to obtain the value of the class mark as ${x_i}$ which can be obtained by dividing the sum of upper limit and lower limit by 2 as mentioned in the solution hint.

Class	Frequency	${x_i}$
$0 - 20$	$5$	$\dfrac{{0 + 20}}{2} = 10$
$20 - 40$	${f_1}$	$\dfrac{{20 + 40}}{2} = 30$
$40 - 60$	$10$	$\dfrac{{40 + 60}}{2} = 50$
$60 - 80$	${f_2}$	$\dfrac{{60 + 80}}{2} = 70$
$80 - 100$	$7$	$\dfrac{{80 + 100}}{2} = 90$
$100 - 120$	$8$	$\dfrac{{100 + 120}}{2} = 110$

Step 2: Now, we have to find ${f_i}{x_i}$ which can be obtained by multiplying the given frequencies to the class mark ${x_i}$ as obtained in the step 1.

Class	Frequency ${f_i}$	${x_i}$	${f_i}{x_i}$
$0 - 20$	$5$	$\dfrac{{0 + 20}}{2} = 10$	$5 \times 10 = 50$
$20 - 40$	${f_1}$	$\dfrac{{20 + 40}}{2} = 30$	$30 \times {f_1} = 30{f_1}$
$40 - 60$	$10$	$\dfrac{{40 + 60}}{2} = 50$	$10 \times 50 = 500$
$60 - 80$	${f_2}$	$\dfrac{{60 + 80}}{2} = 70$	${f_2} \times 70 = 70{f_2}$
$80 - 100$	$7$	$\dfrac{{80 + 100}}{2} = 90$	$7 \times 90 = 630$
$100 - 120$	$8$	$\dfrac{{100 + 120}}{2} = 110$	$8 \times 110 = 880$

Step 3: Now, as we know that the sum of all the frequency is $\sum {f = 50}$hence,
$\Rightarrow 5 + {f_1} + 10 + {f_2} + 7 + 8 = 50 \\\ \Rightarrow {f_1} + {f_2} = 50 - 30 \\\ \Rightarrow {f_1} + {f_2} = 20..............(1)$
Step 4: Now, we have to substitute all the obtained values into formula (a) as mentioned in the solution hint.
$\Rightarrow 6.28 = \dfrac{{50 + 30{f_1} + 500 + 70{f_2} + 630 + 880}}{{30 + {f_1} + {f_2}}}$
$\Rightarrow 6.28 = \dfrac{{2060 + 30{f_1} + 70{f_2}}}{{30 + {f_1} + {f_2}}}$
Applying cross-multiplication in the expression obtained just above,
$\Rightarrow 2060 + 30{f_1} + 70{f_2} = 1884 + 62.8{f_1} + 62.8{f_2} \\\ \Rightarrow 62.8{f_1} + 62.8{f_2} - 30{f_1} - 70{f_2} = 2060 - 1884 \\\ \Rightarrow 32.8{f_1} - 7.2{f_2} = 176$
On simplifying the expression obtained just above,
$\Rightarrow 4.1{f_1} - 0.9{f_2} = 22.....................(2)$
Step 5: Now, to obtain the value of missing frequencies ${f_1}$ and ${f_2}$ by solving the expressions as obtained in step 3 and step 4.
So, first of all we have to multiply the expression with 0.9 and then subtract it with the expression (2),
$\Rightarrow 0.9{f_1} + 0.9{f_2} = 1.8...........(3)$
Now, on subtracting obtained expression (3) by expression (2),
Hence,
${f_1} = 8,$ and ${f_2} = 12$

Hence, with the help of formula (a) we have obtain the values of missing frequencies ${f_1} = 8,$ and ${f_2} = 12$

Note: Mean can be determined by arranging the given data in ascending order and then we have to divide the sum of all the given data with the total number of given data.
The mean of the absolute values of the numerical differences between the numbers of a set such as, a static data and their mean and median.