Question

The mean of the following frequency distribution is 57.6 and the sum of the frequency is 50. Find the missing frequency \[{{f}_{1}}\]and \[{{f}_{2}}\]:

\[\text{Class interval:}\]	\[0-20\]	\[20-40\]	\[40-60\]	\[60-80\]	\[80-100\]	\[100-120\]
\[\text{Frequency:}\]	\[7\]	\[{{f}_{1}}\]	\[12\]	\[{{f}_{2}}\]	\[8\]	\[5\]

Answer 1

Hint: To find the missing values of frequencies we will generate two equations in terms of frequency. One of the equations by applying the condition of sum of the frequency is given 50 and other by applying the mean of the grouped data is 57.6. We will get two equations and two variables just solve the two generated equations and then find the value of missing frequencies.

Complete step-by-step solution:
First, we have to understand what is meant by the mean of the given data. Now, sample mean is the mean of the collected data. To find the mean or the average of a collection of the sample, then we use the sample mean formula. The sample mean of the collected data is calculated by adding the numbers and then dividing the number of data collected.
Mean formula is given below:
Mean \[=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\]
Where, \[\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}={{x}_{1}}{{f}_{1}}+{{x}_{2}}{{f}_{2}}+{{x}_{3}}{{f}_{3}}+{{x}_{4}}{{f}_{4}}+....+{{x}_{n}}{{f}_{n}}\] and \[\sum\limits_{i=1}^{n}{{{f}_{i}}}={{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+....+{{f}_{n}}\]
Here, \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},....,{{x}_{n}}\,\] are the different values and \[{{f}_{1}},{{f}_{2}},{{f}_{3}},{{f}_{4}},....,{{f}_{n}}\] are also different values.
Now, we have to understand what is meant by frequency and frequency table. Now, frequency refers to the number of times an event or a value occurs. A frequency table is a table that lists items and shows the number of times the items occur.
We represent the frequency by the English alphabet \['f'\]
Now, we have to consider the question which is given in the following table that is

\[\text{Class interval:}\]	\[0-20\]	\[20-40\]	\[40-60\]	\[60-80\]	\[80-100\]	\[100-120\]
\[\text{Frequency:}\]	\[7\]	\[{{f}_{1}}\]	\[12\]	\[{{f}_{2}}\]	\[8\]	\[5\]

Now, we have to find the missing frequency for that first of all we need to make the following table

\[\text{Class}\]	\[\text{Frequency}\]\[\left( {{f}_{i}} \right)\]	\[\text{Class Mark}\]\[\left( {{x}_{i}} \right)\]	\[{{x}_{i}}{{f}_{i}}\]
\[0-20\]	\[7\]	\[10\]	\[70\]
\[20-40\]	\[{{f}_{1}}\]	\[30\]	\[30{{f}_{1}}\]
\[40-\text{8}0\]	\[12\]	\[50\]	\[600\]
\[60-\text{80}\]	\[{{f}_{2}}\]	\[70\]	\[70{{f}_{2}}\]
\[80-100\]	\[8\]	\[90\]	\[720\]
\[100-\text{1}20\]	\[5\]	\[110\]	\[550\]
\[\text{Total}\]	\[\sum\limits_{i=1}^{n}{{{f}_{i}}}=32+{{f}_{1}}+{{f}_{2}}\]		\[\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}=1940+30{{f}_{1}}+70{{f}_{2}}\]

According to the question it is given that
Sum of frequency is 50.
So, adding all the frequency we will get:
\[\Rightarrow 7+{{f}_{1}}+12+{{f}_{2}}+8+5=50\]
After simplifying this equation, we get:
\[\Rightarrow {{f}_{1}}+{{f}_{2}}=18---(1)\]
The mean of the grouped data is given as 57.6 so we will calculate the mean and equate it with the given value of mean such that we will get another equation.
We know mean is defined as
Mean \[=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\]
On putting the given values in the above formula, we get​​:
\[\Rightarrow 57.6=\dfrac{30{{f}_{1}}+70{{f}_{2}}+1940}{50}\]
By multiplying the 50 on both sides on above equation we get:
\[\Rightarrow 57.6\times 50=30{{f}_{1}}+70{{f}_{2}}+1940\]
By simplifying this we get:
\[\Rightarrow 30{{f}_{1}}+70{{f}_{2}}+1940=2880\]
By rearranging the term, we get:
\[\Rightarrow 30{{f}_{1}}+70{{f}_{2}}=940\]
Dividing by 10 on both sides we get:
\[\Rightarrow 3{{f}_{1}}+7{{f}_{2}}=94---(2)\]
From the equation (1) we get:
\[\Rightarrow {{f}_{2}}=18-{{f}_{1}}--(3)\]
Substitute this value of equation (3) on equation (2)
\[\Rightarrow 3{{f}_{1}}+7(18-{{f}_{1}})=94\]
By simplifying this bracket, we get:
\[\Rightarrow 3{{f}_{1}}+126-7{{f}_{1}}=94\]
By rearranging the term, we get:
\[\Rightarrow 3{{f}_{1}}-7{{f}_{1}}=94-126\]
By further solving this we get:
\[\Rightarrow -4{{f}_{1}}=-32\]
Dividing by \[-4\] on both sides we get:
\[\Rightarrow {{f}_{1}}=8\]
Substitute this above value on above equation (2) we get:
\[\Rightarrow 3(8)+7{{f}_{2}}=94\]
By rearranging this term, we get:
\[\Rightarrow 7{{f}_{2}}=94-24\]
By further solving this we get:
\[\Rightarrow 7{{f}_{2}}=70\]
Divide by 7 on both sides on this above equation we get:
\[\Rightarrow {{f}_{2}}=10\]
Thus, the value of \[{{f}_{1}}\] is 8 and the value of \[{{f}_{2}}\] is 10.

Note: This is solved by a simple mean calculation method. It can be solved by assuming a mean method also. Generally, there must at least be two equations to find two unknowns. In the mean formula, while computing \[\sum{f_x}\], don’t take the sum of \[f\] and \[x\] separately and then multiply them. It will be challenging. Students should take their time when creating the frequency distribution table because there is a considerable risk of making mistakes when copying and processing data.