Question

The mean of the following frequency table is $50$. But the frequencies ${{f}_{1}}$ and ${{f}_{2}}$ , in class $20-40$ and $60-80$, respectively, are missing. Find the missing frequencies.

Classes	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	${{f}_{1}}$	32	${{f}_{2}}$	19	120

Answer 1

Hint: First, we will start by defining what a mean is and then we will define the frequency and after that we will see from the table and take a value as assumed mean and then we will see what a class fit is, then we will draw a table with class intervals, mid values, frequency, \[{{u}_{i}}=\dfrac{{{x}_{i}}-a}{h}\] where $a$ is assumed mean and $h$ is class fit and ${{f}_{i}}{{u}_{i}}$ . And then we will form two equations in ${{f}_{1}}$ and ${{f}_{2}}$ and ultimately find the values.

Complete step by step answer:
First, let’s 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.
Sample mean formula is given below:
$\overline{x}=\dfrac{\sum\nolimits_{i=1}^{n}{{{x}_{i}}}}{n}$
Where,
$\overline{x}=$ sample mean, \[\sum\nolimits_{i=1}^{n}{{{x}_{i}}}={{x}_{1}}+{{x}_{2}}+......+{{x}_{n}}\] , $n=$ Total number of terms.
Here, ${{x}_{1}},{{x}_{2}},.......,{{x}_{n}}$ are different values.
Now, let’s see 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, let’s take our question we have been given the following table:

Classes	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	${{f}_{1}}$	32	${{f}_{2}}$	19	120

Let the assumed mean be $a=50$ and the class fit $h=20$ :
Now, let’s make the following table:

Class Interval	Mid-Values (\[{{x}_{i}}\])	Frequency (\[{{f}_{i}}\])	\[{{u}_{i}}=\dfrac{{{x}_{i}}-a}{h}\]	\[{{f}_{i}}{{u}_{i}}\]
0-20	10	17	-2	-34
20-40	30	\[{{f}_{1}}\]	-1	\[-{{f}_{1}}\]
40-60	50	32	0	0
60-80	70	\[{{f}_{2}}\]	1	\[{{f}_{2}}\]
80-100	90	19	2	38
Total		\[\sum{{{f}_{i}}=68+{{f}_{1}}+{{f}_{2}}}\]		\[\sum{{{f}_{i}}{{u}_{i}}=4-{{f}_{1}}+{{f}_{2}}}\]

It is given in the question that the total $\sum{{{f}_{i}}=120}$ , now as we saw above: $\begin{align}
  & \Rightarrow 68+{{f}_{1}}+{{f}_{2}}=120 \\
 & \Rightarrow {{f}_{1}}+{{f}_{2}}=52\text{ }..............\text{ Equation}\left( 1 \right) \\
\end{align}$
Now, $\text{Mean}=50$ :
$\begin{align}
  & \Rightarrow \overline{X}=A+h\left( \dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right) \\
 & \Rightarrow 50=50+20\left( \dfrac{4-{{f}_{i}}+{{f}_{2}}}{120} \right) \\
 & \Rightarrow 50=50+\left( \dfrac{4-{{f}_{i}}+{{f}_{2}}}{6} \right)\Rightarrow 0=\left( \dfrac{4-{{f}_{i}}+{{f}_{2}}}{6} \right) \\
 & \Rightarrow {{f}_{1}}-{{f}_{2}}=4\text{ }...........\text{ Equation }\left( 2 \right). \\
\end{align}$
Now, we will solve equation 1 and 2 ,
\[\begin{align}
  & \underline{\begin{matrix}
   {{f}_{1}} & + & {{f}_{2}} & = & 52 \\
   {{f}_{1}} & - & {{f}_{2}} & = & 4 \\
\end{matrix}} \\
 & \begin{matrix}
   2{{f}_{1}} & {} & {} & = & 56 \\
\end{matrix} \\
\end{align}\]
$\Rightarrow {{f}_{1}}=28$
Now, we will put the value of ${{f}_{1}}$ in equation 1, we get:
$28+{{f}_{2}}=52\Rightarrow {{f}_{2}}=24$

Hence, the missing frequencies are: ${{f}_{1}}=28\text{ and }{{f}_{2}}=24$ .

Note: In questions like these always draw a table so that we will have a better understanding and the calculations will be easy. Try and analyze the given conditions carefully and then make the equations in the unknown variables.