Question

The mean of the following frequency distribution is $62.8$and the sum of all the frequencies is $50$. Compute the missing frequency ${{f}_{1}}$ and ${{f}_{2}}$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
Frequency	5	${{f}_{1}}$	$10$	${{f}_{2}}$	$7$	$8$

Answer 1

Hint: In this question, we are given a continuous series, and thus a frequency distribution table. We are also given the mean of this series and sum of all frequencies. We have to find two missing frequencies ${{f}_{1}}$ and ${{f}_{2}}$. For this, we will first draw a frequency distribution table. As we know that the mean of continuous series is given by \[\dfrac{\sum{fx}}{\sum{f}}\] where \[fx\] is the product of frequency and $\sum{f}$ is the class mark. Hence we will first evaluate the class mark of every interval using formula : $\dfrac{upper~limit+lower~limit}{2}$. Then we will find products of particular class marks with frequencies and add them to get $\sum{f}x$. Solving for mean will give us one equation and solving for sum of frequencies will give us a second equation. From these two equations, we will find ${{f}_{1}}$ and ${{f}_{2}}$.

Complete step-by-step answer:
We first have to draw a frequency distribution table with columns as class interval, frequency$\left( f \right)$, class mark $\left( x \right)$ and\[fx\]. For computing class marks for every interval we will use the formula $x=\dfrac{upper~limit+lower~limit}{2}$.

Class Interval	Frequency $\left( f \right)$	Class mark $\left( x \right)$	$fx$
\[0-20\]	$5$	$10$	$50$
\[20-40\]	${{f}_{1}}$	$30$	$30{{f}_{1}}$
$40-60$	$10$	$50$	$500$
$60-80$	${{f}_{2}}$	$70$	$70{{f}_{2}}$
$80-100$	$7$	$90$	$630$
$100-120$	$8$	$110$	$880$

From the table, sum of frequencies is \[5+{{f}_{1}}+10+{{f}_{2}}+7+8=30+{{f}_{1}}+{{f}_{2}}\].
But we are given a sum of frequencies as $50$. Hence,
\[30+{{f}_{1}}+{{f}_{2}}=50\]
Simplifying, we get –
\[{{f}_{1}}+{{f}_{2}}=20~~~~~~~~~~...(1)\]
Now as we know, $mean=\dfrac{\sum{fx}}{\sum{f}}$, so we need sum of all $fx's$, which will be as,
\[50+30{{f}_{1}}+500+70{{f}_{2}}+630+880=2060+30{{f}_{1}}+70{{f}_{2}}\]
Sum of all $f$ is the sum of frequencies which is given as $50$. Hence, we get –
\[Mean=\dfrac{2060+30{{f}_{1}}+70{{f}_{2}}}{50}\]
As we are given, mean as $62.8$, therefore –
\[\begin{align}
  & 62.8=\dfrac{2060+30{{f}_{1}}+70{{f}_{2}}}{50} \\
 & 2060+30{{f}_{1}}+70{{f}_{2}}=3140 \\
 & 30{{f}_{1}}+70{{f}_{2}}=1080 \\
\end{align}\]
Dividing by $10$ on both sides, we get –
\[3{{f}_{1}}+7{{f}_{2}}=108~~~~~~~~~~...(2)\]
Now we have got two equations, solving them will give us the value of \[{{f}_{1}}\] and \[{{f}_{2}}\].
On multiplying equation (1) by $3$, we get –
\[3{{f}_{1}}+3{{f}_{2}}=60\]
Now subtracting this obtained equation from equation (2), we get –
\[\begin{align}
  & 3{{f}_{1}}+7{{f}_{2}}-3{{f}_{1}}-3{{f}_{2}}=108-60 \\
 & 4{{f}_{2}}=48 \\
 & {{f}_{2}}=\dfrac{48}{4} \\
 & {{f}_{2}}=12 \\
\end{align}\]
Putting the value of \[{{f}_{2}}\] in equation (1), we get –
$\begin{align}
  & 12+{{f}_{1}}=20 \\
 & {{f}_{1}}=8 \\
\end{align}$

Hence, the value of ${{f}_{1}}=8$. So we have found both values which are ${{f}_{1}}=8$ and \[{{f}_{2}}=12\].

So, the correct answer is “Option A”.

Note: Students should carefully calculate the class mark for every interval. Class mark is just the value of the interval. 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 difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data.