Question

Find p for the given table if $\overline{x}=54$.

class limit	frequency
0-20	7
20-40	$p$
40-60	10
60-80	9
80-100	13

Answer 1

Hint: We first explain the concept of class mark and then find the formula for weighted mean. We take the formula for $\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}$ and put the values to get linear equation. We solve it to find the value of the variable.

Complete step by step answer:
We use the class limits of the given classes to find the class marks. It is the midpoint of the upper and lower limit of a particular class.
So, if we have \[{{u}_{i}},{{l}_{i}}\] as the upper and lower limit of a particular class then the class mark will be \[{{x}_{i}}=\dfrac{{{u}_{i}}+{{l}_{i}}}{2}\].
Then we use the formula of mean with given frequencies as $\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}$.
We now complete the table adding \[{{x}_{i}}\] for class mark and \[{{x}_{i}}{{f}_{i}}\] for the summation.

class limit	class mark (\[{{x}_{i}}\])	frequency	\[{{x}_{i}}{{f}_{i}}\]
0-20	10	7	$10\times 7=70$
20-40	30	$p$	$30\times p=30p$
40-60	50	10	$50\times 10=500$
60-80	70	9	$70\times 9=630$
80-100	90	13	$90\times 13=1170$

Now we find the summation of $\sum{{{x}_{i}}{{f}_{i}}}=70+30p+500+630+1170=2370+30p$.
We also find the summation of the frequencies $\sum{{{f}_{i}}}=7+p+10+9+13=39+p$.
It is given that $\overline{x}=54$. So, $\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}\Rightarrow 54=\dfrac{2370+30p}{39+p}$.
Simplifying we get
$\begin{align}
  & 2370+30p=54\left( 39+p \right) \\
 & \Rightarrow 54p-30p=2370-2106 \\
 & \Rightarrow 24p=264 \\
 & \Rightarrow p=\dfrac{264}{24}=11 \\
\end{align}$
The value of $p$ is 11.

Note: We need to consider the frequencies as the weight of particular \[{{x}_{i}}\]. The general mean formula without weight $\overline{x}=\dfrac{\sum{{{x}_{i}}}}{n}$ changes to $\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{N}$ where we consider $N=\sum{{{f}_{i}}}$. Also weighted averages assign importance (or weight) to each number. A weighted average can be more useful than a regular average because it offers more nuance.