Question

The mean of the following distribution is 50 and the sum of all the frequencies is 120. Find the values of \[p\] and \[q\].

Class-interval	Frequency
0-20	17
20-40	\[p\]
40-60	32
60-80	\[q\]
80-100	19

Answer 1

Hint: First we will assume that \[{f_i}\] be represents the frequency column and \[{x_i}\] is the division of sum of the class intervals by 2. Then we will use formula to calculate mean is \[Mean = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\], where \[\sum {{f_i}{x_i}} \] is the total sum of \[{f_i}{x_i}\] and \[\sum {{f_i}} \] is the total sum of \[{f_i}\].

Complete step-by-step answer:
Let us assume that \[{f_i}\] represents the frequency column and \[{x_i}\] is the division of sum of the class intervals by 2.
We are given that \[\sum {{f_i}} = 120\] and mean is 50.
We will now form a table to find the value of the product \[{f_i}{x_i}\] for mean.

Class Intervals	\[{x_i}\]	\[{f_i}\]	\[{f_i}{x_i}\]
0-20	\[\dfrac{{0 + 20}}{2} = 10\]	17	170
20-40	\[\dfrac{{20 + 40}}{2} = 30\]	\[p\]	\[30p\]
40-60	\[\dfrac{{40 + 60}}{2} = 50\]	32	1600
60-80	\[\dfrac{{60 + 80}}{2} = 70\]	\[q\]	\[70q\]
80-100	\[\dfrac{{90 + 100}}{2} = 90\]	19	1710
Total		\[\sum {{f_i}} = 68 + p + q\]	\[\sum {{f_i}{x_i}} = 3480 + 30p + 70q\]

Substituting the value of \[\sum {{f_i}} \] from the above table, we get

\[ \Rightarrow 120 = 68 + p + q\]

Subtracting the above equation by \[68 + q\] on both sides, we get
\[
   \Rightarrow 120 - 68 - q = 68 + p + q - 68 - q \\
   \Rightarrow 52 - q = p \\
   \Rightarrow p = 52 - q{\text{ ......eq.(1)}} \\
 \]
We know that the formula to calculate mean is \[Mean = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\], where \[\sum {{f_i}{x_i}} \] is the total sum of \[{f_i}{x_i}\] and \[\sum {{f_i}} \] is the total sum of \[{f_i}\].
Substituting the values of \[\sum {{f_i}{x_i}} \] and \[\sum {{f_i}} \] in the above formula, we get
\[ \Rightarrow 50 = \dfrac{{3480 + 30p + 70q}}{{120}}\]
Cross-multiplying the above equation, we get
\[ \Rightarrow 6000 = 3480 + 30p + 70q\]
Substituting the value of \[p\] in the above equation, we get
\[
   \Rightarrow 6000 = 3480 + 30\left( {52 - q} \right) + 70q \\
   \Rightarrow 6000 = 3480 + 1560 - 30q + 70q \\
   \Rightarrow 6000 = 4040 + 40q \\
 \]
Subtracting the above equation by 4040 on both sides, we get
\[
   \Rightarrow 6000 - 4040 = 4040 + 40q - 4040 \\
   \Rightarrow 960 = 40q \\
 \]
Dividing the above equation by 40 on both sides, we get
\[
   \Rightarrow \dfrac{{960}}{{40}} = \dfrac{{40q}}{{40}} \\
   \Rightarrow 24 = q \\
   \Rightarrow q = 24 \\
 \]
Hence, the value of \[q\] is 24.
Substituting the value of \[q\] in the equation (1), we get
\[
   \Rightarrow p = 52 - 24 \\
   \Rightarrow p = 28 \\
 \]
Hence, the value of \[p\] is 28.

Note: In solving these types of questions, students should know the formula of mean. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong.