Question

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Find the values of P and Q.

Classes	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	5	P	10	Q	7	8

Answer 1

Hint: Make a table having four columns which are classes, mid values \[({{x}_{1}})\] , frequency \[({{f}_{1}})\] , and \[{{\operatorname{f}}_{1}}{{x}_{1}}\] . The mid-values can be found by taking the half of sum of each class interval. The mean of the given frequency distribution table is calculated by dividing \[\sum{{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}}\] by the total number of frequencies. Here the total number of frequencies is 50.

Complete step-by-step answer:
First of all, calculating the mid values( \[{{x}_{1}}\] ) of each class interval.
For the class 0-20, we have the mid-value = \[\dfrac{0+20}{2}=10\] .
For the class 20-40, we have the mid-value = \[\dfrac{40+20}{2}=30\] .
For the class 40-60, we have the mid-value = \[\dfrac{40+60}{2}=50\] .
For the class 60-80, we have the mid-value = \[\dfrac{60+80}{2}=70\] .
For the class 80-100, we have the mid-value = \[\dfrac{80+100}{2}=90\] .
For the class 100-120, we have the mid-value = \[\dfrac{100+120}{2}=110\] .
Now, reconstruct the table and add two more columns that are mid value \[({{x}_{1}})\] and \[{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}\] .

Class	Frequency(f1)	Mid value (x1)	f1x1
0-20	5	10	50
20-40	P	30	30P
40-60	10	50	500
60-80	Q	70	70Q
80-100	7	90	630
100-120	8	110	880
	N = 50		\[\sum{{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}}\text{=30P+70Q+2060}\]

According to the question, it is given that the sum of all frequencies is 50 i.e, N=50.
\[\begin{align}
  & \text{5+P+10+Q+7+8=50} \\
 & \Rightarrow \text{P+Q=50-30} \\
\end{align}\]
\[\Rightarrow \text{P+Q=20}\]
\[\Rightarrow P=20-Q\] …………………….(1)
We know the formula of mean that,
Mean = \[\dfrac{\sum{{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}}}{\text{N}}\] ………………………..(2)
According to the question, we have
Mean = 62.8 …………………….(3)
From equation (2) and equation (3), we get
62.8 = \[\dfrac{\sum{{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}}}{\text{N}}\]
From the table we have, \[\sum{{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}}\text{=30P+70Q+2060}\] and N = 50.
\[\dfrac{\text{30P+70Q+2060}}{50}=62.8\]
\[\begin{align}
  & \Rightarrow 30P+70Q+2060=3140 \\
 & \Rightarrow 30P+70Q=1080 \\
\end{align}\]
\[\Rightarrow 3P+7Q=108\] …………………….(4)
From equation (1) and equation (4), we get
\[\Rightarrow 3P+7Q=108\]
\[\Rightarrow 3(20-Q)+7Q=108\]
\[\begin{align}
  & \Rightarrow 60-3Q+7Q=108 \\
 & \Rightarrow 4Q=48 \\
 & \Rightarrow Q=12 \\
\end{align}\]
Now, putting the value of Q in equation (1), we get
\[\begin{align}
  & \Rightarrow P=20-12 \\
 & \Rightarrow P=8 \\
\end{align}\]
Hence, the value of P and Q is 8 and 12.

Note: In this question, one may calculate the mean by dividing the total number of classes by the total number of frequencies. This is wrong. We know that Mean = \[\dfrac{\sum{{{\text{f}}_{\text{1}}}{{\text{x}}_{\text{1}}}}}{\text{N}}\] , where \[{{\text{f}}_{\text{1}}}\] , and \[{{\operatorname{x}}_{1}}\] are the frequency and mid-values respectively. N is the total number of frequencies.