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.

& \text{5+P+10+Q+7+8=50} \\\ & \Rightarrow \text{P+Q=50-30} \\\ \end{aligned}$$ $$\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{aligned} & \Rightarrow 30P+70Q+2060=3140 \\\ & \Rightarrow 30P+70Q=1080 \\\ \end{aligned}$$ $$\Rightarrow 3P+7Q=108$$ …………………….(4) From equation (1) and equation (4), we get $$\Rightarrow 3P+7Q=108$$ $$\Rightarrow 3(20-Q)+7Q=108$$ $$\begin{aligned} & \Rightarrow 60-3Q+7Q=108 \\\ & \Rightarrow 4Q=48 \\\ & \Rightarrow Q=12 \\\ \end{aligned}$$ Now, putting the value of Q in equation (1), we get $$\begin{aligned} & \Rightarrow P=20-12 \\\ & \Rightarrow P=8 \\\ \end{aligned}$$ Hence, the value of P and Q is 8 and 12. Note: In this question, one may calculate the mean by dividing the total number of classes by the total number of frequencies. This is wrong. We know that Mean = $$\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.