Question

The mean of the following frequency table is 53. But the frequencies ${f_1}$ and ${f_2}$ in the classes 20-40 and 60-80 are missing. Find the missing frequencies.
Age (in years) 0-20 20-40 40-60 60-80 80-100 Total
No. of people 15 ${f_1}$ 21 ${f_2}$ 17 100
A) ${f_1} = 20,{f_2} = 28$
B) ${f_1} = 17,{f_2} = 19$
C) ${f_1} = 19,{f_2} = 25$
D) ${f_1} = 18,{f_2} = 29$

Answer 1

Hint: To find the mean number of people, the first step is to determine the midpoint (also called a class mark) of each class interval. These midpoints should then be multiplied by the frequencies of the corresponding classes. The sum of the products divided by the total number of missing people will be the value of the mean number of days. Equate it to the value 53 to solve for the missing frequencies.

Complete step-by-step answer:
Given
Age (in years) 0-20, 20-40, 40-60, 60-80, 80-100 , Total
No. of people 15, ${f_1}$, 21, ${f_2}$, 17, 100
We have to find the missing frequencies of 20-40 and 60-80 intervals.
$Mean = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{x_i}} }}$
$\begin{array}{*{20}{c}}
  {class}&{{f_i}}&{{x_i}}&{{f_i}{x_i}} \\
  {0 - 20}&{15}&{10}&{150} \\
  {20 - 40}&{{f_1}}&{30}&{30{f_1}} \\
  {40 - 60}&{21}&{50}&{1050} \\
  {60 - 80}&{{f_2}}&{70}&{70{f_2}} \\
  {80 - 100}&{17}&{90}&{1530}
\end{array}$
Where ${x_i}$ is the class mark (midpoint of class interval) and ${f_i}$ is the frequency and ${f_i}{x_i}$ is the product of class mark and frequency.
We are given a sum of the frequencies is 100.
$
  15 + {f_1} + 21 + {f_2} + 17 = 100 \\
  {f_1} + {f_2} + 53 = 100 \\
  {f_1} + {f_2} = 100 - 53 \\
  {f_1} + {f_2} = 47 \\
  {f_1} = 47 - {f_2} \to eq(1) \\
 $
Sum of the ${f_i}{x_i}$ is $2730 + 30{f_1} + 70{f_2}$ .
$
  mean = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{x_i}} }} \\
  mean = 53 \\
  53 = \dfrac{{2730 + 30{f_1} + 70{f_2}}}{{100}} \\
  53 \times 100 = 2730 + 30{f_1} + 70{f_2} \\
  30{f_1} + 70{f_2} = 5300 - 2730 \\
  30{f_1} + 70{f_2} = 2570 \\
  3{f_1} + 7{f_2} = 257 \to eq(2) \\
 $
Substitute equation 1 in equation 2.
$
  3\left( {47 - {f_2}} \right) + 7{f_2} = 257 \\
  141 - 3{f_2} + 7{f_2} = 257 \\
  4{f_2} = 257 - 141 \\
  4{f_2} = 116 \\
  {f_2} = \dfrac{{116}}{4} \\
  {f_2} = 29 \\
 $
Substitute ${f_2} = 29$ in equation 1.
$
  {f_1} + {f_2} = 47 \\
  {f_2} = 29 \\
  {f_1} + 29 = 47 \\
  {f_1} = 47 - 29 = 18 \\
 $
The missing values are 18 and 29.
Therefore, from among the options given in the question Option D is correct, which is ${f_1} = 18,{f_2} = 29$

Note: Mean of any element is the average of its values. An interval is a range of values for a statistic. In grouped data, we have to find the class mark manually to find the mean. We solved the above problem using direct method. Other methods to find mean are Assumed mean method and Step Deviation method.