
Calculate the missing frequency from the following distribution, it is given that the median of the distribution is 24.
Age in years 0-10 10-20 20-30 30-40 40-50 No. of persons 5 25 ? 18 7
| Age in years | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| No. of persons | 5 | 25 | ? | 18 | 7 |
Answer
558.3k+ views
Hint: First assume the number of persons for range 20-30. After that, form a frequency table with three columns, having column 1 as age, column 2 as frequency i.e. number of the person, and column 3 as cumulative frequency. Then apply the formula of the median $L + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right)h$. Then do the simplification to find the missing frequency.
Complete step-by-step answer:
Let us assume that, number of persons having age in the range 20 – 30 be $x$. Using the above table let us form a frequency table, having three columns. Column 1 will have age, column 2 will contain frequency or number of persons, column 3 will contain cumulative frequency which is calculated by adding frequencies in each step. Therefore, the frequency table will look like: -
As the median is 24 which lies in the median class is 20 – 30.
Now the value of $\dfrac{N}{2}$ is,
$ \Rightarrow \dfrac{N}{2} = \dfrac{{55 + x}}{2}$
The lower limit of the median class is,
$ \Rightarrow l = 20$
Cumulative frequency of class preceding the median class is,
$ \Rightarrow cf = 30$
The frequency of the median class is,
$ \Rightarrow f = x$
The height of the class is,
$ \Rightarrow h = 40 - 30 = 10$
Then the value of the median is given by,
Median $ = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right]h$
Substitute the values,
$ \Rightarrow 24 = 20 + \left( {\dfrac{{\dfrac{{55 + x}}{2} - 30}}{x}} \right) \times 10$
Take LCM in the numerator and move 2 in the denominator,
\[ \Rightarrow 24 = 20 + \left( {\dfrac{{55 + x - 60}}{{2x}}} \right) \times 10\]
Simplify the terms,
\[ \Rightarrow 4 = \left( {\dfrac{{x - 5}}{x}} \right) \times 5\]
Cross-multiply the terms,
$ \Rightarrow 4x = 5x - 25$
Move variable part on one side,
$ \Rightarrow - x = - 25$
Multiply both sides by -1,
$\therefore x = 25$
Hence, the missing frequency is 25.
Note: One may note that the class interval (h) is the same for all the rows and therefore we can calculate it by subtracting the lower limit from the upper limit by choosing any one of the rows. In the above solution, we choose row 1, i.e. 10 – 0 = 10. Now, the most important thing is the formula and its terms. We must know about all the terms in the formula and how to calculate it.
Complete step-by-step answer:
Let us assume that, number of persons having age in the range 20 – 30 be $x$. Using the above table let us form a frequency table, having three columns. Column 1 will have age, column 2 will contain frequency or number of persons, column 3 will contain cumulative frequency which is calculated by adding frequencies in each step. Therefore, the frequency table will look like: -
| Age (in years) | Frequency ($f$) | Cumulative Frequency ($cf$) |
| 0 – 10 | 5 | 5 |
| 10 – 20 | 25 | 30 |
| 20 – 30 | $x$ | $30 + x$ |
| 30 – 40 | 18 | $48 + x$ |
| 40 – 50 | 7 | $55 + x$ |
As the median is 24 which lies in the median class is 20 – 30.
Now the value of $\dfrac{N}{2}$ is,
$ \Rightarrow \dfrac{N}{2} = \dfrac{{55 + x}}{2}$
The lower limit of the median class is,
$ \Rightarrow l = 20$
Cumulative frequency of class preceding the median class is,
$ \Rightarrow cf = 30$
The frequency of the median class is,
$ \Rightarrow f = x$
The height of the class is,
$ \Rightarrow h = 40 - 30 = 10$
Then the value of the median is given by,
Median $ = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right]h$
Substitute the values,
$ \Rightarrow 24 = 20 + \left( {\dfrac{{\dfrac{{55 + x}}{2} - 30}}{x}} \right) \times 10$
Take LCM in the numerator and move 2 in the denominator,
\[ \Rightarrow 24 = 20 + \left( {\dfrac{{55 + x - 60}}{{2x}}} \right) \times 10\]
Simplify the terms,
\[ \Rightarrow 4 = \left( {\dfrac{{x - 5}}{x}} \right) \times 5\]
Cross-multiply the terms,
$ \Rightarrow 4x = 5x - 25$
Move variable part on one side,
$ \Rightarrow - x = - 25$
Multiply both sides by -1,
$\therefore x = 25$
Hence, the missing frequency is 25.
Note: One may note that the class interval (h) is the same for all the rows and therefore we can calculate it by subtracting the lower limit from the upper limit by choosing any one of the rows. In the above solution, we choose row 1, i.e. 10 – 0 = 10. Now, the most important thing is the formula and its terms. We must know about all the terms in the formula and how to calculate it.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

