Question

10. The age wise participation of students in the Annual Function of a
school is shown in the following distribution.
[CBSE 2014]
Age (in years) 5-7
7-9 9-11 11-13 13-15 15-17 17-19
Number of
students
X
15
18
30
50
48
X
Find the missing frequencies when the sum of frequencies is 181. Also,
find the mode of the data.​

Answer 1

$\large\underline{\underline{\sf{\maltese\:\: \red{Question \: :}}}}$

The agewise participation of students in the Annual Function of a

school is shown in the given distribution.

Find the missing frequencies when the sum of frequencies is 181. Also,

find the mode of the data.

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$\large\underline{\underline{\sf{\maltese\:\: \red{Answer \: :}}}}$

• Missing Frequencies = 10

• Mode = 14.82

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$\large\underline{\underline{\sf{\maltese\:\: \red{Concept \: Used \: : }}}}$

» Mode = $\bold{l \: + \: h(\frac {f_1 \: - \: f_0}{2f_1 \: - \: f_0 \: - f_2})}$

» Class with maximum frequency is called Modal Class.

» Here ,

• l = Lower Limit of Modal Class

• h = Class Size of Modal Class

• f₁ = Frequency of Modal Class

• f₀ = Frequency of Class preceding the Modal Class

• f₂ = Frequency of Class succeeding the Modal Class

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large\underline{\underline{\sf{\maltese\:\: \red{Solution \: :}}}}$

Age (in years) | Number of Students

—————————————————————

5 - 7⠀⠀⠀‎‏‏⠀⠀⠀⠀| x

—————————————————————

7 - 9⠀⠀⠀‎‏‏⠀⠀⠀⠀| 15

—————————————————————

9 - 11⠀⠀⠀‎‏‏⠀⠀⠀⠀| 18

—————————————————————

11 - 13⠀⠀⠀‎‏‏⠀⠀⠀⠀| 30

—————————————————————

13 - 15⠀⠀⠀‎‏‏⠀⠀⠀⠀| 50

—————————————————————

15 - 17⠀⠀⠀‎‏‏⠀⠀⠀⠀| 48

—————————————————————

17 - 19⠀⠀⠀‎‏‏⠀⠀⠀⠀| x

—————————————————————

★ $\bold{NOTE \: :}$ You can also refer to the attached image.

It is given that the sum of frequencies is 181.

∴ x + 15 + 18 + 30 + 50 + 48 + x = 181

⇒ 2x + 161 = 181

⇒ 2x = 181 – 161

⇒ 2x = 20

⇒ x = $\frac{20}{2}$

⇒ x = 10

$\boxed{\bold{ \underline{\therefore \: x \: = \: 10}}}$

Here the maximum class frequency is 50 so the corresponding class 13 - 15 is the Modal Class.

• l = 13

• h = 15 - 13 = 2

• f₁ = 50

• f₀ = 30

• f₂ = 40

$\bold{Mode = {l \: + \: h(\dfrac {f_1 \: - \: f_0}{2f_1 \: - \: f_0 \: - f_2})}}$

$\bold{Mode = 13 \: + \: 2(\dfrac{50 \: -\: 30}{2*50 \: - \: 30\: - \: 48})}$

$\bold{Mode = 13 \: + \: 2(\dfrac{50 \: -\: 30}{100 \: - \: 30\: - \: 48})}$

$\bold{Mode = 13 \: + \: 2(\dfrac{20}{22})}$

$\bold{Mode =13 \: + \: 1.82 }$

$\bold{Mode = 14.82}$

$\boxed{\bold{\underline{\therefore \: Mode \: = \: 14.82}}}$