Question

10. 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.​

Answer 1

Missing frequencies = 10  & Mode = 14.82

Step-by-step explanation:

Age                 Number of students          

5 - 7                   x

7 - 9                  15

9 - 11                  18

11 - 13                 30

13 - 15                50

15 - 17                48

17 - 19                x

the sum of frequencies is 181

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

=> 2x = 20

=> x = 10

Maximum frequency  50 lies in 13 - 15

Mode lies in 13 - 15

Mode = 13  +  2 * (50 - 30)/(50 - 30 + 50 - 48)

= 13 + 2 * 20/22

= 14.82

Learn more:

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Answer 2

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

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.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

• Missing Frequencies = 10

• Mode = 14.82

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

» 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

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

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

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

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

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

$\bold{Mode = 14.82}$

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