Let M denote the median of the following frequency distribution. cases Class & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 Frequency & 3 & 9 & 10 & 8 & 6 cases Then 20M is equal to :

Correct answer: D. 208 ( Class amp; Frequency amp; Cumulative frequency 0-4 amp; 3 amp; 3 4-8 amp; 9 amp; 12 8-12 amp; 10 amp; 22 12-16 amp; 8 amp; 30 16-20 amp; 6 amp; 36 ) M = l + (((N)/(2) - C)/(f))h Here, N = 3 + 9 + 10 + 8 + 6 = 36 ⇒ (N)/(2) = 18 So, median class is (8 - 12) ⇒ l = 8, C = 22, f = 10, h = 4 ⇒ M = 8 + (18 - 12)/(10) × 4 ⇒ M = 10 . 4 ⇒ 20M = 208

JEE Main 202430 JanuaryMorning Shift

Exam: JEE Main
Year: 2024
Date: 30 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2024 · 30 January · Morning Shift · Mathematics · Statistics

Let M denote the median of the following frequency distribution....

Let $M$ denote the median of the following frequency distribution.

$\begin{cases}\text{Class} & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 \\ \text{Frequency} & 3 & 9 & 10 & 8 & 6\end{cases}$

Then $20M$ is equal to :

Options

A $416$
B $104$
C $52$
D $208$

Correct answer

D. $208$

Step-by-step solution

(

Class	amp; Frequency	amp; Cumulative frequency
0-4	amp; 3	amp; 3
4-8	amp; 9	amp; 12
8-12	amp; 10	amp; 22
12-16	amp; 8	amp; 30
16-20	amp; 6	amp; 36

)

M = l + \left(\dfrac{\dfrac{N}{2} - C}{f}\right)h

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Here,

N = 3 + 9 + 10 + 8 + 6 = 36

⇒ \dfrac{N}{2} = 18

So, median class is \left(8 - 12\right)

</td></tr>

⇒ l = 8, C = 22, f = 10, h = 4

<tr><td>

⇒ M = 8 + \dfrac{18 - 12}{10} \times 4

</td><td>

⇒ M = 10 . 4

</td><td>

⇒ 20M = 208