Consider the following frequency distribution : class 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency α 110 54 30 β If the sum of all frequencies is 584 and median is 45 , then | α - β | is equal to .

Correct answer: 0 ∵ Sum of frequencies = 584 α + 110 + 54 + 30 + β = 584 ⇒ α + β = 390 . . . (1) Now, Median is at (584)/(2) = 292 ^ th term ∵ Median = 45 (lies in class 40 - 50 ) ⇒ α + 110 + 54 + 15 = 292 ⇒ α = 113 . . . (2) On solving, equations (1) and (2) , we get ⇒ α = 113 and β = 390 - 113 = 277 ⇒ | α - β | = 164

JEE Main 202125 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2021 · 25 July · Morning Shift · Mathematics · Statistics

Consider the following frequency distribution : class 10 - 20 20 -...

Consider the following frequency distribution :

class	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$
Frequency	$\alpha$	$110$	$54$	$30$	$\beta$

If the sum of all frequencies is $584$ and median is $45$ , then $| \alpha - \beta |$ is equal to .

Correct answer

Step-by-step solution

$∵$ Sum of frequencies $= 584$

$\alpha + 110 + 54 + 30 + \beta = 584$

$⇒ \alpha + \beta = 390 . . . \left(1\right)Now, Median is at \dfrac{584}{2} =$ {292}^{th} $term$ ∵ $Median$ = 45 $(lies in class$ 40 - 50 $)$ ⇒ \alpha + 110 + 54 + 15 = 292 $$ ⇒ \alpha = 113 . . . $\left(2\right)$

On solving, equations \left(1\right) $and \left(2\right)$ , we get

$⇒ \alpha = 113 \text{\text{ and } } \beta = 390 - 113 = 277$

$⇒ | \alpha - \beta | = 164$

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