JEE Main 2025 Jan 23 Shift 1, Mathematics Q16: Basic Concepts of Statistics

JEE Main2025Jan 23, Shift 1Mathematics

Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is

A
$48$
B
$44$
C
$40$
D
$52$

Solution

Median formula for grouped data: $Median = ℓ + \frac{N /2 - F}{f} \cdot h$ , where $ℓ$ is the lower class boundary of the median class, $F$ is the cumulative frequency just before the median class, $f$ is the median-class frequency, and $h$ is the class width.

Here $ℓ = 12$ , $F = 18$ , $f = 12$ , $h = 6$ :

$14 = 12 + \frac{N /2 - 18}{12} \cdot 6 \Rightarrow 2 = \frac{N /2 - 18}{2} \Rightarrow N /2 - 18 = 4 \Rightarrow N = 44$ .

Concept behind this question

Basic Concepts of StatisticsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 1, Mathematics Q16: Basic Concepts of Statistics

Concept behind this question

More previous year questions on Basic Concepts of Statistics

Practice more Mathematics