JEE Main 2025 23 January Morning Shift
Exam JEE Main
Year 2025
Date 23 January
Shift Morning Shift
Subject Mathematics
Chapter Statistics
JEE Main 2025 · 23 January · Morning Shift · Mathematics · Statistics
Marks obtains by all the students of class 12 are presented in a...
Marks obtains by all the students of class 12 are presented in a freqency 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
Options A 52 52 52 B 48 48 48 C 44 44 44 D 40 40 40
Step-by-step solution
median = ℓ + ( N 2 − F f ) × h = 12 + ( N 2 − 18 12 ) × 6 = 14 ⇒ ( N 2 − 18 12 ) × 6 = 2 N 2 − 18 = 4 ⇒ N = 44 \begin{aligned} & \text{ median }=\ell+\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{\mathrm{f}}\right) \times \mathrm{h} \\ & =12+\left(\frac{\frac{\mathrm{N}}{2}-18}{12}\right) \times 6=14 \\ & \Rightarrow\left(\frac{\frac{\mathrm{ N}}{2}-18}{12}\right) \times 6=2 \\ & \frac{\mathrm{ N}}{2}-18=4 \Rightarrow \mathrm{ N}=44\end{aligned} median = ℓ + ( f 2 N − F ) × h = 12 + ( 12 2 N − 18 ) × 6 = 14 ⇒ ( 12 2 N − 18 ) × 6 = 2 2 N − 18 = 4 ⇒ N = 44
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