Statistics – Median – 23 January 2025 (Shift 1)

Question ID: #248
JEE Main23 January Shift 1, 2025Statistics & Probability

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

  • (1) 48
  • (2) 44
  • (3) 40
  • (4) 52

Solution:

Median formula for grouped data:
$$\text{Median} = l + \frac{\left(\frac{N}{2} – cf\right)}{f} \times h$$

Here,
$l = 12,\; h = 6,\; f = 12,\; cf = 18,\; \text{Median} = 14$

Substitute values:
$$14 = 12 + \frac{\left(\frac{N}{2} – 18\right)}{12} \times 6$$

$$2 = \frac{\frac{N}{2} – 18}{2}$$
$$\frac{N}{2} – 18 = 4$$
$$\frac{N}{2} = 22$$
$$N = 44$$

Ans. (2)

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