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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 :

JEE · Math · previous-year question

  1. A.52
  2. B.44correct
  3. C.40
  4. D.48

Answer

B. 44

Explanation

The median for grouped data is given by: $$ \text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h $$ where $$L$$ is the lower limit (or boundary) of the median class. $$CF$$ is the cumulative frequency of all classes preceding the median class. $$f$$ is the frequency of the median class. $$h$$ is the class width. $$n$$ is the total number of students. Given: Median $$= 14$$ Median class interval is $$12-18$$, so $$L = 12$$ and the class width $$h = 18 - 12 = 6$$. Frequency of median class $$f = 12$$. Cumulative frequency below the median class $$= 18$$ (i.e., $$CF = 18$$). Plugging these into the formula: $$ 14 = 12 + \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6 $$ Step 1: Subtract 12 from both sides: $$ 2 = \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6 $$ Step 2: Simplify the multiplication factor: $$ \left(\frac{6}{12}\right) = \frac{1}{2} $$ So the equation becomes: $$ 2 = \frac{1}{2}\left(\frac{n}{2} - 18\right) $$ Step 3: Multiply both sides by 2 to remove the fraction: $$ 4 = \frac{n}{2} - 18 $$ Step 4: Solve for $$\frac{n}{2}$$: $$ \frac{n}{2} = 4 + 18 = 22 $$ Step 5: Multiply both sides by 2 to find $$n$$: $$ n = 44 $$ Thus, the total number of students is $$44$$.

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