Q43 / 90 · JEE Main 2024 (30 Jan Shift 1) · Mathematics

Let M denote the median of the following frequency distribution.

 Class 0-44-88-1212-1616-20 Frequency 391086

Then 20M is equal to :

  • A
    416
  • B
    104
  • C
    52
  • Correct answer
    D
    208

Solution

\(\begin{array}{|c|c|c|} \hline \text {Class } & \text {Frequency } & \text {Cumulative frequency } \\ \hline 0-4 & 3 & 3 \\ \hline 4-8 & 9 & 12 \\ \hline 8-12 & 10 & 22 \\ \hline 12-16 & 8 & 30 \\ \hline 16-20 & 6 & 36 \\ \hline \end{array}\)

M=l+N2-Cfh

Here, N=3+9+10+8+6=36

⇒N2=18

So, median class is 8-12

⇒l=8, C=22, f=10, h=4

⇒M=8+18-1210×4

⇒M=10.4

⇒20 M =208

Hence, the correct answer is option: 'D'.

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