Statistics
Median of Frequency Distribution
nta_pyq_2024_jan
Grade 11

Question:

Let $M$ denote the median of the following frequency distribution: \begin{array}{|c|c|c|c|c|c|}\hline\text{Class}&0-4&4-8&8-12&12-16&16-20\\\hline\text{Frequency}&3&9&10&8&6\\\hline\end{array} Then $20M$ is equal to:
416
104
52
208

Step-by-Step Solution

Key Concept: Total $N=36$. Median class: $N/2=18$, cumulative frequencies: 3, 12, 22. So median class is $8-12$ (cf before $=12$, $f=10$). $M=8+\frac{18-12}{10}\times4=8+2.4=10.4$.
$M=8+\frac{6}{10}\times4=10.4$. $20M=208$.
Correct Answer: 4

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