Question

Let M denote the median of the following frequency distribution.$x_i$	$f_i$
0 - 4	2
4 - 8	4
8 - 12	7
12 - 16	8
16 - 20	6
Then 20M is equal to:

• A. 416
• B. 104
• C. 52
• D. 208

Answer 1

Answer: (D)

208

Answer 2

Solution: First, calculate the cumulative frequency.

Class	Frequency	Cumulative Frequency
0-4	3	3
4-8	9	12
8-12	10	22
12-16	8	30
16-20	6	36

The total frequency $N = 36$, so $\frac{N}{2} = 18$.

The median class is 8-12, as it is the class where the cumulative frequency first exceeds 18.

Lower limit $l = 8$ Frequency $f = 10$ Cumulative frequency of the class before the median class $C = 12$ Class width $h = 4$

Using the median formula:

$M = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h$

Substitute the values:

$M = 8 + \left( \frac{18 - 12}{10} \right) \times 4$ $= 8 + \left( \frac{6}{10} \right) \times 4$ $= 8 + 0.6 \times 4$ $= 8 + 2.4 = 10.4$

Then,

$20M = 20 \times 10.4 = 208$.