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

The Correct Option is D

Calculate the cumulative frequency first.

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 \( N = 36 \), resulting in \( \frac{N}{2} = 18 \).

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

Identify the following parameters: Lower limit \( l = 8 \), Frequency \( f = 10 \), Cumulative frequency of the class preceding the median class \( C = 12 \), and Class width \( h = 4 \).

Apply the median formula:

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

Substitute the values into the formula:

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

Subsequently,

\[ 20M = 20 \times 10.4 = 208 \].