Question

The variance of the following continuous frequency distribution is:

class interval	Frequency
0 -- 4	2
4 -- 8	3
8 -- 12	2
12 -- 16	1

 

• A. \(\frac{128}{7}\)
• B. 15
• C. 19
• D. \(\frac{130}{7}\)

Hint:

In grouped frequency data, use the midpoints for each class interval to approximate the mean and variance.
- Variance formula: \(\sigma^2 = E[X^2] - (E[X])^2\).

Answer 1

The Correct Option is B

Step 1: Class midpoints and frequencies.
For each class interval, the midpoint \((m)\) is: \[ 0\!-\!4:\,m=2, \quad 4\!-\!8:\,m=6, \quad 8\!-\!12:\,m=10, \quad 12\!-\!16:\,m=14. \] Frequencies \((f)\) are 2, 3, 2, 1 respectively. The total frequency \(N = 2+3+2+1=8.\) 

Step 2: Calculate the mean \(\overline{x}\).
\[ \overline{x} = \frac{\sum f\,m}{N} = \frac{2\cdot 2 + 3\cdot 6 + 2\cdot 10 + 1\cdot 14}{8} = \frac{4 + 18 + 20 + 14}{8} = \frac{56}{8} = 7. \] 

Step 3: Compute \(\sum f\,m^2\) and then variance.
\[ \sum f\,m^2 = 2\cdot 2^2 \;+\; 3\cdot 6^2 \;+\; 2\cdot 10^2 \;+\; 1\cdot 14^2 = 2\cdot 4 + 3\cdot 36 + 2\cdot 100 + 196 = 8 + 108 + 200 + 196 = 512. \] Hence \[ E[X^2] = \frac{\sum f\,m^2}{N} = \frac{512}{8} = 64. \] The variance is \[ \sigma^2 = E[X^2] - (\overline{x})^2 = 64 \;-\; 7^2 = 64 -49 = 15. \] Therefore, the variance is \(\boxed{15}.\)