Question

The standard deviation for the following frequency is 

• A. 3.20
• B. 3.22
• C. 3.26
• D. 3.28

Hint:

To find the standard deviation for a frequency distribution, first calculate the midpoints of each class, then apply the formula for variance, and finally take the square root to get the standard deviation.

Answer 1

The Correct Option is C

To calculate the standard deviation for a frequency distribution, we use the following steps:

(1) Find the midpoints of each class interval:
- For 0-4, midpoint = \( \frac{0 + 4}{2} = 2 \)
- For 4-8, midpoint = \( \frac{4 + 8}{2} = 6 \)
- For 8-12, midpoint = \( \frac{8 + 12}{2} = 10 \)
- For 12-16, midpoint = \( \frac{12 + 16}{2} = 14 \)

(2) Now, calculate the mean \( \mu \):
- \( \mu = \frac{\sum (f \cdot x)}{\sum f} \)
- Here, \( f \) is the frequency and \( x \) is the midpoint.
First calculate \( f \cdot x \) for each class:
- For 0-4: \( 4 \times 2 = 8 \)
- For 4-8: \( 8 \times 6 = 48 \)
- For 8-12: \( 2 \times 10 = 20 \)
- For 12-16: \( 1 \times 14 = 14 \)
Now, sum these values:
- \( \sum f = 4 + 8 + 2 + 1 = 15 \)
- \( \sum (f \cdot x) = 8 + 48 + 20 + 14 = 90 \)
Thus, the mean \( \mu = \frac{90}{15} = 6 \).

(3) Next, calculate the variance \( \sigma^2 \):
- Use the formula \( \sigma^2 = \frac{\sum f \cdot (x - \mu)^2}{\sum f} \).
First, calculate \( (x - \mu)^2 \) for each class:
- For 0-4: \( (2 - 6)^2 = 16 \)
- For 4-8: \( (6 - 6)^2 = 0 \)
- For 8-12: \( (10 - 6)^2 = 16 \)
- For 12-16: \( (14 - 6)^2 = 64 \)
Now, calculate \( f \cdot (x - \mu)^2 \):
- For 0-4: \( 4 \times 16 = 64 \)
- For 4-8: \( 8 \times 0 = 0 \)
- For 8-12: \( 2 \times 16 = 32 \)
- For 12-16: \( 1 \times 64 = 64 \)
Sum these values: - \( \sum f \cdot (x - \mu)^2 = 64 + 0 + 32 + 64 = 160 \) The variance is: - \( \sigma^2 = \frac{160}{15} \approx 10.67 \) 
(4) Finally, calculate the standard deviation \( \sigma \):
- \( \sigma = \sqrt{10.67} \approx 3.26 \) 
Conclusion: The standard deviation for the given frequency distribution is \( \boxed{3.26} \).