Question

Variance of the following discrete frequency distribution is: 
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]

• A. \( \frac{463}{15} \)
• B. \( \frac{838}{15} \)
• C. \( \frac{44}{5} \)
• D. \( \frac{88}{15} \) Correct Answer

Hint:

For a grouped frequency distribution: 1. Find mid-points (\(x_i\)) of class intervals. 2. Calculate the mean \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \). 3. Calculate the variance \( \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \bar{x}^2 \) or \( \sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} \). The first formula for variance is often computationally easier.

Answer 1

The Correct Option is D

Step 1: Find the mid-points (\(x_i\)) of each class interval.
\begin{itemize} \item 0-2: \(x_1 = (0+2)/2 = 1\) \item 2-4: \(x_2 = (2+4)/2 = 3\) \item 4-6: \(x_3 = (4+6)/2 = 5\) \item 6-8: \(x_4 = (6+8)/2 = 7\) \item 8-10: \(x_5 = (8+10)/2 = 9\) \end{itemize}
Step 2: Calculate the total frequency \( N = \sum f_i \).
\( N = 2+3+5+3+2 = 15 \).

Step 3: Calculate \( \sum f_i x_i \).
\begin{itemize} \item \(f_1 x_1 = 2 \times 1 = 2\) \item \(f_2 x_2 = 3 \times 3 = 9\) \item \(f_3 x_3 = 5 \times 5 = 25\) \item \(f_4 x_4 = 3 \times 7 = 21\) \item \(f_5 x_5 = 2 \times 9 = 18\) \end{itemize} \( \sum f_i x_i = 2+9+25+21+18 = 75 \).

Step 4: Calculate the mean \( \bar{x} \).
\( \bar{x} = \frac{\sum f_i x_i}{N} = \frac{75}{15} = 5 \).

Step 5: Calculate \( \sum f_i x_i^2 \).
\begin{itemize} \item \(x_1^2 = 1^2 = 1 \implies f_1 x_1^2 = 2 \times 1 = 2\) \item \(x_2^2 = 3^2 = 9 \implies f_2 x_2^2 = 3 \times 9 = 27\) \item \(x_3^2 = 5^2 = 25 \implies f_3 x_3^2 = 5 \times 25 = 125\) \item \(x_4^2 = 7^2 = 49 \implies f_4 x_4^2 = 3 \times 49 = 147\) \item \(x_5^2 = 9^2 = 81 \implies f_5 x_5^2 = 2 \times 81 = 162\) \end{itemize} \( \sum f_i x_i^2 = 2+27+125+147+162 = 463 \).

Step 6: Calculate the variance \( \sigma^2 \).
The formula for variance is \( \sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 \).
\[ \sigma^2 = \frac{463}{15} - (5)^2 = \frac{463}{15} - 25 \] \[ \sigma^2 = \frac{463}{15} - \frac{25 \times 15}{15} = \frac{463 - 375}{15} \] \[ \sigma^2 = \frac{88}{15} \] This matches option (4).