We are given the following information:
We can calculate the mean \( \bar{x} \) using the formula:
\[ \bar{x} = \frac{\sum_{i=1}^{50} x_i}{50} = \frac{650}{50} = 13 \]
Next, we use the formula for variance:
\[ \text{Variance} = \frac{1}{n} \left( \sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n} \right) \] Substitute the known values into the formula: \[ \text{Variance} = \frac{1}{50} \left( 10000 - \frac{650^2}{50} \right) \] First, calculate \( \frac{650^2}{50} \): \[ \frac{650^2}{50} = \frac{422500}{50} = 8450 \] Now, substitute this into the variance formula: \[ \text{Variance} = \frac{1}{50} \left( 10000 - 8450 \right) = \frac{1}{50} \times 1550 = 31 \]
Answer: 31
Variance of the following discrete frequency distribution is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]