Answer: 0.001824

## Explanation This is a **population variance** calculation since we have data for the entire population (all 5 managers). ### Step 1: Calculate the population mean (μ) \[ \mu = \frac{\sum X_i}{N} = \frac{0.24 + 0.26 + 0.30 + 0.18 + 0.20}{5} = \frac{1.18}{5} = 0.236 \] ### Step 2: Calculate squared deviations from the mean \[ (0.24 - 0.236)^2 = (0.004)^2 = 0.000016 \] \[ (0.26 - 0.236)^2 = (0.024)^2 = 0.000576 \] \[ (0.30 - 0.236)^2 = (0.064)^2 = 0.004096 \] \[ (0.18 - 0.236)^2 = (-0.056)^2 = 0.003136 \] \[ (0.20 - 0.236)^2 = (-0.036)^2 = 0.001296 \] ### Step 3: Sum the squared deviations \[ 0.000016 + 0.000576 + 0.004096 + 0.003136 + 0.001296 = 0.00912 \] ### Step 4: Calculate population variance \[ \sigma^2 = \frac{\sum (X_i - \mu)^2}{N} = \frac{0.00912}{5} = 0.001824 \] **Key Points:** - This is **population variance** (divide by N = 5) - For **sample variance**, we would divide by (n-1) = 4 - The result 0.001824 corresponds to option A

Author: Tanishq Prabhu