1. The problem asks to calculate the variance by computing $ (x - \mu)^2 \times P(x) $ for each row and summing them up. 2. The variance formula is: $$\sigma^2 = \sum (x - \mu)^2 P(x)$$ where $x$ is each value, $\mu$ is the mean, and $P(x)$ is the probability of $x$. 3. Given values are already squared differences multiplied by probabilities: - $6641 \times 0.30 = 0.49923$ - $0841 \times 0.28 = 0.023548$ - $5041 \times 0.25 = 0.126025$ - $9241 \times 0.17 = 0.4971$ 4. Sum these values to find the variance: $$0.49923 + 0.023548 + 0.126025 + 0.4971 = 1.145903$$ 5. Therefore, the variance is approximately: $$\boxed{1.1459}$$ 6. Your calculations are correct and the sum gives the variance as expected.