1. **State the problem:** We want to find the probability that the mean number of hours 20 children watch television is greater than 26.3 hours, given the average is 25 hours and the standard deviation is 3 hours. 2. **Identify the distribution and parameters:** The variable is normally distributed with mean $\mu = 25$ hours and standard deviation $\sigma = 3$ hours. 3. **Calculate the standard error of the mean (SEM):** Since the sample size is $n=20$, the SEM is $$ SEM = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{20}} = \frac{3}{\cancel{\sqrt{20}}} = 0.6708 $$ 4. **Find the z-score for the sample mean 26.3:** $$ z = \frac{\bar{x} - \mu}{SEM} = \frac{26.3 - 25}{0.6708} = \frac{1.3}{0.6708} = 1.937 $$ 5. **Find the probability that the mean is greater than 26.3:** Using standard normal distribution tables or a calculator, the probability that $Z > 1.937$ is approximately 0.0264. **Final answer:** The probability that the mean number of hours watched is greater than 26.3 is **0.0264**.