1. **Stating the problem:** We have a population with unknown mean $\mu$ and variance $\sigma^2 = 400$. We want to test the null hypothesis $H_0: \mu = 100$ against the alternative hypothesis $H_a: \mu > 100$ at a 95% significance level. Given the true mean is $\mu = 105$, we want to find the probability of rejecting $H_0$ (i.e., the power of the test). 2. **Formula and important rules:** Since the sample is large, by the Central Limit Theorem, the sample mean $\overline{X}$ is approximately normally distributed with mean $\mu$ and variance $\sigma^2/n$, where $n$ is the sample size. The test statistic for the hypothesis test is: $$Z = \frac{\overline{X} - 100}{\sigma/\sqrt{n}}$$ We reject $H_0$ if $Z > z_{\alpha}$, where $z_{\alpha}$ is the critical value for the significance level $\alpha = 0.05$ in a one-tailed test. 3. **Find the critical value:** For $\alpha = 0.05$, the critical value from the standard normal distribution is: $$z_{0.05} = 1.645$$ 4. **Calculate the rejection region in terms of $\overline{X}$:** $$Z > 1.645 \implies \frac{\overline{X} - 100}{\sigma/\sqrt{n}} > 1.645$$ $$\overline{X} > 100 + 1.645 \times \frac{\sigma}{\sqrt{n}}$$ 5. **Calculate the power of the test:** The power is the probability of rejecting $H_0$ when $\mu = 105$: $$P\left(\overline{X} > 100 + 1.645 \times \frac{\sigma}{\sqrt{n}} \mid \mu = 105 \right)$$ Since $\overline{X} \sim N\left(105, \frac{400}{n}\right)$, standardize: $$P\left(Z > \frac{100 + 1.645 \times \frac{20}{\sqrt{n}} - 105}{20/\sqrt{n}}\right) = P\left(Z > \frac{100 - 105}{20/\sqrt{n}} + 1.645\right) = P\left(Z > -\frac{5\sqrt{n}}{20} + 1.645\right)$$ 6. **Final expression for power:** $$\text{Power} = 1 - \Phi\left(1.645 - \frac{5\sqrt{n}}{20}\right)$$ where $\Phi$ is the standard normal cumulative distribution function. **Note:** The sample size $n$ is not given, so the power depends on $n$. If $n$ is known, substitute it to compute the numerical value. **Summary:** The probability of rejecting $H_0$ given $\mu=105$ is the power of the test, calculated by the formula above.