= Solution
{c}
An independent sum of normal variables is normal, with the means and variances added. Therefore $\overline X\sim N(\mu,\sigma^2/n)$ and \b[$Z\sim N(0,1)$] exactly.
Under the null, $Z_0=\sqrt n(\overline X-\mu_0)/\sigma$ is standard normal. Reject for $Z_0>w$, where $w=\Phi^{-1}(1-\alpha)$, giving size $\alpha$. For $\alpha=0.05$, the supplied table gives \b[the critical region $\overline X>\mu_0+1.64\sigma/\sqrt n$], using its rounded quantile.
The <power function> is the probability of rejection when the true parameter is $\mu$. Recenter by that true <mean>:
$$
Q(\mu)=\mathbb P_\mu\left(\frac{\sqrt n(\overline X-\mu)}\sigma>w-\frac{\sqrt n(\mu-\mu_0)}\sigma\right)
=1-\Phi\left(w-\frac{\sqrt n(\mu-\mu_0)}\sigma\right).
$$
For $n=100$, $\sigma=2$, power at least $0.70$ means $w-5(\mu-\mu_0)\leq-0.52$. Hence, with the supplied rounding,
$$
\boxed{\mu\geq\mu_0+0.432.}
$$
The exact unrounded threshold is $\mu_0+(\Phi^{-1}(0.95)-\Phi^{-1}(0.30))/5$.
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