We are dealing with proportions, so the data follows a binomial distribution:
$$
X \sim \operatorname{Binomial}(n, p)
$$
where
- $n=100$ (number of fuses tested),
- $p=$ true proportion of functioning fuses.
From binomial theory:
$$
E[\hat{p}]=p \quad \text { and } \quad \operatorname{Var}(\hat{p})=\frac{p(1-p)}{n}
$$
When $n$ is large (say $n p>5$ and $n(1-p)>5$ ), the sampling distribution of $\hat{p}$ can be approximated by a normal distribution:
$$
\hat{p} \sim N\left(p, \frac{p(1-p)}{n}\right)
$$
The general $z$-formula is:
$$
\begin{aligned}
& z=\frac{x-\mu}{\sigma / \sqrt{n}} \\
& z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}
\end{aligned}
$$
$$
\begin{aligned}
& p_0\left(1-p_0\right)=0.9(0.1)=0.09 \\
& \frac{0.09}{n}=\frac{0.09}{100}=0.0009 \\
& \sqrt{0.0009}=0.03 \\
& \hat{p}-p_0=0.92-0.90=0.02 \\
& z=\frac{0.02}{0.03}=0.67 \\
& 0.67<1.28 \text { (for } \alpha=0.10) \rightarrow \text { Fail to reject } H_0 \\
& 0.67<1.645 \text { (for } \alpha=0.05 \text { ) } \rightarrow \text { Fail to reject } H_0
\end{aligned}
$$
Hence, A and C are correct
