Can someone please explain the answers for 2019 nht exam 2 mcq 19 and 20?
Answers are D and B
Question 19This is equivalent to having a margin of error of no more than 1 g. Where \(a>0\), \[\Pr(-a<Z<a)=0.98\implies a=-2.3263...,\] and so \[2.3263...\frac{3}{\sqrt{n}}\leq 1\implies n_{\text{min}}=49.\]
Question 20Let \(X_1,\,X_2\sim\mathcal{N}(0.875,\ 0.188^2)\) be independent random variables that denote the nitrogen oxide emissions of cars. We're looking for \[\Pr(|X_1-X_2|\geq 0.5).\] We'll find the distribution of \(X_1-X_2\): \[\text{E}(X_1-X_2)=0.875-0.875=0\]\[\text{sd}(X_1-X_2)=\sqrt{0.188^2+0.188^2}=0.26587...\]Thus, we have \begin{align*}\Pr(|X_1-X_2|\geq 0.5)&=\Pr\left(\left|\frac{(X_1-X_2)-0}{0.26587...}\right|\geq\frac{0.5-0}{0.26587...}\right)\\ &=\Pr(|Z|\geq 1.88060...)\\ &=1-\Pr(-1.88060...\leq Z\leq 1.88060...)\\ &=0.060026...\end{align*}