= Solution
{c}
Let $A=\{|Z|>2\}$ and $B=\{Z>0\}$. The <standard normal distribution> is symmetric and assigns no <probability mass> to zero. Thus $\mathbb P(B)=1/2$, and its equal two-sided tails give
$$
\mathbb P(A)=2\mathbb P(Z>2),\qquad
\mathbb P(A\cap B)=\mathbb P(Z>2).
$$
Therefore
$$
\boxed{\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B),}
$$
so \b[the two <events> are <independent>]. The permitted approximation gives $\mathbb P(A)\approx0.05$ and $\mathbb P(A\cap B)\approx0.025$, but <independence> is exact.
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