= Solution
Let $I_j$ indicate a point outside its pointwise 95% limits. Under exact 95% marginal calibration, linearity gives
$$
\boxed{\mathbb E\sum_{j=1}^8I_j=8(0.05)=0.4.}
$$
<Independence> is not needed for that expected count. If the eight hospital observations are independent, the <probability> of at least one false alarm is
$$
\boxed{1-(1-0.05)^8=1-0.95^8\approx0.3366.}
$$
Thus nominal pointwise 95% coverage is quite different from 95% simultaneous coverage of the whole display, a <familywise error rate> issue. Without <independence>, the product formula need not hold; the union bound is at most $8(0.05)=0.4$.
Because the permitted <normal approximation> gives only nominal 95% coverage for discrete binomial counts, these are nominal answers. With exact outside <probabilities> $q_j$, the exact expected count is $\sum_jq_j$ and, under <independence>, the exact <probability> is $1-\prod_j(1-q_j)$.
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