 Solution

Let $I_j$ indicate a point outside its pointwise 95% limits. Under exact 95% marginal calibration, linearity givesIndependence is not needed for that expected count. If the eight hospital observations are independent, the probability of at least one false alarm isThus 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 ${\sum }_j{q}_j$ and, under independence, the exact probability is $1-{\prod }_j(1-q_j)$.