BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: You are given:
(i) \(\quad X_{\text {partial }}=\) pure premium calculated from partially credible data
(ii) \(\quad \mu=E\left[X_{\text {partial }}\right]\)
(iii) Fluctuations are limited to \(\pm k \mu\) of the mean with probability \(P\)
(iv) \(Z=\) credibility factor

Determine which of the following is equal to \(P\).

Options:
A
\(\operatorname{Pr}\left[\mu-k \mu \leq X_{\text {partial }} \leq \mu+k \mu\right]\)
B
\(\quad \operatorname{Pr}\left[\mathrm{Z} \mu-k \leq Z X_{\text {partial }} \leq Z \mu+k\right]\)
C
\(\operatorname{Pr}\left[\mathrm{Z} \mu-\mu \leq Z X_{\text {partial }} \leq Z \mu+\mu\right]\)
D
\(\quad \operatorname{Pr}\left[1-k \leq Z X_{\text {partial }}+(1-Z) \mu \leq 1+k\right]\)
E
\(\operatorname{Pr}\left[\mu-k \mu \leq Z X_{\text {partial }}+(1-Z) \mu \leq \mu+k \mu\right]\)

Correct answer: E

Solution: The absolute difference of the credibility estimate from its expected value is to be less than or equal to \(k \mu\) (with probability \(P\) ). That is,
\(\left|\left[Z X_{\text {partial }}+(1-Z) M\right]-[Z \mu+(1-Z) M]\right| \leq k \mu\)
\(-k \mu \leq Z X_{\text {partial }}-Z \mu \leq k \mu\).
Adding \(\mu\) to all three sides produces answer choice (E).
Question: In a clinic, physicians volunteer their time on a daily basis to provide care to those who are not eligible to obtain care otherwise. The number of physicians who volunteer in any day is uniformly distributed on the integers 1 through 5 . The number of patients that can be served by a given physician has a Poisson distribution with mean 30.

Determine the probability that 120 or more patients can be served in a day at the clinic, using the normal approximation with continuity correction.

Options:
A
\(1-\Phi(0.68)\)
B
\(1-\Phi(0.72)\)
C
\(1-\Phi(0.93)\)
D
\(1-\Phi(3.13)\)
E
\(1-\Phi(3.16)\)

Correct answer: A

Solution: \(N=\) number of physicians \(\quad E(N)=3 \quad \operatorname{Var}(N)=2\)
\(X=\) visits per physician \(\quad \mathrm{E}(X)=30 \quad \operatorname{Var}(X)=30\)
\(S=\) total visits
\(E(S)=E(N) E(X)=3(30)=90\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=3(30)+900(2)=1890\)
\(S D(S)=\sqrt{1890}=43.5\)
\(\operatorname{Pr}(S>119.5)=\operatorname{Pr}\left(Z>\frac{119.5-90}{43.5}=0.68\right)=1-\Phi(0.68)\)