Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-2 Probability and Statistics — National Examination, May 2018. Closed book (approved calculator + one hand-written information sheet); any 5 of 8 questions constitute a complete paper (only 5 marked) — all 8 are solved below as a complete study resource.
Reference texts: Walpole, Myers, Myers & Ye, Probability and Statistics for Engineers and Scientists (9th ed.), Pearson — used throughout for point-estimation, confidence-interval, and hypothesis-testing formulas; Montgomery & Runger, Applied Statistics and Probability for Engineers, for the regression and correlation-inference formulas in Question 8.
Given. (A) Daily claim count $X\sim\text{Poisson}(\lambda=4)$; a two-day period has $\lambda_2=8$. (B) Support for the fee increase is Binomial: (a) $n=15,\ p=0.55$; (b) $n=1{,}000,\ p=0.55$.
Find. (A)(a) $P(X<4)$; (A)(b) $P(6
Approach. Sum Poisson pmf terms for the exact discrete probabilities in (A); sum Binomial pmf terms for the small-$n$ case in (B)(a); switch to the normal approximation with continuity correction for the large-$n$ case in (B)(b).
A(a) Fewer than four claims in a day. With $\lambda=4$, $p(x)=e^{-\lambda}\lambda^x/x!$:
$$P(X<4)=\sum_{x=0}^{3}\frac{e^{-4}4^x}{x!}=e^{-4}\left(1+4+8+\tfrac{64}{6}\right)=\boxed{0.4335}.$$
A(b) 6–9 claims over a two-day period. Over 2 independent days, $\lambda_2=2(4)=8$:
$$P(6
B(a) 7–9 in favour out of 15. With $Y\sim\text{Bin}(15,0.55)$:
$$P(6
B(b) Fewer than 530 of 1,000 in favour (normal approximation). Since $np=550$ and $n(1-p)=450$ are both large, $W\approx N(np,\ np(1-p))$:
$$\mu_W=1{,}000(0.55)=550,\qquad \sigma_W=\sqrt{1{,}000(0.55)(0.45)}=15.73.$$
With the continuity correction, $P(W<530)\approx P(W\le 529.5)$:
$$z=\frac{529.5-550}{15.73}=-1.30,\qquad P(W<530)\approx\Phi(-1.30)=\boxed{0.0963}.$$
