Question 2 of 8: Binomial, Normal and Poisson Approximations

Question 2: Binomial, Normal and Poisson Approximations (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (A) X = number in favour of the fee increase, $X\sim\text{Binomial}(n,p{=}0.65)$ for varying $n$. (B) probability of malpractice suit $=0.0015$, $n=2{,}000$.

Find. (A)(a) $P(6\lt X\lt 10)$, $n=15$. (b) $P(X\gt 5)$, $n=8$. (c) $P(X\gt 1{,}030)$, $n=1{,}600$, via a suitable approximation. (B) $P(X\gt 2)$, $n=2{,}000$, via a suitable approximation, with justification.

Approach. Exact binomial sums for small $n$; a normal approximation with continuity correction for $n=1{,}600$ (np and n(1−p) both large); a Poisson approximation for the rare-event part B (n large, p small, np moderate).

1. (A)(a) n=15, more than six but fewer than 10 in favour. "More than six but fewer than 10" means $X\in\{7,8,9\}$. $$P(7\le X\le 9)=\sum_{k=7}^{9}\binom{15}{k}(0.65)^k(0.35)^{15-k}=0.0710+0.1319+0.1906$$ $$\boxed{P(6\lt X\lt 10)=0.3935}$$
2. (A)(b) n=8, more than five in favour. "More than five" means $X\in\{6,7,8\}$: $$P(X\gt 5)=\sum_{k=6}^{8}\binom{8}{k}(0.65)^k(0.35)^{8-k}=0.4278$$ $$\boxed{P(X\gt 5)=0.4278}$$
3. (A)(c) n=1,600, normal approximation with continuity correction. $\mu=np=1{,}600(0.65)=1{,}040$, $\sigma=\sqrt{np(1-p)}=\sqrt{1{,}600(0.65)(0.35)}=19.079$. Using the continuity correction, $$z=\dfrac{1{,}030.5-1{,}040}{19.079}=-0.498$$ $$P(X\gt 1{,}030)\approx P(Z\gt -0.498)=0.6907$$ $$\boxed{P(X\gt 1{,}030)\approx 0.6907}$$
4. (B) n=2,000, Poisson approximation for a rare event. Here $n=2{,}000$ is large and $p=0.0015$ is small, with $np=3.0$ a moderate value — the classic regime where the Binomial$(n,p)$ is well-approximated by a Poisson with $\lambda=np=3.0$ (the normal approximation would be poor here since $np=3$ is far below the usual $np\ge 5$ rule of thumb). $$P(X\gt 2)=1-\sum_{k=0}^{2}\dfrac{e^{-3}3^k}{k!}=1-0.4232=0.5768$$ $$\boxed{P(X\gt 2)\approx 0.5768}$$