Question 2 of 8: Binomial and Poisson Approximations
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2015 — 04-BS-2 Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator). Statistical tables of the Normal, t, chi-square and F distributions are supplied with the exam. The rubric states "Any 5 questions constitute a complete paper"; every question is answered.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (Pearson) — used throughout for normal/binomial/Poisson/hypergeometric probability, confidence intervals, hypothesis tests, and simple linear regression.
Check — source defect in Question 4. As literally printed, K/y cannot integrate
to a finite value over (1, ∞), so the density as given is not valid. This solution
reconstructs the intended density as f(y) = 1/y² for y > 1 — the unique
minimal correction (recovering one dropped exponent) that makes the function integrate to 1 on the stated
domain, and it is the standard "infinite mean" example from this course's own textbook (Walpole). This choice is disclosed here rather than presented as certain.
Question 2: Binomial and Poisson Approximations (20 marks)
Given. (A) Population proportion in favour $p=0.6$. (B) $p=0.0015$, $n=2000$.
Given data
Part
n
p
2(A)(a)
15
0.6
2(A)(b)
12
0.6 (favour) / 0.4 (not favour)
2(A)(c)
4,000
0.6
2(B)
2,000
0.0015
Find. Exact/approximate binomial probabilities for each scenario, and a justification of the Poisson approximation in 2(B).
Approach. Small $n$ → exact binomial sum; large $n$ with $np,\,n(1-p)$ both moderate → normal approximation with continuity correction; large $n$ with $p$ small and $np$ moderate → Poisson approximation.
2(A)(a) 6–8 in favour, exact binomial. "More than five but fewer than nine" means $X=6,7,8$ with $X\sim\text{Bin}(15,0.6)$:
$$P(6\le X\le8)=\sum_{k=6}^{8}\binom{15}{k}(0.6)^k(0.4)^{15-k}=0.1782+0.1782+0.1188=\boxed{0.3564}.$$
2(A)(b) fewer than three NOT in favour. Let $Y=$ number NOT in favour in a sample of 12, so $Y\sim\text{Bin}(12,\,0.4)$ (complementary probability). "Fewer than three" means $Y=0,1,2$:
$$P(Y\le2)=\binom{12}{0}(0.4)^0(0.6)^{12}+\binom{12}{1}(0.4)^1(0.6)^{11}+\binom{12}{2}(0.4)^2(0.6)^{10}=\boxed{0.0834}.$$
2(A)(c) normal approximation, n=4,000. Since $np=2400$ and $n(1-p)=1600$ are both large, $X\sim\text{Bin}(4000,0.6)\approx N\big(np,\ \sqrt{np(1-p)}\big)=N(2400,\ 30.98^2)$. With the continuity correction (P(X<2450) → $x=2449.5$):
$$z=\dfrac{2449.5-2400}{30.98}=1.598,\qquad P(X<2450)\approx\Phi(1.598)=\boxed{0.9446}.$$
2(B) Poisson approximation, n=2,000, p=0.0015. Here $p$ is small and $n$ large with $np=3$ of moderate size, so $X\sim\text{Bin}(2000,0.0015)\approx\text{Poisson}(\lambda=3)$. Then
$$P(X>2)=1-P(X\le2)=1-e^{-3}\!\left(1+3+\dfrac{3^2}{2}\right)=1-e^{-3}(8.5)=\boxed{0.5768}.$$
The Poisson approximation is appropriate here (and not the normal one) because $p=0.0015$ is very small while $n=2000$ is large and $np=3$ stays moderate — exactly the regime ("rare event, many trials") the Poisson limit theorem covers; a normal approximation would need $np$ and $n(1-p)$ both reasonably large, but $n(1-p)\approx1997$ is fine while $np=3$ is far too small for the normal shape to be accurate in the tail.
