Binomial, Normal and Poisson Approximations: 04-BS-2, May 2013, Q2

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. Part A: p(against) = 0.15. Part B: p(sued) = 0.001, n = 3,000.

Find. (A)(a) P(X>2), n=14, exact binomial; (A)(b) P(Y>9), n=12, p(favour)=0.85, exact binomial; (A)(c) P(X<5,256), n=6,150, normal approximation; (B) P(X>2), n=3,000, Poisson approximation, and a justification.

Approach. Use the exact Binomial pmf when n is small enough to sum directly (A(a),(b)); switch to the Normal approximation to the Binomial (with continuity correction) once n is large and np, nq are both >5 (A(c)); switch to the Poisson approximation when n is large but p is small so np stays moderate (B).

2A(a) Exact binomial, n=14, p=0.15. $X\sim\text{Bin}(14,0.15)$. $$P(X>2)=1-P(X\le 2)=1-\sum_{k=0}^{2}\binom{14}{k}(0.15)^k(0.85)^{14-k}=1-0.6479=\boxed{0.3521}.$$
2A(b) Exact binomial, n=12, "in favour" p=0.85. $Y\sim\text{Bin}(12,0.85)$, favour-probability $q=1-0.15=0.85$. $$P(Y>9)=P(Y=10)+P(Y=11)+P(Y=12)=\boxed{0.7358}.$$
2A(c) Normal approximation, n=6,150. With $p=0.85$ (favour), $\mu=np=6150(0.85)=5227.5$ and $\sigma=\sqrt{npq}=\sqrt{6150(0.85)(0.15)}=28.00$. Applying a continuity correction for "fewer than 5,256" ($X\le 5255$): $$Z=\dfrac{5255.5-5227.5}{28.00}=1.00,\qquad P(X<5256)\approx\Phi(1.00)=\boxed{0.8413}.$$
2B Poisson approximation, n=3,000, p=0.001. $\lambda=np=3000(0.001)=3.0$. Using $X\sim\text{Poisson}(3)$, $$P(X>2)=1-P(X\le 2)=1-e^{-3}\left(1+3+\dfrac{3^2}{2!}\right)=1-e^{-3}(8.5)=1-0.4232=\boxed{0.5768}.$$ This approximation is appropriate because n = 3,000 is large while p = 0.001 is very small, giving a moderate mean $\lambda=np=3$; the Poisson is the limiting form of the Binomial exactly under $n\to\infty,\,p\to 0,\,np$ fixed, and reproduces the exact Binomial probability to four decimal places here while being far cheaper to sum than $\binom{3000}{k}$ terms.

Question 2 — final results
Part	Quantity	Value
A(a)	P(X>2), n=14,p=.15	0.3521
A(b)	P(Y>9), n=12,p=.85	0.7358
A(c)	P(X<5,256), n=6,150	0.8413
B	P(X>2), λ=3	0.5768

Topic: Discrete-distribution approximations — when to switch from exact Binomial to Normal or Poisson.

Key relations: Binomial pmf $\binom{n}{k}p^kq^{n-k}$; Normal approx $Z=\dfrac{X-np}{\sqrt{npq}}$ (valid once $np,nq>5$); Poisson approx with $\lambda=np$ (valid once $n\ge 20$ and $p\le 0.05$, or $n\ge100,\,np\le10$).

Why this works: the Binomial is a sum of n independent Bernoulli trials; when n is large and p moderate the CLT drives it toward Normal, but when p is small the distribution stays right-skewed and the Poisson limit (which preserves that skew) is the better match.

Common pitfall: using the Normal approximation with a small np or nq (skew still large, poor fit) or omitting the continuity correction, which shifts the boundary by 0.5 and materially changes borderline probabilities like 2A(c).

Source: Walpole, Myers, Myers & Ye, Ch. 5 (Discrete Probability Distributions), Ch. 6 (Normal Approximation to the Binomial).

Question 2 of 8: Binomial, Normal and Poisson Approximations

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