Binomial/Normal-Approximation Survey and Hypergeometric Sampling: 04-BS-2, December 2018, Q2

Question 2: Binomial/Normal-Approximation Survey and Hypergeometric Sampling (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: $p=0.70$ satisfied; sample sizes $n=12$ and $n=2000$. Part B: lot of $N=16$ solar lights; (a) $K=4$ substandard, sample $n=8$; (b) $K=6$ substandard (10 standard), sample $n=4$.

Find. A(a) P(6<X<10), Binomial(12,0.7). A(b) P(X>1430), Binomial(2000,0.7) via normal approximation. B(a) P(at most 2 substandard in 8), Hypergeometric(16,4,8). B(b) P(more than 2 standard in 4), Hypergeometric(16,10,4).

Approach. A(a) sum individual Binomial terms; A(b) apply the normal approximation to the binomial with a continuity correction; B(a)/(b) sum individual Hypergeometric terms (sampling without replacement from a finite lot).

A(a) Binomial, exact sum. $X\sim\text{Binomial}(n=12,p=0.7)$. “More than six but fewer than ten” means $X\in\{7,8,9\}$: $$P(7\le X\le9)=\sum_{k=7}^{9}\binom{12}{k}(0.7)^{k}(0.3)^{12-k}=0.1585+0.2311+0.2397=0.6293$$ $$\boxed{P(6\lt X\lt 10)=0.6293}$$
A(b) Normal approximation to the Binomial. $n=2000,\ p=0.7$: $\mu=np=1400$, $\sigma^2=np(1-p)=420$, $\sigma=20.494$. With the continuity correction for $P(X\gt 1430)$: $$z=\dfrac{1430.5-1400}{20.494}=1.488$$ $$P(X\gt 1430)\approx P(Z\gt 1.488)=1-0.9316=0.0683$$ $$\boxed{P(X\gt 1430)\approx0.0683}$$
B(a) Hypergeometric — at most 2 non-functioning. Non-functioning = substandard; $N=16$, $K=4$ substandard, $n=8$ drawn. $$P(X\le2)=\sum_{k=0}^{2}\dfrac{\binom{4}{k}\binom{12}{8-k}}{\binom{16}{8}}$$ Evaluating term by term: $P(0)=0.0385$, $P(1)=0.2462$, $P(2)=0.4308$: $$\boxed{P(X\le2)=0.7154}$$
B(b) Hypergeometric — more than 2 standard. Now $N=16$ with $K=10$ standard units, sample $n=4$: $$P(X\gt 2)=P(3)+P(4)=\dfrac{\binom{10}{3}\binom{6}{1}}{\binom{16}{4}}+\dfrac{\binom{10}{4}\binom{6}{0}}{\binom{16}{4}}=0.3956+0.1154$$ $$\boxed{P(X\gt 2)=0.5110}$$

Question 2 — final results
Part	Quantity	Result
A(a)	P(6<X<10), n=12,p=0.7	0.6293
A(b)	P(X>1430), n=2000,p=0.7 (normal approx.)	0.0683
B(a)	P(at most 2 substandard of 8, from 16 with 4 bad)	0.7154
B(b)	P(more than 2 standard of 4, from 16 with 10 good)	0.5110

Topic: Binomial, its normal approximation, and the Hypergeometric distribution.

Key relations: Binomial pmf $\binom{n}{k}p^k(1-p)^{n-k}$; normal approximation $Z=\dfrac{X-np\pm0.5}{\sqrt{np(1-p)}}$; Hypergeometric pmf $\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$.

Why this works: Binomial sampling assumes replacement/independence (valid for a large population like city adults); the normal approximation is accurate once $np$ and $n(1-p)$ are both comfortably above 5. The Hypergeometric instead models sampling WITHOUT replacement from a small finite lot, so the success probability changes after each draw — it must be used whenever the sample is a non-trivial fraction of a small population.

Common pitfall: Forgetting the continuity correction in the normal approximation (shifts the answer by a few hundredths); using the Binomial instead of the Hypergeometric for the solar-light lot, which over/understates the true probability because it ignores the shrinking population.

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, Ch. 5 (Discrete Distributions) & Ch. 6 (Normal Approximation).

Question 2 of 8: Binomial/Normal-Approximation Survey and Hypergeometric Sampling

Question 2: Binomial/Normal-Approximation Survey and Hypergeometric Sampling (20 marks)