04-BS-2, May 2016, Q2 | Nivaar Exam Prep

Question 2 (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) $p=0.25$ (no car); (a) $n=10$; (b) $n=1{,}800$. (B) a lot of $N=12$ lawnmowers, 4 substandard (8 good); (a) sample of 6 sold, without replacement; (b) $X$ = number substandard in a random sample of 6.

Find. (A)(a) $P(X>3)$; (A)(b) $P(X>420)$; (B)(a) $P(\text{at least 3 good of 6})$; (B)(b) the full pmf of $X$.

Approach. Small independent-trial counts use the exact Binomial model; the large-$n$ proportion problem uses the normal approximation to the binomial with a continuity correction; the lot of 12 lawnmowers is finite and sampled without replacement, so both parts of (B) use the Hypergeometric distribution.

(A)(a) Exact binomial, n=10. $X\sim\text{Bin}(10,0.25)$. $P(X>3)=1-P(X\le 3)=1-\sum_{k=0}^{3}\binom{10}{k}(0.25)^k(0.75)^{10-k}=1-0.7759=\boxed{0.2241}$.
(A)(b) Normal approximation to the binomial, n=1,800. $\mu=np=1{,}800(0.25)=450$, $\sigma=\sqrt{np(1-p)}=\sqrt{1{,}800(0.25)(0.75)}=18.371$. With the continuity correction, $P(X>420)=P\!\left(Z>\dfrac{420.5-450}{18.371}\right)=P(Z>-1.605)=\Phi(1.605)=\boxed{0.9458}$.
(B)(a) Hypergeometric, "at least 3 good of 6". Good count $G\sim\text{Hyper}(N=12,K=8\text{ good},n=6)$. $P(G\ge 3)=1-P(G\le 2)=1-\left[\dfrac{\binom{8}{0}\binom{4}{6}}{\binom{12}{6}}+\dfrac{\binom{8}{1}\binom{4}{5}}{\binom{12}{6}}+\dfrac{\binom{8}{2}\binom{4}{4}}{\binom{12}{6}}\right]$. The first two terms are structurally zero ($\binom{4}{6}=\binom{4}{5}=0$, since only 4 substandard units exist), leaving $P(G\le 2)=\dfrac{\binom{8}{2}\binom{4}{4}}{\binom{12}{6}}=\dfrac{28(1)}{924}=0.0303$, so $\boxed{P(G\ge3)=0.9697}$.
(B)(b) Hypergeometric pmf of X (substandard count). $X\sim\text{Hyper}(N=12,K=4\text{ substandard},n=6)$, $P(X=x)=\dfrac{\binom{4}{x}\binom{8}{6-x}}{\binom{12}{6}}$ for $x=0,1,2,3,4$ (max$(0,6-8)\le x\le\min(4,6)$).

Q2(B)(b) — probability distribution of X (substandard lawnmowers in a sample of 6)
x	0	1	2	3	4
P(X=x)	0.0303	0.2424	0.4545	0.2424	0.0303

Question 2 — final results
Part	Quantity	Value
(A)(a)	P(X > 3), n=10	0.2241
(A)(b)	P(X > 420), n=1,800	0.9458
(B)(a)	P(at least 3 of 6 not substandard)	0.9697
(B)(b)	pmf of X (see table above)	sums to 1.0000

Topic: Binomial, normal-approximation, and Hypergeometric counting models.

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

Why this works: the binomial applies to independent Bernoulli trials with a fixed success probability (households sampled from a huge population, so "without replacement" is negligible); for large $np(1-p)$ the CLT lets the binomial be approximated by a normal curve with the same mean/variance; the lawnmower lot, however, is a small finite population sampled without replacement, so the trials are dependent and the Hypergeometric model is the exact one.

Common pitfall: applying the binomial to the lawnmower lot (double-counting probability because trials are not independent once a unit is removed), and omitting the 0.5 continuity correction in the normal approximation.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 5 (Discrete Distributions).

Question 2 of 8

Question 2 (20 marks)