04-BS-2, December 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) $p=0.65$ in favour; (a) $n=14$; (b) $n=5{,}000$. (B) a lot of $N=15$ snow blowers, $K=5$ substandard (10 good); a random sample of $n=8$ sold in November.

Find. (A)(a) $P(5<X<9)$; (A)(b) $P(X<3{,}200)$; (B)(a) $P(\text{substandard}<3\text{ of }8)$; (B)(b) the full pmf, mean and variance of $X$ (substandard count).

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 15 snow blowers is finite and sampled without replacement, so (B) uses the Hypergeometric distribution.

(A)(a) Exact binomial, n=14. $X\sim\text{Bin}(14,0.65)$. $P(5<X<9)=P(X=6)+P(X=7)+P(X=8)=\binom{14}{6}(0.65)^6(0.35)^8+\binom{14}{7}(0.65)^7(0.35)^7+\binom{14}{8}(0.65)^8(0.35)^6=0.1043+0.1552+0.1757=\boxed{0.3352}$.
(A)(b) Normal approximation to the binomial, n=5,000. $\mu=np=5{,}000(0.65)=3{,}250$, $\sigma=\sqrt{np(1-p)}=\sqrt{5{,}000(0.65)(0.35)}=33.727$. With the continuity correction, $P(X<3{,}200)=P\!\left(Z<\dfrac{3199.5-3250}{33.727}\right)=P(Z<-1.497)=\boxed{0.0672}$.
(B)(a) Hypergeometric, "fewer than 3 substandard of 8". $X\sim\text{Hyper}(N=15,K=5\text{ substandard},n=8)$. $P(X<3)=P(X=0)+P(X=1)+P(X=2)=\dfrac{\binom50\binom{10}8}{\binom{15}8}+\dfrac{\binom51\binom{10}7}{\binom{15}8}+\dfrac{\binom52\binom{10}6}{\binom{15}8}=0.0070+0.0932+0.3263=\boxed{0.4266}$.
(B)(b) Hypergeometric pmf, mean and variance. $P(X=x)=\dfrac{\binom5x\binom{10}{8-x}}{\binom{15}8}$ for $x=0,\dots,5$ (support limited to $\max(0,8-10)\le x\le\min(5,8)$, i.e. $x=0,\dots,5$). Mean $=n\dfrac{K}{N}=8\!\left(\dfrac{5}{15}\right)=\boxed{2.667}$. Variance $=n\dfrac{K}{N}\!\left(1-\dfrac{K}{N}\right)\!\dfrac{N-n}{N-1}=8\!\left(\dfrac13\right)\!\left(\dfrac23\right)\!\left(\dfrac{7}{14}\right)=\boxed{0.8889}$.

Q2(B)(b) — probability distribution of X (substandard snow blowers in a sample of 8)
x	0	1	2	3	4	5
P(X=x)	0.0070	0.0932	0.3263	0.3916	0.1632	0.0186

Question 2 — final results
Part	Quantity	Value
(A)(a)	P(5 < X < 9), n=14	0.3352
(A)(b)	P(X < 3,200), n=5,000	0.0672
(B)(a)	P(fewer than 3 substandard of 8)	0.4266
(B)(b)	Mean, variance of X (see table above)	2.667, 0.8889

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 $P(X=x)=\dfrac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}$, mean $n K/N$, variance $n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}$.

Why this works: the binomial applies to independent Bernoulli trials with a fixed success probability (adult inhabitants sampled from a huge population); the snow-blower lot is a small finite population sampled without replacement, so trials are dependent and the Hypergeometric model is the exact one.

Common pitfall: assuming a hypergeometric tail term is automatically zero without checking the binomial-coefficient support — here $\binom{10}{8}=45\neq0$, so $P(X{=}0)$ is small but nonzero.

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)