04-BS-2, December 2017, Q2 | Nivaar Exam Prep

Question 2 (20 marks: (A)(a) 5, (b) 5; (B)(a) 5, (b) 5)

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 = 0.70 of households own exactly one car. (B) A lot of N = 15 lawnmowers contains 4 substandard units; 6 are drawn without replacement.

Find. (A)(a) P(more than 8 of 12), (A)(b) P(fewer than 1,240 of 1,800, via Normal approximation). (B)(a) P(at least 4 of 6 sold are good), (B)(b) the pmf and mean of the number substandard in a sample of 6.

Approach. Part (A) uses the Binomial distribution directly for n = 12, and the Normal approximation to the Binomial (continuity-corrected) for the large n = 1,800. Part (B) is sampling without replacement from a finite lot — the Hypergeometric distribution.

2(A)(a) Binomial, n = 12, p = 0.70. "More than eight" means $X \ge 9$: $$P(X>8) = 1 - P(X\le 8) = 1 - \sum_{k=0}^{8}\binom{12}{k}(0.70)^k(0.30)^{12-k} = 1-0.5075 = \boxed{0.4925}.$$
2(A)(b) Normal approximation, n = 1,800, p = 0.70. $$\mu = np = 1{,}260, \qquad \sigma=\sqrt{npq}=\sqrt{1800(0.7)(0.3)}=19.442.$$ With the continuity correction, "fewer than 1,240" → $X \le 1239.5$: $$Z = \frac{1239.5-1260}{19.442} = -1.054, \qquad P(X<1240) \approx \Phi(-1.054) = \boxed{0.1458}.$$
2(B)(a) Hypergeometric, N = 15, K = 4 substandard, n = 6. "At least 4 not substandard" (of the 6 sold) means at most 2 substandard, i.e. $X \le 2$ where $X$ counts substandard units in the sample: $$P(X\le 2) = \sum_{k=0}^{2}\frac{\binom{4}{k}\binom{11}{6-k}}{\binom{15}{6}} = \boxed{0.8571}.$$
2(B)(b) Full distribution and mean. $X$ ranges over $k=0,\dots,4$ with $P(X=k)=\dfrac{\binom{4}{k}\binom{11}{6-k}}{\binom{15}{6}}$: $P(0)=0.0923$, $P(1)=0.3692$, $P(2)=0.3956$, $P(3)=0.1319$, $P(4)=0.0110$ (sums to 1.000). The mean of a Hypergeometric variable is $$E(X) = \frac{nK}{N} = \frac{6(4)}{15} = \boxed{1.6 \text{ substandard units}}.$$

Question 2 — final results
Part	Quantity	Value
(A)(a)	P(X > 8), n=12	0.4925
(A)(b)	P(X < 1,240), n=1,800	0.1458
(B)(a)	P(at least 4 not substandard)	0.8571
(B)(b)	E(X), n=6 sample	1.6 substandard units

Topic: Binomial, Normal approximation to the Binomial, and Hypergeometric distributions.

Key relations: Binomial pmf $\binom{n}{k}p^k q^{n-k}$; Normal approximation $Z=(x\pm 0.5-np)/\sqrt{npq}$; Hypergeometric pmf $\binom{K}{k}\binom{N-K}{n-k}/\binom{N}{n}$, mean $nK/N$.

Why this works: The Normal approximation is valid here because $np$ and $nq$ are both far above the usual rule-of-thumb of 5; the continuity correction accounts for approximating a discrete distribution with a continuous one. The Hypergeometric applies whenever sampling is without replacement from a small, finite, two-category lot — the trial outcomes are not independent, unlike the Binomial.

Common pitfall: Using the Binomial (independent trials) for the lawnmower lot, which double-counts probability because the lot is small and sampled without replacement; omitting the continuity correction in the Normal approximation.

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

Question 2 of 8

Question 2 (20 marks: (A)(a) 5, (b) 5; (B)(a) 5, (b) 5)