Binomial Safety Glasses & Committee Decision: 04-BS-2, May 2014, Q3

Question 3: Binomial Safety Glasses & Committee Decision (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) Defect rate p=0.25 in safety-glass production. (B) Four independent Bernoulli decisions with probabilities 0.40, 0.80, 0.60, 0.90.

Find. (A)(a) P(fewer than 3 defective of 10); (A)(b) P(more than 1,230 of 1,600 meet standard); (B)(a) the pmf of X = number in favour; (B)(b) E(X) and Var(X).

Approach. (A)(a) is an exact Binomial(10, 0.25) calculation. (A)(b) uses the Normal approximation to the Binomial with a continuity correction, since n=1,600 is far too large for exact Binomial computation. (B) is a Poisson-binomial (sum of independent, non-identical Bernoulli trials): enumerate all 2⁴=16 outcomes to build the exact pmf of X, since the four probabilities are unequal so X is not Binomial.

(A)(a) Exact Binomial, n=10, p=0.25 defective. Let $D\sim\text{Binomial}(10,0.25)$ count defective glasses. "Fewer than three fail" means $D<3$: $$P(D<3)=\sum_{k=0}^{2}\binom{10}{k}(0.25)^k(0.75)^{10-k}=0.0563+0.1877+0.2816=\boxed{0.5256}.$$
(A)(b) Normal approximation, n=1,600. Let $S$ = number meeting the standard, $S\sim\text{Binomial}(1600,\,0.75)$ since 75% meet the standard. Then $$\mu_S=np=1200,\qquad \sigma_S=\sqrt{np(1-p)}=\sqrt{1600(0.75)(0.25)}=17.32.$$ With a continuity correction for "more than 1,230" (i.e. $S\ge1231\Leftrightarrow S>1230.5$ on the continuous scale), $$Z=\dfrac{1230.5-1200}{17.32}=1.76\ \Rightarrow\ P(S>1230)\approx 1-\Phi(1.76)=1-0.9608=\boxed{0.0392}.$$
(B)(a) Poisson-binomial pmf by full enumeration. Let $A,B,C,D$ be independent Bernoulli indicators with $P(A{=}1){=}0.40$, $P(B{=}1){=}0.80$, $P(C{=}1){=}0.60$, $P(D{=}1){=}0.90$, and $X=A+B+C+D$. Summing the products of the sixteen $\{0,1\}^4$ outcomes grouped by their total gives $$P(X{=}0){=}\boxed{0.0048},\ P(X{=}1){=}\boxed{0.0728},\ P(X{=}2){=}\boxed{0.3128},\ P(X{=}3){=}\boxed{0.4368},\ P(X{=}4){=}\boxed{0.1728}.$$ (For example, $P(X{=}4)=(0.40)(0.80)(0.60)(0.90)=0.1728$, and $P(X{=}0)=(0.60)(0.20)(0.40)(0.10)=0.0048$.) The five terms sum to 1.0000, confirming the enumeration.
(B)(b) Mean and variance of X. By linearity of expectation and independence of the four terms, $$E(X)=\sum p_i=0.40+0.80+0.60+0.90=\boxed{2.70},$$ $$\mathrm{Var}(X)=\sum p_i(1-p_i)=0.24+0.16+0.24+0.09=\boxed{0.73}.$$

Final results — Question 3
Part	Quantity	Result
(A)(a)	P(D < 3), n=10, p=0.25	0.5256
(A)(b)	P(S > 1,230), n=1,600	0.0392
(B)(a)	pmf of X (k=0..4)	0.0048, 0.0728, 0.3128, 0.4368, 0.1728
(B)(b)	E(X), Var(X)	2.70, 0.73

Topic: The Binomial distribution, its Normal approximation for large n, and the Poisson-binomial (sum of unequal independent Bernoulli trials).

Key relations: Binomial $P(k)=\binom{n}{k}p^k(1-p)^{n-k}$, $\mu=np$, $\sigma^2=np(1-p)$; continuity-corrected Normal approximation $Z=\dfrac{(k+0.5)-\mu}{\sigma}$; for independent but non-identical Bernoulli $X_i$, $E(\sum X_i)=\sum p_i$ and $\mathrm{Var}(\sum X_i)=\sum p_i(1-p_i)$ regardless of whether the $p_i$ are equal.

Why this works: the Central Limit Theorem makes a Binomial count approximately Normal once np and n(1−p) are both reasonably large (here 1200 and 400); the continuity correction accounts for approximating a discrete count with a continuous density. The committee sum is not Binomial because the four "success" probabilities differ, but expectation and variance still add because covariance terms vanish under independence.

Common pitfall: forgetting the continuity correction in (A)(b) (using Z=1.7609 without the +0.5 shifts the tail probability noticeably), or trying to force the committee sum in (B) into a Binomial(4, p̄) with an averaged p — that would badly distort the shape of the pmf, especially at the tails k=0 and k=4.

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

Question 3 of 8: Binomial Safety Glasses & Committee Decision

Question 3: Binomial Safety Glasses & Committee Decision (20 marks)