23-Ind-B1 Reliability and Maintainability · December 2017

Question 3 of 10: Halifax Ferry — Joint Distribution, Expected Profit, Marginals, Conditional Expectation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2017 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 (30 marks); Section B: do 2 of 3 (30 marks); Section C: do 2 of 3 (40 marks) — a 6-question, 100-mark paper as printed. All ten questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — discrete/continuous distributions (ch. 3–4), joint distributions (ch. 5), point/interval estimation and sample size (ch. 8), hypothesis testing incl. two-sample tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor experiments (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — multi-factor and $2^k$ factorial designs (ch. 5–6).

Question 3 (Section A.3): Halifax Ferry — Joint Distribution, Expected Profit, Marginals, Conditional Expectation (15 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. Joint distribution $P(X{=}x,Y{=}y)$, regular fare \$3/head, priority fare \$8/head, operating cost \$450/trip:

$P(X=x,\ Y=y)$
$Y\backslash X$	100	150	200
0	0.01	0.01	0.03
10	0.03	0.08	0.07
20	0.03	0.06	0.06
30	0.07	0.07	0.13
40	0.12	0.04	0.03
50	0.08	0.06	0.02

Find. (a) $E[\text{profit}]$; (b) $p_X(x)$; (c) $p_Y(y)$; (d) $E[\text{profit}\mid Y=20]$.

Approach. Profit per trip is the linear function $\text{Profit}=3X+8Y-450$, so $E[\text{Profit}]=3E[X]+8E[Y]-450$ follows directly from the marginal means; the marginals themselves are row/column sums of the joint table, and the conditional expectation in (d) uses the row $Y{=}20$ renormalized to sum to 1.

(a) Expected profit. Summing $x\cdot p_X(x)$ and $y\cdot p_Y(y)$ (or equivalently $\sum_{x,y}xy$-weighted terms) over the joint table gives $E[X]=150.0$ and $E[Y]=28.5$ passengers. $$E[\text{Profit}]=3(150.0)+8(28.5)-450=450+228-450=\boxed{\$228}$$
(b) Marginal pmf of $X$ (sum each column). $$p_X(100)=0.34,\quad p_X(150)=0.32,\quad p_X(200)=0.34$$
(c) Marginal pmf of $Y$ (sum each row). $$p_Y(0)=0.05,\ p_Y(10)=0.18,\ p_Y(20)=0.15,\ p_Y(30)=0.27,\ p_Y(40)=0.19,\ p_Y(50)=0.16$$
(d) Expected profit given $Y=20$ booked. Renormalizing the $Y{=}20$ row ($0.03,0.06,0.06$, summing to $p_Y(20)=0.15$) gives the conditional pmf of $X$: $$p_{X\mid Y=20}(100,150,200)=(0.20,\ 0.40,\ 0.40)$$ $$E[X\mid Y=20]=100(0.20)+150(0.40)+200(0.40)=\boxed{160}$$ With $Y$ fixed at the known 20 priority bookings: $$E[\text{Profit}\mid Y=20]=3(160)+8(20)-450=480+160-450=\boxed{\$190}$$

Summary
Quantity	Value
(a) $E[\text{Profit}]$	\$228
(b) $p_X$ (100,150,200)	0.34, 0.32, 0.34
(c) $p_Y$ (0,10,...,50)	0.05, 0.18, 0.15, 0.27, 0.19, 0.16
(d) $E[\text{Profit}\mid Y{=}20]$	\$190

Topic: Discrete joint distributions — marginals, linear functions of random variables, and conditional expectation.

Key relations: $p_X(x)=\sum_y p(x,y)$; for a linear payoff $aX+bY+c$, $E[aX+bY+c]=aE[X]+bE[Y]+c$ regardless of any dependence between $X$ and $Y$; $E[X\mid Y=y]=\sum_x x\,p(x,y)/p_Y(y)$.

Why this works: Linearity of expectation holds unconditionally, so profit's expectation can be computed from the marginal means alone in part (a) without ever needing the full joint table; part (d) is a genuinely different (smaller, conditional) calculation because fixing $Y=20$ changes the effective distribution of $X$ from its marginal to the renormalized row.

Common pitfall: In part (d), reusing the unconditional $E[X]=150$ instead of recomputing $E[X\mid Y=20]=160$ from the renormalized row — the two differ here because $X$ and $Y$ are not independent (row 20's shape is not proportional to the marginal $p_X$).

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed.), ch. 5 (joint probability distributions, conditional expectation).

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