23-Ind-A6 Systems Simulation · December 2018

Question 1 of 8: Extra-Bag Survey — Mean, Variance, Distribution Hypothesis

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

Notes on this paper

National Exams — December 2018 — 17-Ind-A6 Systems Simulation. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides). Format: three sections — Part A (Input Modelling): do 2 of 3 questions, 30 marks total; Part B (Modelling Concepts): do 1 of 2, 20 marks; Part C (Output Analysis): do 1 of 2, 20 marks — plus a 1-mark trivia bonus. All seven graded questions and the bonus are solved below for completeness. The exam's own front matter has two internal quirks, transcribed as printed: NOTES item "4." appears twice on page 1, and every page footer reads "17-Ind-A6/Dec. 2019" against a "December 2018" masthead.

Reference texts: Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed., Pearson) — input data analysis (ch. 9), random-variate generation, output analysis for a single system and comparing alternative systems (ch. 11–12), verification and validation (ch. 10).

Question 1 (Part A.1): Extra-Bag Survey — Mean, Variance, Distribution Hypothesis (10 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 survey of 100 Nova Scotia patrols, tallied by number of extra bags brought:

Extra bags $x$	0	1	2	3	4	5	6	7	8	9	10
Count $f$	1	1	2	5	12	21	24	19	11	4	0

Find. (a) sample mean; (b) sample variance; (c) a candidate distribution family, with logical rationale; (d) that distribution's parameter(s).

Approach. Compute $\bar x=\sum xf/n$ and the sample variance $s^2=\sum f(x-\bar x)^2/(n-1)$ directly from the frequency table, then reason about the data's shape (discrete non-negative counts, single interior mode) to hypothesize a distribution family.

Sample size and mean. $n=\sum f = 1+1+2+5+12+21+24+19+11+4+0=100$ patrols. $$\bar x = \frac{\sum xf}{n} = \frac{0(1)+1(1)+2(2)+\cdots+9(4)+10(0)}{100} = \boxed{5.74\ \text{extra bags}}.$$
Sample variance. $$s^2 = \frac{\sum f(x-\bar x)^2}{n-1} = \boxed{3.02},\qquad s = \sqrt{3.02} = 1.74\ \text{bags}.$$

(c) Distribution hypothesis. The data are a count of a discrete event with no natural upper bound printed in the survey (the table's tail simply thins out rather than hitting a hard ceiling), the counts are non-negative integers, and the shape is a single interior mode around $x=6$ with roughly exponential-looking decay on both sides — the classic signature of a Poisson distribution. The underlying mechanism also fits: each of the 10 patrol members independently decides whether to bring a second bag, so the patrol total is a sum of many roughly-independent, individually rare 0/1 decisions — exactly the setting (many trials, each with small probability of "success") in which a Binomial count is well approximated by a Poisson. A Poisson hypothesis is also attractive because it needs only one parameter, letting Robert get a testable model quickly before committing to more machinery.

(d) Parameter. The Poisson distribution has a single parameter $\lambda$ = the mean rate, so $\hat\lambda = \bar x = \boxed{5.74}$ (Robert's own hypothesis in Question 2 rounds this to $\lambda=6$ — a reasonable simplification, formally tested there).

Quantity	Result
Sample mean $\bar x$	5.74 extra bags/patrol
Sample variance $s^2$ (sd $s$)	3.02 (1.74)
Hypothesized distribution	Poisson
Parameter	$\lambda=\bar x=5.74$

Check: note for later — the sample variance (3.02) is well below the sample mean (5.74). A true Poisson requires mean $=$ variance, so this under-dispersion is a visible early warning that the Poisson hypothesis may not survive a formal goodness-of-fit test; Question 2 checks this directly.

Concept & Theory

Topic. Input-data summary statistics and distribution hypothesis selection.

Key relations. $\bar x=\sum xf/n$; $s^2=\sum f(x-\bar x)^2/(n-1)$; Poisson has mean $=$ variance $=\lambda$.

Why this works. Hypothesizing a distribution from the shape of the data (discrete, unbounded tail, single mode, generated as a sum of many small independent decisions) is the standard first step of Banks' input-modelling cycle, done before any goodness-of-fit machinery; the formal test in Question 2 is what actually accepts or rejects the hypothesis.

Common pitfall. Assuming any mound-shaped count data must be Poisson without checking mean $\approx$ variance — a Binomial($n=10,p$) count is naturally under-dispersed (variance $=np(1-p)

Source. Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation, ch. 9 (input data analysis — hypothesizing families of distributions).

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