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23-Ind-A6 Systems Simulation · December 2017

Question 5 of 8: Absenteeism Data: Mean, Variance, Distribution Hypothesis

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

Notes on this paper

National Exams — December 2017 — 98-Ind-A6 Systems Simulation. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), two 8.5″×11.0″ aid sheets (both sides). Format: eight questions of equal value (10 marks each); candidates complete SIX of the EIGHT (only the first six as they appear in the answer book are marked). All eight questions are solved below for completeness (the paper's own Q6 is printed with an orphan leading "6." followed by "2. Consider an M/M/1 system…" and is answered as the paper's sixth question). Common Discrete/Continuous Distribution tables, Student-t and Chi-square tables were supplied with the exam; the values below are the same table values obtained by direct computation.

Reference texts: Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed., Pearson) — random-number/random-variate generation, input data analysis, output analysis (replications), comparing alternative systems, queueing simulation, and the simulation study life cycle.

Question 5 — Absenteeism Data: 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. Number of employees absent per day, over 100 observed days:

Absent012345678910
Frequency035921222411410

Find. (a) sample mean; (b) sample variance; (c) a defensible input distribution family with rationale.

Approach. Compute the mean and (unbiased, $n-1$) variance directly from the frequency table, then compare the mean/variance relationship against the candidate discrete distributions on the supplied reference sheet.

  1. Sample mean. $$\bar x = \frac{\sum x_i f_i}{100} = \frac{0(0)+1(3)+2(5)+\cdots+9(1)+10(0)}{100} = \frac{496}{100} = \boxed{4.96\ \text{absences/day}}.$$
  2. Sample variance. $$s^2 = \frac{\sum f_i(x_i-\bar x)^2}{n-1} = \frac{270.04}{99} = \boxed{2.726}.$$
QuantityResult
(a) Mean absences/day4.96
(b) Sample variance2.726 (SD $=1.651$)
(c) Hypothesized distributionBinomial($n\approx11$, $p\approx0.45$)

(c) Distribution hypothesis. A Poisson model would require the variance to equal the mean; here $\bar x = 4.96 \gg s^2=2.726$ (ratio $\approx1.82$), a clear under-dispersion relative to Poisson, which rules it out. Under-dispersion this pronounced, together with the fact that daily absence counts are bounded above (there are only so many staff who could possibly be absent), points to a Binomial($n,p$) model instead: matching moments, $s^2=np(1-p)$ and $\bar x=np$ give $\hat p = 1-s^2/\bar x = 1-2.726/4.96 \approx 0.450$ and $\hat n = \bar x/\hat p \approx 11.0$. An $n\approx11$ Binomial is physically sensible — it is consistent with roughly an 11-person crew, each independently absent on a given day with probability $\approx0.45$ — and correctly reproduces the observed bell-shaped, right-truncated-at-10 histogram, unlike the unbounded, equidispersed Poisson.

Check: the Binomial parameters ($n\approx11$, $p\approx0.45$) are recovered by the method of moments, not a formal goodness-of-fit test (not requested); a chi-square GOF test against Binomial(11, 0.45) would be the natural next step before using this distribution in the production-process model.