23-Ind-B1 Reliability and Maintainability · December 2018

Question 1 of 9: Normal-Distribution Assembly Times

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

Notes on this paper

National Exams — December 2018 — 17-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 (20 marks); Section B: do 2 of 3 (40 marks); Section C: do 1 of 2 (20 marks) — a 9-question, printed-total-80-mark paper as actually structured (the front page's own summary line states “Exam: 5 Questions. Total marks: 100”, but 20+40+20=80 by the section instructions and per-question mark values printed beside each question — the front page's 100 is internally inconsistent with its own section table; the 80-mark reading is used throughout, consistent with the section-by-section split printed on pages 2, 4 and 6). All nine questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — normal distribution and sums of normals (ch. 4–5), point/interval estimation (ch. 8), two-sample hypothesis testing (ch. 9–10), simple linear regression (ch. 11), single-factor ANOVA and multiple comparisons (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — Bartlett's test, $2^k$ factorial designs (ch. 3, 6).

Question 1 (Section A.1): Normal-Distribution Assembly Times (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. Step 1 time $X_1\sim N(\mu_1=12,\sigma_1=3)$ s. Step 2 time $X_2\sim N(\mu_2=10,\sigma_2=5)$ s, independent of $X_1$.

Find. (a) $P(X_1>15)$; (b) whether $P(T>28)$, $T=X_1+X_2$, is “unusual”; (c) whether $N(\cdot)$ is a reasonable model for a repeated task time.

Approach. Standardize each normal variable ($z=(x-\mu)/\sigma$) and read the tail area from the standard-normal table; for the sum, use $T\sim N(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2})$ since $X_1,X_2$ are independent normals.

Standardize and find $P(X_1>15)$. $z=\dfrac{15-12}{3}=1.00$. From the standard-normal table, $\Phi(1.00)=0.8413$, so $$P(X_1>15)=1-\Phi(1.00)=1-0.8413=\boxed{0.1587}$$
Distribute the sum $T=X_1+X_2$. Independent normals sum to a normal with $\mu_T=\mu_1+\mu_2$ and $\sigma_T=\sqrt{\sigma_1^2+\sigma_2^2}$: $$\mu_T=12+10=22\text{ s},\qquad \sigma_T=\sqrt{3^2+5^2}=\sqrt{34}=5.831\text{ s}$$
Standardize and find $P(T>28)$. $z=\dfrac{28-22}{5.831}=1.029$. From the table, $\Phi(1.03)\approx0.8485$, so $$P(T>28)=1-\Phi(1.029)=\boxed{0.1517}$$ Taking “unusual” as an event with probability below about 5% (the conventional exam threshold, roughly $|z|>1.645$–$2$), $P(T>28)=15.2\%$ is well above that cutoff — a total time over 28 s is not unusual; it is well within one standard deviation of the mean total time.
Assess the normality assumption (essay). A task-completion time built from several sequential physical sub-steps (positioning, fastening, verifying) is the sum of a number of small, roughly-independent sources of variation (operator dexterity, part fit-up, tool response), and the Central Limit Theorem gives a solid theoretical basis for a normal approximation once the task is repeated many times, as stated. Two caveats are worth flagging: (i) a normal distribution has support on all of $(-\infty,\infty)$, while a real task time cannot be negative — here $\mu_1=12,\sigma_1=3$ puts $P(X_1<0)=\Phi(-4)\approx3\times10^{-5}$, negligible, so the approximation is safe in the body of the distribution even though it is technically wrong in the extreme left tail; (ii) if the process is subject to occasional large disruptions (a dropped part, a jam) the true distribution would be right-skewed with a heavier tail than the normal, which the given $N(12,3)$ model would then understate. For a well-controlled, repetitive manual assembly step with no such disruptions, normal is a reasonable working assumption.

Final Results
Quantity	Value
$P(X_1>15\text{ s})$	0.1587
$\mu_T,\ \sigma_T$ for $T=X_1+X_2$	22 s, 5.831 s
$P(T>28\text{ s})$	0.1517 — not unusual

Topic: Normal distribution and sums of independent normal random variables.

Key relations: $Z=\dfrac{X-\mu}{\sigma}$; if $X_1\sim N(\mu_1,\sigma_1)$, $X_2\sim N(\mu_2,\sigma_2)$ independent, then $X_1+X_2\sim N(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2})$.

Why this works: The normal family is closed under summation of independent members — variances (not standard deviations) add because $\mathrm{Var}(X_1+X_2)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)$ for independent variables, and the sum of two normal random variables is itself exactly normal (not just approximately, by moment-generating-function closure).

Common pitfall: Adding the standard deviations directly ($\sigma_T=3+5=8$) instead of adding variances and taking the square root ($\sigma_T=\sqrt{34}=5.83$) — this overstates the spread of the sum substantially.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th ed., §4.6–4.7 (normal distribution, linear combinations of normal variables).

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