23-Ind-B1 Reliability and Maintainability · December 2017

Question 1 of 10: Oil-Well Drilling — Binomial, Geometric, Normal Probability, and a Sum of Normals

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 1 (Section A.1): Oil-Well Drilling — Binomial, Geometric, Normal Probability, and a Sum of Normals (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. Independent Bernoulli drilling trials, success probability $p=0.25$; drilling time per well $\sim N(\mu=72,\ \sigma=3)$ days.

Find. (a) $P(X=1)$ for $n=10$; (b) $P(\text{first success on trial}\ge 9)$; (c) $P(75\lt T\lt 80)$; (d) is $T=68.3$ unusual; (e) 95% interval for the total time of 10 sequential wells.

Approach. (a) is a binomial pmf on the success/failure of each location; (b) reframes "9 or more drills before the first success" as the geometric tail $P(N\ge 9)$; (c)–(d) standardize the drilling-time normal directly; (e) sums 10 i.i.d. drilling times, whose distribution is itself normal by the additivity of independent normals.

(a) Exactly 1 success in 10 independent drills. $X\sim\mathrm{Binomial}(n=10,p=0.25)$: $$P(X=1)=\binom{10}{1}(0.25)^1(0.75)^9=\boxed{0.1877}$$
(b) Probability of bankruptcy. "9 or more drills before the first success" means the trial $N$ of the first success satisfies $N\ge 9$, i.e. the first 8 drills are all failures: $$P(N\ge 9)=(1-p)^{8}=(0.75)^{8}=\boxed{0.1001}$$ There is roughly a 10% chance the firm exhausts 8 straight failures and faces bankruptcy before its first success. (This counts the successful drill itself as the 9th drill, the usual geometric-distribution convention for this textbook problem. If instead “9 or more times before” is read as 9 or more failed drills preceding the first success, the probability is $(0.75)^9=0.0751$; state whichever reading you adopt.)
(c) Drilling time between 75 and 80 days. With $T\sim N(72,3)$: $$z_1=\frac{75-72}{3}=1.000,\qquad z_2=\frac{80-72}{3}=2.667$$ $$P(75\lt T\lt80)=\Phi(2.667)-\Phi(1.000)=0.9962-0.8413=\boxed{0.1548}$$
(d) Is 68.3 days unusual? $$z=\frac{68.3-72}{3}=\boxed{-1.233}$$ Using the conventional $|z|\gt 2$ (roughly the 95% two-tailed cutoff) as the "unusual" threshold, $|-1.233|\lt 2$, so a single well finishing in 68.3 days is not unusual — it sits well within the typical spread of a $N(72,3)$ process (about 1.2 standard deviations below the mean, which happens for roughly 1 well in 9 by chance alone).
(e) 95% interval for the total drilling-program time. Drilling one well at a time means the 10 individual times are summed sequentially, not run in parallel. For 10 independent, identically distributed drilling times $T_i\sim N(72,3)$, the total $S=\sum_{i=1}^{10}T_i$ is itself normal: $$E[S]=10(72)=720\ \text{days},\qquad \mathrm{Var}(S)=10(3)^2=90\ \Rightarrow\ \mathrm{sd}(S)=\sqrt{90}=9.487\ \text{days}$$ $$720\pm 1.96(9.487)=\boxed{(701.4,\ 738.6)\ \text{days}}$$

Summary
Part	Result
(a) $P(X=1\mid n=10)$	0.1877
(b) $P(\text{bankrupt})$	0.1001
(c) $P(75\lt T\lt80)$	0.1548
(d) $z$ for 68.3 days	−1.233 — not unusual
(e) 95% interval, total program time	(701.4, 738.6) days

Topic: Binomial and geometric probability from repeated Bernoulli trials, and the normal distribution of a sum of i.i.d. normal random variables.

Key relations: $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$; $P(N\ge m)=(1-p)^{m-1}$ (geometric survival); if $T_i\overset{iid}{\sim}N(\mu,\sigma^2)$ then $\sum_{i=1}^n T_i\sim N(n\mu,\ n\sigma^2)$ — variances add, not standard deviations.

Why this works: "9 or more drills before the first success" is exactly the event that the first 8 attempts all fail, which is the survival function of the geometric distribution and needs no explicit sum; part (e) relies on the fact that a sum of independent normals is itself exactly normal (not merely approximately, by any CLT argument), so the interval can be built directly from $N(720,90)$.

Common pitfall: In part (e), scaling the single-well standard deviation by $\sqrt{10}$ times itself (i.e. using $10\sigma$ instead of $\sqrt{10}\,\sigma$) — variances of independent sums add linearly, so the sum's standard deviation grows only with $\sqrt{n}$, not $n$.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed.), ch. 3 (discrete distributions) & ch. 5 (functions of random variables, sums of independent normals).

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