Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2015 — 04-BS-2 Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator). Statistical tables of the Normal, t, chi-square and F distributions are supplied with the exam. The rubric states "Any 5 questions constitute a complete paper"; every question is answered.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (Pearson) — used throughout for normal/binomial/Poisson/hypergeometric probability, confidence intervals, hypothesis tests, and simple linear regression.
Check — source defect in Question 4. As literally printed, K/y cannot integrate
to a finite value over (1, ∞), so the density as given is not valid. This solution
reconstructs the intended density as f(y) = 1/y² for y > 1 — the unique
minimal correction (recovering one dropped exponent) that makes the function integrate to 1 on the stated
domain, and it is the standard "infinite mean" example from this course's own textbook (Walpole). This choice is disclosed here rather than presented as certain.
Question 1: Tire Life — Normal Distribution (20 marks)
Given. Tire life $X\sim N(\mu=175{,}000\text{ km},\ \sigma=20{,}000\text{ km})$.
Given data
Quantity
Value
Mean, μ
175,000 km
Standard deviation, σ
20,000 km
Sample size (part c), n
4
Number of tires summed (part d), n
36
Find. (a) P(X<190,000); (b) P(|X−μ|<10,000); (c) the sampling distribution of the mean M of 4 tires and P(M>170,000); (d) the distribution of the sum T of 36 tires and P(T>6,300,000).
Approach. Standardize every event with $Z=(x-\mu)/\sigma$ (or the appropriate sampling-distribution standard error), then read the Normal table.
(a) Density and P(X<190,000). The pdf of X is
$$f(x)=\dfrac{1}{20{,}000\sqrt{2\pi}}\exp\!\left[-\dfrac{(x-175{,}000)^2}{2(20{,}000)^2}\right],\qquad -\infty<x<\infty.$$
Standardizing, $z=(190{,}000-175{,}000)/20{,}000=0.75$. From the Normal table, $\Phi(0.75)=0.7734$, so
$$P(X<190{,}000)=\boxed{0.7734}.$$
Fig. 1 — N(175,000, 20,000²) with the shaded area P(Z<0.75)=0.7734.
(b) P(|X−μ|<10,000). This is $P(-0.5<Z<0.5)$ since $10{,}000/20{,}000=0.5$.
$$P(-0.5<Z<0.5)=\Phi(0.5)-\Phi(-0.5)=2\Phi(0.5)-1=2(0.6915)-1=\boxed{0.3829}.$$
Fig. 2 — shaded central area P(−0.5<Z<0.5)=0.3829.
(c) Distribution of M and P(M>170,000). For $n=4$ tires, the sampling distribution of the mean is
$$M\sim N\!\left(\mu,\ \dfrac{\sigma}{\sqrt n}\right)=N(175{,}000,\ 10{,}000^2),\qquad \sigma_M=\dfrac{20{,}000}{\sqrt4}=10{,}000.$$
Its density is $f(m)=\dfrac{1}{10{,}000\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-175{,}000)^2}{2(10{,}000)^2}\right]$ — the same bell shape as X's but half as wide (M is less variable than a single tire, as averaging reduces spread by $\sqrt n$). Standardizing $M$, $z=(170{,}000-175{,}000)/10{,}000=-0.5$, so
$$P(M>170{,}000)=1-\Phi(-0.5)=\Phi(0.5)=\boxed{0.6915}.$$
Fig. 3 — X (wide curve) and M (narrow curve, half the standard deviation) on one axis in units of X's z-score; shaded area is P(M>170,000)=0.6915.
(d) Distribution of T and P(T>6,300,000). T is the sum of 36 independent tire lives, so
$$T\sim N(n\mu,\ \sqrt n\,\sigma)=N(36\times175{,}000,\ \sqrt{36}\times20{,}000)=N(6{,}300{,}000,\ 120{,}000^2).$$
Since $6{,}300{,}000$ is exactly the mean of T, $z=0$ and
$$P(T>6{,}300{,}000)=1-\Phi(0)=\boxed{0.5}.$$
