Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-2 Probability and Statistics — National Examination, May 2018. Closed book (approved calculator + one hand-written information sheet); any 5 of 8 questions constitute a complete paper (only 5 marked) — all 8 are solved below as a complete study resource.
Reference texts: Walpole, Myers, Myers & Ye, Probability and Statistics for Engineers and Scientists (9th ed.), Pearson — used throughout for point-estimation, confidence-interval, and hypothesis-testing formulas; Montgomery & Runger, Applied Statistics and Probability for Engineers, for the regression and correlation-inference formulas in Question 8.
Given. The lifetime $X$ of a single mechanical component is normally distributed, $X \sim N(\mu, \sigma^2)$ with $\mu = 10{,}600.0$ hours and $\sigma = 400.0$ hours.
Given data
Symbol
Meaning
Value
$\mu$
Population mean lifetime
10,600.0 h
$\sigma$
Population standard deviation
400.0 h
$n_M$
Sample size for $M$
16
$n_T$
Number of components summed for $T$
100
Find. (a) $P(X>10{,}000)$ and $f_X(x)$; (b) the lower quartile and upper decile of $X$; (c) the mean, sd and pdf of $M=\bar X_{16}$, and $P(|M-\mu|>200)$; (d) the mean, variance of $T=\sum_{i=1}^{100}X_i$, and $P(T>1{,}066{,}000)$.
Approach. Standardize every event with $Z=(x-\mu)/\sigma$ (or the appropriate scaled $\sigma$ for $M$ and $T$), read the normal table, and use the linear-combination rules $E[\bar X_n]=\mu,\ \mathrm{sd}(\bar X_n)=\sigma/\sqrt n$ and $E[\sum X_i]=n\mu,\ \mathrm{Var}(\sum X_i)=n\sigma^2$.
a) pdf and $P(X>10{,}000)$. The density is
$$f_X(x)=\frac{1}{400\sqrt{2\pi}}\exp\!\left[-\frac{(x-10{,}600)^2}{2(400)^2}\right],\qquad x\in(-\infty,\infty).$$
Standardizing, $Z=(10{,}000-10{,}600)/400=-1.50$, so
$$P(X>10{,}000)=P(Z>-1.50)=0.5+\Phi(1.50)=0.5+0.4332=\boxed{0.9332}.$$
f(x) for X ~ N(10,600, 400²); shaded region = P(X>10,000) = 0.9332.
b) Lower quartile and upper decile.(i) The lower quartile satisfies $P(Z\le z_{0.25})=0.25\Rightarrow z_{0.25}=-0.6745$, so
$$Q_1=10{,}600+(-0.6745)(400)=\boxed{10{,}330.2\text{ h}}.$$
(ii) The upper decile satisfies $P(Z\le z_{0.90})=0.90\Rightarrow z_{0.90}=1.2816$, so
$$D_9=10{,}600+(1.2816)(400)=\boxed{11{,}112.6\text{ h}}.$$
c) Distribution of $M$ (n=16) and $P(|M-\mu|>200)$. By the sampling-distribution-of-the-mean rule, $M\sim N(\mu,\ \sigma^2/n)$. (i)
$$E[M]=10{,}600\text{ h},\qquad \mathrm{sd}(M)=\frac{400}{\sqrt{16}}=100\text{ h}.$$
(ii) $f_M(m)=\dfrac{1}{100\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-10{,}600)^2}{2(100)^2}\right]$ — the same bell shape as $f_X$ but four times narrower (sd shrinks by $\sqrt{16}=4$). (iii) The two curves are plotted together below. (iv) For the tail probability, $z=200/100=2.00$, so
$$P(|M-10{,}600|>200)=2[1-\Phi(2.00)]=2(1-0.9772)=\boxed{0.0456}.$$
Sampling distribution of M (n=16): M ~ N(10,600, 100²); shaded = P(|M-10,600|>200).
d) Distribution of $T$ (sum of 100) and $P(T>1{,}066{,}000)$. For an independent sum, $E[T]=n\mu$ and $\mathrm{Var}(T)=n\sigma^2$:
$$E[T]=100(10{,}600)=1{,}060{,}000\text{ h},\qquad \mathrm{Var}(T)=100(400)^2=16{,}000{,}000\text{ h}^2,\qquad \mathrm{sd}(T)=4{,}000\text{ h}.$$
Standardizing, $z=(1{,}066{,}000-1{,}060{,}000)/4{,}000=1.50$, so
$$P(T>1{,}066{,}000)=P(Z>1.50)=0.5-0.4332=\boxed{0.0668}.$$
Distribution of T = sum of 100 lifetimes: T ~ N(1,060,000, 4,000²); shaded = P(T>1,066,000).
