Question 1 of 8: Sampling Distributions of a Normal Component Lifetime
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2013 — 04-BS-2 Probability and Statistics, 2 hours, closed book except one hand-written information sheet and an approved calculator. Statistical tables (Normal, t, chi-square, F) are supplied with the paper. The instructions state that any 5 of the 8 questions constitute a complete paper; every question is answered in full below.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists (9th ed.) — sampling distributions Ch. 8, estimation Ch. 9, hypothesis testing Ch. 10, ANOVA/chi-square Ch. 13–16.
Question 1: Sampling Distributions of a Normal Component Lifetime (20 marks)
Find. $P(Y>55{,}000)$; the lower quartile and upper decile of $Y$; the sampling distribution of $M$ and $P(|M-\mu|<300)$; $E(T)$, $Var(T)$ and $P(T>444{,}000)$.
Approach. Standardize each event with $Z=(x-\mu)/\sigma$ (or the appropriate propagated standard deviation for $M$ and $T$) and read the tail area from the Normal table.
Check: part (c)(iv) as printed asks for the probability that $M$ differs from the mean "by less than 300.0 hours" — inconsistent with every other quantity in Question 1, which is in km. Treated as a typesetting slip and solved as 300.0 km, consistent with the rest of the question.
(a) Tail probability for a single component. The pdf is $f(y)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\!\left[-\dfrac{(y-\mu)^2}{2\sigma^2}\right]$, $-\infty<y<\infty$. Standardizing, $z=\dfrac{55{,}000-50{,}000}{4{,}000}=1.25$. From the Normal table, $\Phi(1.25)=0.8944$, so
$$P(Y>55{,}000)=1-\Phi(1.25)=\boxed{0.1056}$$
The shaded upper tail beyond $Y=55{,}000$ is drawn below.
Fig. 1(a) — $N(50{,}000,\,4{,}000^2)$ pdf of $Y$, shaded area = $P(Y>55{,}000)=0.1056$.
(b)(i) Lower quartile of $Y$. The 25th percentile satisfies $\Phi(z_{0.25})=0.25$, i.e. $z_{0.25}=-0.6745$ (Normal table, area 0.25 below the mean). Then
$$y_{0.25}=\mu+z_{0.25}\sigma=50{,}000+(-0.6745)(4{,}000)=\boxed{47{,}302\text{ km}}$$
Fig. 1(b)(i) — shaded area below $Y=47{,}302$ km is the lower quartile, area $=0.25$.
(b)(ii) Upper decile of $Y$. The 90th percentile satisfies $\Phi(z_{0.90})=0.90$, i.e. $z_{0.90}=1.2816$. Then
$$y_{0.90}=50{,}000+(1.2816)(4{,}000)=\boxed{55{,}126\text{ km}}$$
Fig. 1(b)(ii) — shaded area below $Y=55{,}126$ km is the upper decile (90th percentile), area $=0.90$.
(c)(i)–(ii) Sampling distribution of $M$. By the Central Limit Theorem (exact here, since $Y$ is already Normal), the mean of $n=64$ components is
$$M\sim N\!\left(\mu_M=\mu,\ \sigma_M=\dfrac{\sigma}{\sqrt{n}}\right),\qquad \sigma_M=\dfrac{4{,}000}{\sqrt{64}}=\boxed{500\text{ km}}$$
so $\mu_M=50{,}000$ km and the pdf of $M$ is $f(m)=\dfrac{1}{500\sqrt{2\pi}}\exp\!\left[-\dfrac{(m-50{,}000)^2}{2(500)^2}\right]$.
(c)(iii) $Y$ and $M$ on one diagram. $M$ shares the mean of $Y$ but is 8× less spread ($\sigma_M=\sigma/8$), so its curve is much taller and narrower, nested inside the curve of $Y$.
Fig. 1(c)(iii) — $Y\sim N(50{,}000,4{,}000^2)$ (solid, blue) and $M\sim N(50{,}000,500^2)$ (dashed, red) on the same axes; both centred at 50,000 km, $M$ far more concentrated.
(c)(iv) $P(|M-\mu|<300)$. Standardize with $\sigma_M=500$: $z=300/500=0.60$. From the table, $\Phi(0.60)=0.7257$, so
$$P(|M-50{,}000|<300)=2\Phi(0.60)-1=2(0.7257)-1=\boxed{0.4514}$$
(d) Sum $T$ of nine components. $T=\sum_{i=1}^{9}Y_i$ with the $Y_i$ independent, so $E(T)=n\mu=9(50{,}000)=450{,}000$ km and $Var(T)=n\sigma^2=9(4{,}000)^2=144{,}000{,}000$ km$^2$, giving $sd(T)=\sqrt{144{,}000{,}000}=12{,}000$ km. Hence $T\sim N(450{,}000,\,12{,}000^2)$.
Standardizing, $z=\dfrac{444{,}000-450{,}000}{12{,}000}=-0.50$, and $\Phi(-0.50)=1-\Phi(0.50)=1-0.6915=0.3085$, so
$$P(T>444{,}000)=1-0.3085=\boxed{0.6915}$$
Fig. 1(d) — $N(450{,}000,\,12{,}000^2)$ pdf of $T$, shaded area = $P(T>444{,}000)=0.6915$.
