Exam MAS-I — Survival & Reliability Models Practice Flashcards

**Answer: (B).** First integrate the hazard to get the cumulative hazard: $H(t)=\int_{0}^{t}(0.03+0.002u)\,du = 0.03t+0.001t^{2}$. At $t=5$: $H(5)=0.03(5)+0.001(25)=0.15+0.025=0.175$. Then $S(5)=e^{-H(5)}=e^{-0.175}\approx 0.8395$. Dropping the quadratic term (using only $H=0.15$) gives $e^{-0.15}\approx 0.8607$ (C); treating $H$ itself as the failure probability, $1-0.175=0.825$, gives (A). The survival is $e^{-H}$, not $1-H$.

**Answer: (B).** Use $h(t)=-\dfrac{d}{dt}\ln S(t)$. Here $\ln S(t)=-2\ln(1+0.5t)$, so $h(t)=-\dfrac{d}{dt}\big[-2\ln(1+0.5t)\big]=2\cdot\dfrac{0.5}{1+0.5t}=\dfrac{1}{1+0.5t}$. At $t=4$: $h(4)=\dfrac{1}{1+0.5(4)}=\dfrac{1}{3}\approx 0.3333$. Evaluating $S(4)=\dfrac{1}{(1+2)^{2}}=\dfrac{1}{9}\approx 0.1111$ confuses the survival value with the hazard (A); forgetting to bring down the factor $2$ from the exponent but keeping it elsewhere can produce $0.6667$ (D).

**Answer: (B).** The cumulative hazard is $H(t)=\int_{0}^{t}0.003u^{2}\,du = 0.001t^{3}$. At $t=10$: $H(10)=0.001(1000)=1$. Then $S(10)=e^{-H(10)}=e^{-1}\approx 0.3679$. Using the instantaneous hazard $h(10)=0.003(100)=0.3$ as a failure probability and reporting $1-0.3=0.70$ gives (C); $h$ is a rate, not a probability, and survival comes from the integrated $H$.

**Answer: (C).** The density is $f(t)=-S'(t)=-3\left(1-\dfrac{t}{10}\right)^{2}\cdot\left(-\dfrac{1}{10}\right)=\dfrac{3}{10}\left(1-\dfrac{t}{10}\right)^{2}$. Then $h(t)=\dfrac{f(t)}{S(t)}=\dfrac{\frac{3}{10}\left(1-\frac{t}{10}\right)^{2}}{\left(1-\frac{t}{10}\right)^{3}}=\dfrac{3/10}{1-\frac{t}{10}}=\dfrac{3}{10-t}$. At $t=4$: $h(4)=\dfrac{3}{10-4}=\dfrac{3}{6}=0.5$. Reporting the density $f(4)=\frac{3}{10}(0.6)^{2}=0.108$ or the value $S(4)=0.216$ (A) skips the $h=f/S$ step.

**Answer: (B).** The hazard is the derivative of the cumulative hazard: $h(t)=H'(t)=\dfrac{d}{dt}\ln(1+0.4t)=\dfrac{0.4}{1+0.4t}$. At $t=5$: $h(5)=\dfrac{0.4}{1+0.4(5)}=\dfrac{0.4}{3}\approx 0.1333$. Reporting $H(5)=\ln(3)\approx 1.0986$ (E) gives the cumulative hazard, not the instantaneous one; using $\dfrac{0.4}{1+0.4(5)}$ but forgetting the $0.4$ in the numerator yields $\dfrac{1}{3}$-type errors. The corresponding survival, for reference, is $S(5)=e^{-H(5)}=\dfrac{1}{3}$.

**Answer: (A).** For an exponential lifetime the mean is $\dfrac{1}{\lambda}=250$, so the rate is $\lambda=\dfrac{1}{250}=0.004$ per hour and $S(t)=e^{-\lambda t}$. $S(400)=e^{-0.004(400)}=e^{-1.6}\approx 0.2019$. Treating the mean as if it were the time scale, $e^{-400/400}=e^{-1}\approx 0.3679$, gives (B); reporting $1-S(400)=0.798$ gives the failure probability (E). The exponential survival uses $\lambda t=\frac{400}{250}=1.6$.

**Answer: (D).** The Weibull survival function is $S(t)=e^{-(t/\theta)^{\tau}}$. $\left(\dfrac{t}{\theta}\right)^{\tau}=\left(\dfrac{40}{80}\right)^{2}=(0.5)^{2}=0.25$. $S(40)=e^{-0.25}\approx 0.7788$. Forgetting to apply the shape exponent (using $e^{-(40/80)}=e^{-0.5}\approx 0.6065$) gives (B); using $\tau$ as a multiplier of the scale rather than an exponent of the ratio is the most common slip here.

**Answer: (B).** The Weibull hazard is $h(t)=\dfrac{\tau}{\theta}\left(\dfrac{t}{\theta}\right)^{\tau-1}$. With $\tau=2$, $\theta=200$, $t=100$: $h(100)=\dfrac{2}{200}\left(\dfrac{100}{200}\right)^{1}=0.01\times 0.5=0.005$. Forgetting the ratio factor and reporting just $\dfrac{\tau}{\theta}=\dfrac{2}{200}=0.01$ gives (C); since $\tau>1$ the hazard rises with $t$, so this aging component will have a higher hazard later in life.