{
  "deckName": "Exam MAS-I — Survival & Reliability Models",
  "examCode": "Exam MAS-I",
  "cards": [
    {
      "front": "Define the **survival function** $S(t)$ and give its relationship to the cdf $F(t)$ and density $f(t)$.",
      "back": "For a nonnegative lifetime $T$, $S(t)=\\Pr(T>t)=1-F(t)$ is the probability of surviving past time $t$.\nProperties: $S(0)=1$, $S$ is nonincreasing, and $S(\\infty)=0$.\nThe density is $f(t)=-\\frac{d}{dt}S(t)=F'(t)$, so $S(t)=\\int_t^\\infty f(u)\\,du$.",
      "tag": "Survival & hazard"
    },
    {
      "front": "Define the **hazard (failure) rate** $h(t)$ and state the three equivalent forms.",
      "back": "The hazard rate is the instantaneous failure intensity given survival to $t$:\n$h(t)=\\lim_{\\Delta t\\to 0}\\frac{\\Pr(t<T\\le t+\\Delta t\\mid T>t)}{\\Delta t}$.\nEquivalent forms: $h(t)=\\frac{f(t)}{S(t)}=-\\frac{d}{dt}\\ln S(t)$.\nIt is a rate (can exceed $1$), not a probability. In reliability it is called the **force of mortality** or **failure rate**.",
      "tag": "Survival & hazard"
    },
    {
      "front": "Define the **cumulative hazard** $H(t)$ and state how it recovers $S(t)$.",
      "back": "$H(t)=\\int_0^t h(u)\\,du = -\\ln S(t)$.\nInverting, $S(t)=e^{-H(t)}$. Since $S$ decreases from $1$ to $0$, $H$ increases from $0$ to $\\infty$.\nThis is the master identity of survival analysis: any one of $S$, $f$, $h$, $H$ determines the other three.",
      "tag": "Survival & hazard"
    },
    {
      "front": "Given $S(t)=e^{-0.02t}$ for $t\\ge 0$, find $f(t)$, $h(t)$, and $H(t)$.",
      "back": "Density: $f(t)=-S'(t)=0.02e^{-0.02t}$.\nHazard: $h(t)=\\frac{f(t)}{S(t)}=\\frac{0.02e^{-0.02t}}{e^{-0.02t}}=0.02$ (constant).\nCumulative hazard: $H(t)=-\\ln S(t)=0.02t$.\nThis is an exponential lifetime with rate $\\lambda=0.02$; its constant hazard is the signature of the exponential.",
      "tag": "Survival & hazard"
    },
    {
      "front": "A lifetime has hazard $h(t)=0.01+0.002t$ for $t\\ge 0$. Find $S(t)$ and the probability of surviving past $t=10$.",
      "back": "Cumulative hazard: $H(t)=\\int_0^t(0.01+0.002u)\\,du = 0.01t + 0.001t^{2}$.\nSurvival: $S(t)=e^{-H(t)}=e^{-(0.01t+0.001t^{2})}$.\nAt $t=10$: $H(10)=0.01(10)+0.001(100)=0.1+0.1=0.2$, so $S(10)=e^{-0.2}\\approx 0.8187$.\nThe linearly increasing hazard makes this an aging (wear-out) component.",
      "tag": "Survival & hazard"
    },
    {
      "front": "Recover the survival function from a hazard: given $h(t)=\\frac{2t}{1+t^{2}}$ for $t\\ge 0$, find $S(t)$.",
      "back": "$H(t)=\\int_0^t \\frac{2u}{1+u^{2}}\\,du$. With substitution $w=1+u^{2}$, $dw=2u\\,du$:\n$H(t)=\\big[\\ln(1+u^{2})\\big]_0^t=\\ln(1+t^{2})$.\n$S(t)=e^{-H(t)}=e^{-\\ln(1+t^{2})}=\\frac{1}{1+t^{2}}$.\nCheck: $S(0)=1$ and $S(t)\\to 0$, as required.",
      "tag": "Survival & hazard"
    },
    {
      "front": "State the **exponential** lifetime model: survival, hazard, mean, variance, and its defining property.",
      "back": "For rate $\\lambda>0$: $S(t)=e^{-\\lambda t}$, $f(t)=\\lambda e^{-\\lambda t}$, $h(t)=\\lambda$ (constant).\nMean $E[T]=\\frac{1}{\\lambda}$ (the MTTF), variance $\\frac{1}{\\lambda^{2}}$.\nDefining property: **memorylessness** — $\\Pr(T>s+t\\mid T>s)=\\Pr(T>t)$. A used component is stochastically as good as new.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "A component has an exponential lifetime with mean $500$ hours. Find the hazard rate and the probability it survives past $800$ hours.",
      "back": "Mean $=\\frac{1}{\\lambda}=500$, so the hazard $\\lambda=\\frac{1}{500}=0.002$ per hour (constant).\n$S(800)=e^{-\\lambda(800)}=e^{-0.002(800)}=e^{-1.6}\\approx 0.2019$.\nSo about a $20.2\\%$ chance of lasting beyond $800$ hours.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "An exponential component has already survived $300$ hours. Given $\\lambda=0.002$, find the probability it survives an **additional** $400$ hours.",
      "back": "By memorylessness the past survival is irrelevant:\n$\\Pr(T>300+400\\mid T>300)=\\Pr(T>400)=e^{-\\lambda(400)}=e^{-0.002(400)}=e^{-0.8}\\approx 0.4493$.\nThe used component behaves like a brand-new one — a unique feature of the constant-hazard exponential.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "Define **MTTF** (mean time to failure) and give two ways to compute it from the survival function.",
      "back": "MTTF $=E[T]$, the expected lifetime of a component.\nDirect: $E[T]=\\int_0^\\infty t\\,f(t)\\,dt$.\nVia survival (often easier): for $T\\ge 0$, $E[T]=\\int_0^\\infty S(t)\\,dt$.\nFor the exponential, $\\int_0^\\infty e^{-\\lambda t}\\,dt=\\frac{1}{\\lambda}$, recovering MTTF $=1/\\lambda$.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "A lifetime has $S(t)=\\left(\\frac{100}{100+t}\\right)^{3}$ for $t\\ge 0$ (a Pareto/Lomax form). Find the MTTF.",
      "back": "Use MTTF $=\\int_0^\\infty S(t)\\,dt = \\int_0^\\infty 100^{3}(100+t)^{-3}\\,dt$.\n$=100^{3}\\left[\\frac{(100+t)^{-2}}{-2}\\right]_0^\\infty = 100^{3}\\cdot\\frac{(100)^{-2}}{2}=\\frac{100^{3}}{2\\cdot 100^{2}}=\\frac{100}{2}=50$.\nMTTF $=50$. (General Pareto mean $=\\frac{\\theta}{\\alpha-1}=\\frac{100}{3-1}=50$.)",
      "tag": "Lifetime distributions"
    },
    {
      "front": "State the **Weibull** lifetime model and how the shape parameter $\\tau$ controls the hazard.",
      "back": "Weibull survival: $S(t)=e^{-(t/\\theta)^{\\tau}}$, with scale $\\theta>0$ and shape $\\tau>0$.\nHazard: $h(t)=\\frac{\\tau}{\\theta}\\left(\\frac{t}{\\theta}\\right)^{\\tau-1}\\propto t^{\\tau-1}$.\n**$\\tau>1$:** increasing hazard (wear-out / aging).\n**$\\tau<1$:** decreasing hazard (infant mortality / burn-in).\n**$\\tau=1$:** constant hazard — reduces to the exponential.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "A Weibull lifetime has shape $\\tau=2$ and scale $\\theta=100$. Find the hazard at $t=50$ and the probability of survival past $t=100$.",
      "back": "Hazard: $h(t)=\\frac{\\tau}{\\theta}\\left(\\frac{t}{\\theta}\\right)^{\\tau-1}=\\frac{2}{100}\\left(\\frac{t}{100}\\right)^{1}=\\frac{2t}{100^{2}}=\\frac{2t}{10000}$.\nAt $t=50$: $h(50)=\\frac{2(50)}{10000}=0.01$ per unit (rising in $t$ since $\\tau>1$).\nSurvival: $S(100)=e^{-(100/100)^{2}}=e^{-1}\\approx 0.3679$.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "Distinguish **infant mortality**, **random failure**, and **wear-out** on the bathtub hazard curve, and which $\\tau$ models each.",
      "back": "The bathtub curve plots $h(t)$ in three phases:\n**Infant mortality:** decreasing hazard (early defects burn in) — Weibull $\\tau<1$.\n**Useful life:** roughly constant hazard (random shocks) — exponential, $\\tau=1$.\n**Wear-out:** increasing hazard (aging/fatigue) — Weibull $\\tau>1$.\nReliability engineering blends these phases to describe a component's whole life.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "Define the **reliability function** $R(t)$ and how it relates to the survival function and hazard.",
      "back": "In reliability theory $R(t)=\\Pr(T>t)=S(t)$ — the probability a component is still functioning at time $t$. The two terms are interchangeable.\nThus $R(t)=e^{-H(t)}$, the failure rate is $h(t)=-\\frac{R'(t)}{R(t)}$, and MTTF $=\\int_0^\\infty R(t)\\,dt$.",
      "tag": "Reliability & systems"
    },
    {
      "front": "State the **series system** reliability formula for $n$ independent components and explain when it fails.",
      "back": "A series system works **only if every** component works, so it fails when **any** one fails.\nWith independent components of reliability $R_i$, the system reliability is\n$R_S=\\prod_{i=1}^{n} R_i = R_1 R_2 \\cdots R_n$.\nBecause each $R_i\\le 1$, $R_S$ is never greater than the weakest component — series links degrade reliability.",
      "tag": "Reliability & systems"
    },
    {
      "front": "State the **parallel system** reliability formula for $n$ independent components and explain when it fails.",
      "back": "A parallel (redundant) system fails **only if all** components fail, so it works if **any** one works.\nThe system **un**reliability is the product of component unreliabilities:\n$R_P = 1 - \\prod_{i=1}^{n}(1-R_i)$.\nBecause each $(1-R_i)\\le 1$, $R_P$ is never less than the strongest component — redundancy improves reliability.",
      "tag": "Series & parallel"
    },
    {
      "front": "A series system has three independent components with reliabilities $0.95$, $0.90$, and $0.99$. Find the system reliability.",
      "back": "Series: multiply the component reliabilities.\n$R_S = 0.95 \\times 0.90 \\times 0.99$.\n$0.95\\times 0.90 = 0.855$; $0.855\\times 0.99 = 0.84645$.\n$R_S \\approx 0.8465$.\nNote the system is **less** reliable than its weakest part ($0.90$) — the hallmark of a series arrangement.",
      "tag": "Series & parallel"
    },
    {
      "front": "Two independent components, each with reliability $0.80$, are placed in **parallel**. Find the system reliability.",
      "back": "Parallel: the system fails only if both fail. Each failure probability is $1-0.80=0.20$.\n$R_P = 1 - (1-0.80)(1-0.80) = 1 - (0.20)(0.20) = 1 - 0.04 = 0.96$.\nThe redundant pair ($0.96$) is far more reliable than a single unit ($0.80$).",
      "tag": "Series & parallel"
    },
    {
      "front": "A system has two components in **series**, each of which is itself a **parallel** pair of units with unit reliability $0.90$. Find the overall reliability.",
      "back": "First each parallel pair: $R_{\\text{pair}}=1-(1-0.90)^{2}=1-0.01=0.99$.\nThen the two pairs are in series: $R_S = 0.99 \\times 0.99 = 0.9801$.\nOverall reliability $\\approx 0.9801$. Redundancy at the unit level lifts each block to $0.99$ before the series multiplication.",
      "tag": "Series & parallel"
    },
    {
      "front": "Each component of a $3$-component **series** system is exponential with the same rate $\\lambda=0.01$. Find the system survival function and its MTTF.",
      "back": "Series survival: $R_S(t)=\\prod_i e^{-\\lambda t}=e^{-3\\lambda t}=e^{-0.03t}$.\nThe system itself is exponential with combined rate $3\\lambda=0.03$ (hazards of independent series components **add**).\nMTTF $=\\frac{1}{3\\lambda}=\\frac{1}{0.03}\\approx 33.33$, one-third the single-component MTTF of $100$.",
      "tag": "Reliability & systems"
    },
    {
      "front": "Why do the **hazard rates add** for independent components in series, and what does this imply for the system MTTF?",
      "back": "Series survival is $R_S(t)=\\prod_i e^{-H_i(t)}=e^{-\\sum_i H_i(t)}$, so $H_S(t)=\\sum_i H_i(t)$ and therefore $h_S(t)=\\sum_i h_i(t)$ — the hazards add.\nFor identical exponential parts each with rate $\\lambda$, $h_S=n\\lambda$ and the system MTTF $=\\frac{1}{n\\lambda}$, shrinking as components are added in series.",
      "tag": "Reliability & systems"
    },
    {
      "front": "Define a **$k$-out-of-$n$** system and give the reliability formula for identical independent components.",
      "back": "A $k$-out-of-$n$ system works if **at least $k$** of its $n$ components work. (Series is the $n$-out-of-$n$ case; parallel is $1$-out-of-$n$.)\nFor i.i.d. components each with reliability $p$, the number working is Binomial$(n,p)$, so\n$R = \\sum_{j=k}^{n}\\binom{n}{j} p^{j}(1-p)^{n-j}$.",
      "tag": "Reliability & systems"
    },
    {
      "front": "A $2$-out-of-$3$ system has identical independent components each with reliability $0.90$. Find the system reliability.",
      "back": "Works if at least $2$ of $3$ work; let $X\\sim\\text{Bin}(3,0.9)$, need $\\Pr(X\\ge 2)$.\n$\\Pr(X=2)=\\binom{3}{2}(0.9)^{2}(0.1)=3(0.81)(0.1)=0.243$.\n$\\Pr(X=3)=(0.9)^{3}=0.729$.\n$R=0.243+0.729=0.972$.\nThe $2$-of-$3$ majority system ($0.972$) beats a single unit ($0.90$) without full triple redundancy.",
      "tag": "Reliability & systems"
    },
    {
      "front": "A bridge-free system has component A in **series** with a **parallel** block of B and C. Reliabilities: $R_A=0.98$, $R_B=0.85$, $R_C=0.85$. Find the system reliability.",
      "back": "Parallel block B$\\parallel$C: $R_{BC}=1-(1-0.85)(1-0.85)=1-(0.15)^{2}=1-0.0225=0.9775$.\nThen A in series with that block: $R_S = R_A \\times R_{BC}=0.98\\times 0.9775$.\n$=0.95795$.\n$R_S \\approx 0.9580$.",
      "tag": "Series & parallel"
    },
    {
      "front": "Two independent components have exponential lifetimes with rates $\\lambda_1=0.01$ and $\\lambda_2=0.02$, arranged in **parallel**. Find the probability the system survives past $t=50$.",
      "back": "Component survivals at $t=50$: $R_1=e^{-0.01(50)}=e^{-0.5}\\approx 0.6065$, $R_2=e^{-0.02(50)}=e^{-1}\\approx 0.3679$.\nParallel: $R_P = 1-(1-R_1)(1-R_2)=1-(0.3935)(0.6321)$.\n$(0.3935)(0.6321)\\approx 0.2487$.\n$R_P \\approx 1-0.2487 = 0.7513$.",
      "tag": "Series & parallel"
    },
    {
      "front": "Why is a **parallel** system's lifetime distribution generally NOT exponential even when its components are exponential?",
      "back": "For a parallel pair the system lifetime is the **maximum** of the component lifetimes, $T_S=\\max(T_1,T_2)$, with $R_P(t)=1-(1-e^{-\\lambda_1 t})(1-e^{-\\lambda_2 t})$.\nDifferentiating shows the hazard $h_S(t)$ is **not constant** — it rises from $0$ as the system ages. The memoryless property is lost; redundancy creates an increasing failure rate even from constant-hazard parts.",
      "tag": "Reliability & systems"
    },
    {
      "front": "Define the **mean residual life** (mean excess life) $e(t)$ and give its integral formula.",
      "back": "$e(t)=E[T-t\\mid T>t]$ is the expected remaining lifetime given survival to $t$.\nFormula: $e(t)=\\frac{1}{S(t)}\\int_t^\\infty S(u)\\,du$.\nIn loss models this is the **mean excess loss** $e_X(d)=E[X-d\\mid X>d]$ above a deductible $d$. At $t=0$ it equals the MTTF: $e(0)=E[T]$.",
      "tag": "Mean residual life"
    },
    {
      "front": "What is the mean residual life $e(t)$ of an **exponential** lifetime, and why?",
      "back": "For an exponential with rate $\\lambda$, $e(t)=\\frac{1}{S(t)}\\int_t^\\infty e^{-\\lambda u}\\,du = \\frac{1}{e^{-\\lambda t}}\\cdot\\frac{e^{-\\lambda t}}{\\lambda}=\\frac{1}{\\lambda}$.\nThe mean residual life is **constant** at $\\frac{1}{\\lambda}=$ MTTF for every $t$, a direct consequence of memorylessness: a survivor's expected remaining life never changes.",
      "tag": "Mean residual life"
    },
    {
      "front": "A lifetime is uniform on $[0,100]$. Find the mean residual life $e(40)$.",
      "back": "For Uniform$[0,100]$, $S(t)=\\frac{100-t}{100}$ for $0\\le t\\le 100$.\n$\\int_t^{100} S(u)\\,du = \\int_t^{100}\\frac{100-u}{100}\\,du = \\frac{(100-t)^{2}}{200}$.\n$e(t)=\\frac{1}{S(t)}\\cdot\\frac{(100-t)^{2}}{200}=\\frac{100}{100-t}\\cdot\\frac{(100-t)^{2}}{200}=\\frac{100-t}{2}$.\n$e(40)=\\frac{100-40}{2}=30$. (Expected remaining life of a uniform survivor is half the time left.)",
      "tag": "Mean residual life"
    },
    {
      "front": "How does the **sign of the hazard's trend** relate to whether mean residual life increases or decreases (DFR vs IFR)?",
      "back": "**Increasing hazard (IFR, e.g. Weibull $\\tau>1$):** an aging component's expected remaining life **decreases** with $t$.\n**Decreasing hazard (DFR, e.g. Weibull $\\tau<1$):** mean residual life **increases** with $t$ — survivors that pass the burn-in phase are sturdier.\n**Constant hazard (exponential):** $e(t)$ is constant.\nThus $e'(t)$ and $h'(t)$ generally have opposite signs.",
      "tag": "Mean residual life"
    },
    {
      "front": "A lifetime has $S(t)=e^{-0.05t}$. A maintenance contract pays for the **expected remaining life** of a unit that has survived to $t=20$. What value does it use?",
      "back": "This is exponential with $\\lambda=0.05$, so by memorylessness $e(20)=\\frac{1}{\\lambda}=\\frac{1}{0.05}=20$.\nThe contract values the expected remaining life at $20$ time units — exactly the original MTTF, regardless of the $20$ units already survived.",
      "tag": "Mean residual life"
    },
    {
      "front": "Relate the mean residual life $e(d)$ to the **expected value with a deductible** in loss models.",
      "back": "Splitting the unconditional mean of losses above $d$:\n$E[(X-d)_+] = E[X-d\\mid X>d]\\cdot \\Pr(X>d) = e(d)\\,S(d)$.\nSo the expected payment per loss with deductible $d$ equals the mean excess loss times the survival probability. Equivalently $e(d)=\\frac{E[(X-d)_+]}{S(d)}$ — the per-payment severity above the deductible.",
      "tag": "Mean residual life"
    },
    {
      "front": "What problem do the **Kaplan-Meier** and **Nelson-Aalen** estimators solve, and what is **right censoring**?",
      "back": "They estimate $S(t)$ (or $H(t)$) **nonparametrically** from lifetime data without assuming a distribution.\n**Right censoring:** for some subjects we only know the lifetime exceeds their last observed time (study ended, lost to follow-up) — the exact failure time is unknown but bounded below. KM and NA correctly use this partial information through the at-risk set, rather than discarding censored observations.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "State the **Kaplan-Meier** (product-limit) estimator and define $r_i$ and $s_i$.",
      "back": "At each distinct **death time** $t_i$, let $r_i$ = number **at risk** (alive and uncensored) just before $t_i$ and $s_i$ = number of **deaths** at $t_i$.\n$\\hat S(t)=\\prod_{t_i\\le t}\\left(1-\\frac{s_i}{r_i}\\right)$.\nThe estimate is a right-continuous step function, dropping only at observed death times; censored times reduce later $r_i$ but cause no drop.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "State the **Nelson-Aalen** estimator of the cumulative hazard and how it gives an alternative estimate of $S(t)$.",
      "back": "$\\hat H(t)=\\sum_{t_i\\le t}\\frac{s_i}{r_i}$ — a sum (not a product) of the per-time hazard contributions.\nAn alternative survival estimate is $\\tilde S(t)=e^{-\\hat H(t)}$.\nNelson-Aalen is preferred for estimating $H$ directly and tends to be more stable with small samples; it generally gives slightly **higher** survival estimates than KM.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "Ten units are on test. Failures occur at $t=2$ ($1$ unit), $t=5$ ($2$ units), and $t=8$ ($1$ unit); no censoring. Compute the Kaplan-Meier estimate $\\hat S(8)$.",
      "back": "Build the risk sets:\n$t=2$: $r_1=10$, $s_1=1 \\Rightarrow 1-\\frac{1}{10}=0.9$.\n$t=5$: $r_2=9$, $s_2=2 \\Rightarrow 1-\\frac{2}{9}=\\frac{7}{9}\\approx 0.7778$.\n$t=8$: $r_3=7$, $s_3=1 \\Rightarrow 1-\\frac{1}{7}=\\frac{6}{7}\\approx 0.8571$.\n$\\hat S(8)=0.9\\times 0.7778\\times 0.8571 \\approx 0.6000$.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "Eight units are tested. Deaths at $t=3,7,9$; a unit is **censored** at $t=5$. Risk counts: $r=8$ at $t=3$, then $r=6$ at $t=7$, $r=5$ at $t=9$ (one death each). Compute $\\hat S(9)$.",
      "back": "Censoring at $t=5$ removes a unit from the risk set without a survival drop.\n$t=3$: $1-\\frac{1}{8}=0.8750$.\n$t=7$: $1-\\frac{1}{6}\\approx 0.8333$ (the death at $t=3$ and the censoring at $t=5$ both leave $6$ at risk just before $t=7$).\n$t=9$: $1-\\frac{1}{5}=0.8000$.\n$\\hat S(9)=0.8750\\times 0.8333\\times 0.8000 \\approx 0.5833$.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "Using the same data — $10$ units, deaths $1$ at $t=2$, $2$ at $t=5$, $1$ at $t=8$, no censoring — compute the Nelson-Aalen $\\hat H(8)$ and the implied $\\tilde S(8)=e^{-\\hat H(8)}$.",
      "back": "Sum the hazard contributions $\\frac{s_i}{r_i}$:\n$\\hat H(8)=\\frac{1}{10}+\\frac{2}{9}+\\frac{1}{7}=0.1000+0.2222+0.1429=0.4651$.\n$\\tilde S(8)=e^{-0.4651}\\approx 0.6281$.\nThis exceeds the Kaplan-Meier $\\hat S(8)\\approx 0.6000$, illustrating that Nelson-Aalen typically gives a slightly higher survival estimate.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "How does a **censored** observation at time $c$ enter the Kaplan-Meier calculation, and why does it not cause $\\hat S$ to drop?",
      "back": "A unit censored at $c$ stays counted in the at-risk set $r_i$ for all death times $t_i\\le c$, then is removed before any later death time. Because no death is recorded at $c$, there is no factor $\\left(1-\\frac{s_i}{r_i}\\right)$ for it — $\\hat S$ does not step down there.\nIts only effect is to shrink the later risk sets, which makes subsequent death-time drops slightly larger. This is how KM extracts information from incomplete (right-censored) lifetimes.",
      "tag": "KM & Nelson-Aalen"
    },
    {
      "front": "Define the **maximum likelihood estimator** of an exponential rate from complete lifetime data, and give it for a sample mean of $250$.",
      "back": "For i.i.d. exponential$(\\lambda)$ data $t_1,\\dots,t_n$, the log-likelihood is $\\ell(\\lambda)=n\\ln\\lambda - \\lambda\\sum t_i$. Setting $\\ell'(\\lambda)=\\frac{n}{\\lambda}-\\sum t_i=0$ gives $\\hat\\lambda=\\frac{n}{\\sum t_i}=\\frac{1}{\\bar t}$.\nWith $\\bar t=250$: $\\hat\\lambda=\\frac{1}{250}=0.004$, so the estimated MTTF is $\\frac{1}{\\hat\\lambda}=250$.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "A test runs $20$ identical exponential units; total observed run-time across all units is $4{,}000$ hours and $8$ failures occur (Type I censoring). Estimate the failure rate $\\hat\\lambda$ and MTTF.",
      "back": "For exponential data with censoring, the MLE of the rate is\n$\\hat\\lambda=\\frac{\\text{number of failures}}{\\text{total time on test}}=\\frac{8}{4000}=0.002$ per hour.\nMTTF $=\\frac{1}{\\hat\\lambda}=\\frac{1}{0.002}=500$ hours.\nOnly the failure count and aggregate exposure are needed — a standard reliability life-test result.",
      "tag": "Lifetime distributions"
    },
    {
      "front": "Compare **series** and **parallel** systems of identical components on reliability and on system MTTF (assume constant-hazard parts).",
      "back": "**Series ($n$ identical, reliability $p$):** $R_S=p^{n}\\le p$; reliability **falls** as parts are added. For exponential parts the MTTF $=\\frac{1}{n\\lambda}$ — adding parts shortens system life.\n**Parallel ($n$ identical):** $R_P=1-(1-p)^{n}\\ge p$; reliability **rises** toward $1$. For exponential parts the MTTF $=\\frac{1}{\\lambda}\\sum_{j=1}^{n}\\frac{1}{j}$ — adding redundancy lengthens system life.\nSeries trades reliability for nothing; parallel buys reliability at the cost of redundant hardware.",
      "tag": "Reliability & systems"
    }
  ]
}