Reliability and Survival Analysis Questions and Answers

Q: Let X be a random variable with distribution function :\(F_{\alpha}(t) = 1 - \exp(-(\lambda t)^{\alpha})\), \(\lambda, \alpha \gt 0\), \(t \ge 0\)Then :

\(F_{\alpha}\) is IFR if \(1 \le \alpha\).This is the Weibull distribution. Its hazard rate is\[ \begin{aligned} h(t) &= \frac{f(t)}{1 - F(t)} \\ &= \alpha\lambda^{\alpha}t^{\alpha - 1} \end{aligned} \]which is increasing in \(t\) when \(\alpha \gt 1\), constant when \(\alpha = 1\) (the exponential case, which counts as both IFR and DFR), and decreasing when \(\alpha \lt 1\). So the distribution is IFR for \(\alpha \ge 1\) and DFR for \(\alpha \le 1\).It is not IFR or DFR for every \(\alpha\), so A and B fail, and C has the range reversed.

Q: Suppose that in a K-unit parallel system each unit has life time distribution with distribution function F. Then reliability function of the system at time \(t\) is given by :

The reliability is \(1 - (F(t))^{K}\).A parallel system works as long as at least one unit works, so it fails only when all K units have failed by time \(t\). With independent units each failing by \(t\) with probability \(F(t)\), that probability is \(F(t)^{K}\), and the reliability is its complement.\((1 - F(t))^{K}\) is the reliability of a series system, which needs every unit to survive. \(F(t)^{K}\) is the failure probability of the parallel system, and \(1 - (1 - F(t))^{K}\) is the failure probability of the series system.

Q: The failure time of a component is exponentially distributed with mean life 200 hours. The design of the component is modified after which the mean life has increased to 400 hours. What is the amount of increase in reliability at 800 hours ? (Let \(a = e^{-2}\).)

The increase in reliability is \(a(1 - a)\).For an exponential life with mean \(\theta\), the reliability at time \(t\) is \(R(t) = e^{-t/\theta}\). At 800 hours:\[ \begin{aligned} R_{\text{old}} &= e^{-800/200} = e^{-4} = a^{2} \\ R_{\text{new}} &= e^{-800/400} = e^{-2} = a \end{aligned} \]The increase is \(a - a^{2} = a(1 - a) \approx 0.117\).\(a\) and \(a^{2}\) are the new and old reliabilities themselves, and \((1 - a)/a\) is the relative increase \((a - a^{2})/a^{2}\), not the amount.

Q: If the failure rate function \(r\) of a component is \(r(t) = \dfrac{t}{1 + t}\), \(t \ge 0\); then its survival function is :

The survival function is \((1 + t)e^{-t}\).The survival function is \(S(t) = \exp\left(-\int_{0}^{t} r(u)\,du\right)\). Since \(\dfrac{u}{1 + u} = 1 - \dfrac{1}{1 + u}\),\[ \int_{0}^{t}\frac{u}{1 + u}du = t - \ln(1 + t) \]so \(S(t) = e^{-t + \ln(1 + t)} = (1 + t)e^{-t}\). Check: \(S(0) = 1\), and \(-S'(t)/S(t) = t/(1 + t)\).\(e^{-t}\) has constant hazard 1, \(te^{-t}\) is 0 at \(t = 0\), and \(1 - e^{-t}\) is a distribution function, not a survival function. (This is the gamma(2, 1) lifetime.)

Q: Suppose X is an absolutely continuous non-negative real-valued random variable whose distribution function is \(F(x)\) and probability density function is \(f(x)\). Suppose it is given that \(f(x)/[1 - F(x)] = k\), a non-negative constant. Then which of the following statements is true ?

The density is \(f(x) = k\,e^{-kx},\ x \gt 0\).\(f(x)/[1 - F(x)]\) is the hazard (failure) rate. Writing \(S = 1 - F\), we have \(f = -S'\), so\[ \begin{aligned} -\frac{S'(x)}{S(x)} &= k \\ S(x) &= e^{-kx} \quad (S(0) = 1) \\ f(x) &= ke^{-kx} \end{aligned} \]A constant hazard rate characterises the exponential distribution (its memoryless property). \(f(x) = 1\) on \((0, \infty)\) is not a density at all, adding a constant C would make the total probability infinite, and the step function in (D) does not integrate to 1 either.

Q: The denominator in infant mortality rate is :

The denominator is the total number of live births in a given period.The infant mortality rate is\[ \begin{aligned} &\text{IMR} \\ &= \frac{\text{deaths under 1 year in the year}}{\text{live births in the year}} \times 1000 \end{aligned} \]The population at risk of dying in infancy is the live-born children, so live births are the denominator. Still births are counted in the perinatal or still-birth rates, pregnancies in fertility measures, and total population in the crude death rate.

Q: What is the failure rate function of the series system of two independent components when the failure rate functions of the two components are respectively \(r_{1}(t)\) and \(r_{2}(t)\) ?

The failure rate is \(r_{1}(t) + r_{2}(t)\).A series system works only while both components work, so its reliability is the product of theirs:\[ \begin{aligned} R(t) &= R_{1}(t)R_{2}(t) \\ &= e^{-\int_{0}^{t} (r_{1} + r_{2})\,du} \end{aligned} \]since each \(R_{i}(t) = \exp(-\int_{0}^{t} r_{i})\). Hence the system's failure rate \(-R'/R\) is \(r_{1}(t) + r_{2}(t)\): hazards add in series. The minimum or maximum would ignore one component's risk, and the product has no reliability meaning here.

Q: A column \(L_{x}\) in the life table denotes :

\(L_{x}\) is the number of years lived by the persons in the age interval \((x, x + 1)\).In a life table, \(L_{x}\) is the total person-years lived between exact ages \(x\) and \(x + 1\) by the cohort, approximately \((l_{x} + l_{x+1})/2\). Summing \(L_{x}\) from age \(x\) onwards gives \(T_{x}\), and \(e_{x}^{0} = T_{x}/l_{x}\).The number surviving to age \(x\) is \(l_{x}\) (B), to age \(x + 1\) is \(l_{x+1}\) (C), and the average remaining lifetime is the expectation of life \(e_{x}^{0}\) (D).

Q: Suppose the lifetimes of systems A and B are independent and follow exponential distributions with failure rates 3 and 2 respectively. Then the probability that system A fails before system B is :

The probability that A fails first is 0.6.For independent exponential lifetimes with rates \(\lambda_{A}\) and \(\lambda_{B}\),\[ \begin{aligned} P(T_{A} \lt T_{B}) &= \int_{0}^{\infty} \lambda_{A} e^{-\lambda_{A}t} e^{-\lambda_{B}t}\, dt \\ &= \frac{\lambda_{A}}{\lambda_{A} + \lambda_{B}} = \frac{3}{3 + 2} = 0.6 \end{aligned} \]The system with the higher failure rate is more likely to fail first, so the answer must exceed 1/2. The value 0.4 is \(\lambda_{B}/(\lambda_{A} + \lambda_{B})\), the probability that B fails first. 0.5 would need equal rates, and 1 would need B never to fail.

Let X be a random variable with distribution function :

\(F_{\alpha}(t) = 1 - \exp(-(\lambda t)^{\alpha})\), \(\lambda, \alpha \gt 0\), \(t \ge 0\)

Then :

A.\(F_{\alpha}\) is IFR
B.\(F_{\alpha}\) is DFR
C.\(F_{\alpha}\) is IFR if \(0 \lt \alpha \le 1\)
D.\(F_{\alpha}\) is IFR if \(1 \le \alpha\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

\(F_{\alpha}\) is IFR if \(1 \le \alpha\).

This is the Weibull distribution. Its hazard rate is

\[ \begin{aligned} h(t) &= \frac{f(t)}{1 - F(t)} \\ &= \alpha\lambda^{\alpha}t^{\alpha - 1} \end{aligned} \]

which is increasing in \(t\) when \(\alpha \gt 1\), constant when \(\alpha = 1\) (the exponential case, which counts as both IFR and DFR), and decreasing when \(\alpha \lt 1\). So the distribution is IFR for \(\alpha \ge 1\) and DFR for \(\alpha \le 1\).

It is not IFR or DFR for every \(\alpha\), so A and B fail, and C has the range reversed.

Discuss in Forum

Suppose that in a K-unit parallel system each unit has life time distribution with distribution function F. Then reliability function of the system at time \(t\) is given by :

A.\((F(t))^{K}\)
B.\((1 - F(t))^{K}\)
C.\(1 - (F(t))^{K}\)
D.\(1 - (1 - F(t))^{K}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The reliability is \(1 - (F(t))^{K}\).

A parallel system works as long as at least one unit works, so it fails only when all K units have failed by time \(t\). With independent units each failing by \(t\) with probability \(F(t)\), that probability is \(F(t)^{K}\), and the reliability is its complement.

\((1 - F(t))^{K}\) is the reliability of a series system, which needs every unit to survive. \(F(t)^{K}\) is the failure probability of the parallel system, and \(1 - (1 - F(t))^{K}\) is the failure probability of the series system.

Discuss in Forum

The failure time of a component is exponentially distributed with mean life 200 hours. The design of the component is modified after which the mean life has increased to 400 hours. What is the amount of increase in reliability at 800 hours ? (Let \(a = e^{-2}\).)

A.\(a\)
B.\(a^{2}\)
C.\(a(1 - a)\)
D.\((1 - a)/a\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The increase in reliability is \(a(1 - a)\).

For an exponential life with mean \(\theta\), the reliability at time \(t\) is \(R(t) = e^{-t/\theta}\). At 800 hours:

\[ \begin{aligned} R_{\text{old}} &= e^{-800/200} = e^{-4} = a^{2} \\ R_{\text{new}} &= e^{-800/400} = e^{-2} = a \end{aligned} \]

The increase is \(a - a^{2} = a(1 - a) \approx 0.117\).

\(a\) and \(a^{2}\) are the new and old reliabilities themselves, and \((1 - a)/a\) is the relative increase \((a - a^{2})/a^{2}\), not the amount.

Discuss in Forum

If the failure rate function \(r\) of a component is \(r(t) = \dfrac{t}{1 + t}\), \(t \ge 0\); then its survival function is :

A.\(e^{-t}\)
B.\(te^{-t}\)
C.\(1 - e^{-t}\)
D.\((1 + t)e^{-t}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The survival function is \((1 + t)e^{-t}\).

The survival function is \(S(t) = \exp\left(-\int_{0}^{t} r(u)\,du\right)\). Since \(\dfrac{u}{1 + u} = 1 - \dfrac{1}{1 + u}\),

\[ \int_{0}^{t}\frac{u}{1 + u}du = t - \ln(1 + t) \]

so \(S(t) = e^{-t + \ln(1 + t)} = (1 + t)e^{-t}\). Check: \(S(0) = 1\), and \(-S'(t)/S(t) = t/(1 + t)\).

\(e^{-t}\) has constant hazard 1, \(te^{-t}\) is 0 at \(t = 0\), and \(1 - e^{-t}\) is a distribution function, not a survival function. (This is the gamma(2, 1) lifetime.)

Discuss in Forum

Suppose X is an absolutely continuous non-negative real-valued random variable whose distribution function is \(F(x)\) and probability density function is \(f(x)\). Suppose it is given that \(f(x)/[1 - F(x)] = k\), a non-negative constant. Then which of the following statements is true ?

A.\(f(x) = 1, 0 \lt x \lt \infty\)
B.\(f(x) = k\, e^{-kx}, x \gt 0\)
C.\(f(x) = k\, e^{-kx} + C, x \gt 0\), C a constant
D.\(f(x) = \begin{cases} 0 & x \lt k, \\ 1 & x \ge k. \end{cases}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The density is \(f(x) = k\,e^{-kx},\ x \gt 0\).

\(f(x)/[1 - F(x)]\) is the hazard (failure) rate. Writing \(S = 1 - F\), we have \(f = -S'\), so

\[ \begin{aligned} -\frac{S'(x)}{S(x)} &= k \\ S(x) &= e^{-kx} \quad (S(0) = 1) \\ f(x) &= ke^{-kx} \end{aligned} \]

A constant hazard rate characterises the exponential distribution (its memoryless property). \(f(x) = 1\) on \((0, \infty)\) is not a density at all, adding a constant C would make the total probability infinite, and the step function in (D) does not integrate to 1 either.

Discuss in Forum

The denominator in infant mortality rate is :

A.total number of live births in a given period
B.total number of live births and still births in a given period
C.total number of pregnancies in a given period
D.total population in a given period

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The denominator is the total number of live births in a given period.

The infant mortality rate is

\[ \begin{aligned} &\text{IMR} \\ &= \frac{\text{deaths under 1 year in the year}}{\text{live births in the year}} \times 1000 \end{aligned} \]

The population at risk of dying in infancy is the live-born children, so live births are the denominator. Still births are counted in the perinatal or still-birth rates, pregnancies in fertility measures, and total population in the crude death rate.

Discuss in Forum

In terms of life table functions, the chance that a child just born to a mother of age 31 and a father of age 33 will be alive till 40 years but will be orphaned by both parents is :

A.\(\left(\dfrac{l_{40}}{l_{0}}\right)\left(\dfrac{l_{71} - l_{31}}{l_{0}}\right)\) \(\left(\dfrac{l_{73} - l_{33}}{l_{0}}\right)\)
B.\(p_{40}\, q_{31}\, q_{33}\)
C.\(\left(\dfrac{L_{40}}{l_{0}}\right)\left(\dfrac{L_{71} - L_{31}}{l_{0}}\right)\) \(\left(\dfrac{L_{73} - L_{33}}{l_{0}}\right)\)
D.\(\left(\dfrac{l_{40}}{l_{0}}\right)\left(\dfrac{l_{71} - l_{31}}{l_{31}}\right)\) \(\left(\dfrac{l_{73} - l_{33}}{l_{33}}\right)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The correct expression is the one in option D, which matches the key.

The child must survive to 40, which has probability \(l_{40}/l_{0}\). During those 40 years the mother (31 at the birth) goes from age 31 to age 71 and the father from 33 to 73. The mother must die in that span, given that she was alive at 31, and the father likewise:

\[ \begin{aligned} &\frac{l_{40}}{l_{0}} \cdot \frac{l_{31} - l_{71}}{l_{31}} \\ &\quad \cdot \frac{l_{33} - l_{73}}{l_{33}} \end{aligned} \]

Option D prints the differences as \(l_{71} - l_{31}\) and \(l_{73} - l_{33}\), each of which is negative, but their product is the same as that of \((l_{31} - l_{71})(l_{33} - l_{73})\), so the value is correct. A divides by \(l_{0}\) instead of conditioning on the parents' ages, C uses \(L_{x}\) (person-years lived) and B is not a probability of this event.

Discuss in Forum

What is the failure rate function of the series system of two independent components when the failure rate functions of the two components are respectively \(r_{1}(t)\) and \(r_{2}(t)\) ?

A.\(\min(r_{1}(t), r_{2}(t))\)
B.\(r_{1}(t) + r_{2}(t)\)
C.\(r_{1}(t) \times r_{2}(t)\)
D.\(\max(r_{1}(t), r_{2}(t))\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The failure rate is \(r_{1}(t) + r_{2}(t)\).

A series system works only while both components work, so its reliability is the product of theirs:

\[ \begin{aligned} R(t) &= R_{1}(t)R_{2}(t) \\ &= e^{-\int_{0}^{t} (r_{1} + r_{2})\,du} \end{aligned} \]

since each \(R_{i}(t) = \exp(-\int_{0}^{t} r_{i})\). Hence the system's failure rate \(-R'/R\) is \(r_{1}(t) + r_{2}(t)\): hazards add in series. The minimum or maximum would ignore one component's risk, and the product has no reliability meaning here.

Discuss in Forum

A column \(L_{x}\) in the life table denotes :

A.the number of years lived by the persons in the age interval \((x, x + 1)\)
B.the number of individuals surviving to age \(x\)
C.the number of individuals surviving to age \(x + 1\)
D.the average number of years of life remaining after age \(x\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

\(L_{x}\) is the number of years lived by the persons in the age interval \((x, x + 1)\).

In a life table, \(L_{x}\) is the total person-years lived between exact ages \(x\) and \(x + 1\) by the cohort, approximately \((l_{x} + l_{x+1})/2\). Summing \(L_{x}\) from age \(x\) onwards gives \(T_{x}\), and \(e_{x}^{0} = T_{x}/l_{x}\).

The number surviving to age \(x\) is \(l_{x}\) (B), to age \(x + 1\) is \(l_{x+1}\) (C), and the average remaining lifetime is the expectation of life \(e_{x}^{0}\) (D).

Discuss in Forum

10.

Suppose the lifetimes of systems A and B are independent and follow exponential distributions with failure rates 3 and 2 respectively. Then the probability that system A fails before system B is :

A.0.4
B.0.5
C.0.6
D.1

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The probability that A fails first is 0.6.

For independent exponential lifetimes with rates \(\lambda_{A}\) and \(\lambda_{B}\),

\[ \begin{aligned} P(T_{A} \lt T_{B}) &= \int_{0}^{\infty} \lambda_{A} e^{-\lambda_{A}t} e^{-\lambda_{B}t}\, dt \\ &= \frac{\lambda_{A}}{\lambda_{A} + \lambda_{B}} = \frac{3}{3 + 2} = 0.6 \end{aligned} \]

The system with the higher failure rate is more likely to fail first, so the answer must exceed 1/2. The value 0.4 is \(\lambda_{B}/(\lambda_{A} + \lambda_{B})\), the probability that B fails first. 0.5 would need equal rates, and 1 would need B never to fail.

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