Reliability and Survival Analysis Questions and Answers

Q: Suppose the lifetime of a system follows the exponential distribution with failure rate 0.8. Then the probability that the system will survive longer than twice its mean time to failure is closest to :

The probability is closest to 0.135.With failure rate \(\lambda = 0.8\), the mean time to failure is \(1/\lambda = 1.25\), and twice that is 2.5. The survival function of the exponential gives\[ \begin{aligned} P(T \gt 2/\lambda) &= e^{-\lambda \cdot 2/\lambda} \\ &= e^{-2} \approx 0.135 \end{aligned} \]The rate cancels: for any exponential lifetime, the chance of surviving beyond twice the mean is \(e^{-2}\), and beyond the mean itself is \(e^{-1} \approx 0.368\).None of 0.331, 0.261 or 0.101 equals \(e^{-2}\); the trap is to evaluate \(e^{-0.8t}\) at a time other than 2.5, for example the mean 1.25 itself, which gives \(e^{-1} \approx 0.368\).

Q: Suppose the lifetimes of two systems, system 1 and system 2, follow exponential distributions with failure rates 2 and ½ respectively. Let the probability that system 'i' fails prior to its mean time to failure be denoted by pi, i = 1, 2. Which of the following is true ?

p1 = p2.For an exponential lifetime with failure rate \(\lambda\), the mean time to failure is \(1/\lambda\), and\[ \begin{aligned} P\left(T \lt \frac{1}{\lambda}\right) &= 1 - e^{-\lambda \cdot 1/\lambda} \\ &= 1 - e^{-1} \\ &\approx 0.632 \end{aligned} \]whatever the rate. So both systems have the same probability, about 0.632, of failing before their own mean life; the rates 2 and ½ change the mean but not this probability. Their product is about 0.4, not 1.

Q: A system consists of 4 components, 3 of which form a parallel subsystem connected in series with the 4th component. The components work independently of each other. If the reliability of each component is 0.8, then the system reliability is :

The system reliability is 0.7936.The parallel block of three components fails only if all three fail, and it is in series with the fourth component:\[ \begin{aligned} R_{\text{parallel}} &= 1 - (0.2)^{3} = 0.992 \\ R &= 0.992 \times 0.8 = 0.7936 \end{aligned} \]0.4096 is \(0.8^{4}\), all four components in series. The other values do not follow from this structure.

Q: Suppose only the first 6 failure times of 18 items kept on test are observed. Suppose the failure times are independent and follow an exponential distribution with mean \(\theta\). If the data are :1.5, 2, 2.5, 3, 4, 5the maximum likelihood estimate of \(\theta\) :

The maximum likelihood estimate is 13.This is type II censoring: r = 6 failures are observed out of n = 18, and the other 12 items are still working at the sixth failure time, 5. The MLE of the exponential mean is the total time on test divided by the number of failures:\[ \begin{aligned} T &= (1.5 + 2 + 2.5 + 3 + 4 + 5) + 12 \times 5 \\ &= 18 + 60 = 78 \\ \hat{\theta} &= \frac{78}{6} = 13 \end{aligned} \]Averaging only the six observed times gives 3, which ignores the 12 surviving items. Censored data still give an MLE, so 'cannot be obtained' is wrong.

Q: Consider a series system with 3 components. Suppose the component lifetimes are independent but with failure rates \(\lambda_{1}\), \(\lambda_{2}\) and \(\lambda_{3}\). Then the failure rate of the system lifetime :

The failure rate of the series system is \(\lambda_{1} + \lambda_{2} + \lambda_{3}\).A series system works only while every component works, so its reliability is the product of the component reliabilities. With independent components,\[ \begin{aligned} R(t) &= R_{1}(t)R_{2}(t)R_{3}(t) \\ h(t) &= -\frac{d}{dt}\ln R(t) = h_{1}(t) + h_{2}(t) + h_{3}(t) \end{aligned} \]so failure rates add. For constant rates the system lifetime is exponential with rate \(\lambda_{1} + \lambda_{2} + \lambda_{3}\).The product of the rates has the wrong units, the minimum would describe the most reliable component alone, and the information given is enough.

Q: Let \(X_{1}, \ldots, X_{n}\) be a random sample from a lifetime distribution F with parameter \(\theta\). Let f denote the corresponding probability density function and let \(T_{0}\) be a fixed time. We observe \(X_{i} = x_{i}\) only if \(x_{i} \le T_{0}\), i = 1, 2, ....., n. Let \(n_{u}\) denote the number of observed lifetimes and \(x_{(1)} \lt x_{(2)} \lt \ldots\) be the ordered observations. Which of the following is the most appropriate 'likelihood' for estimating \(\theta\) ?

The appropriate likelihood is \(\prod_{i=1}^{n_{u}} f(x_{(i)})\,(1 - F(T_{0}))^{n - n_{u}}\).This is Type I censoring at the fixed time \(T_{0}\). Each observed failure contributes its density \(f(x_{(i)})\). Each of the \(n - n_{u}\) items still working at \(T_{0}\) contributes only the information that its lifetime exceeds \(T_{0}\), with probability \(1 - F(T_{0})\), the survival function.Option A uses all n densities as if every lifetime were observed. Option B ignores the censored items, which biases the estimate towards short lifetimes. Option C divides by the density at \(T_{0}\), which is not a probability of surviving past \(T_{0}\).

11.

If the failure rate of a random variable is \(h(t) = 3t + 2\), \(t \ge 0\), then its distribution function :

A.is \(1 - \exp\left(-\frac{3t^{2}}{2} - 2t\right),\ t \ge 0\)
B.is \(1 - \exp\left(-(3t + 2)^{2}\right),\ t \ge 0\)
C.is \(\exp\left(-(3t + 2)^{2}\right),\ t \ge 0\)
D.cannot be obtained from the given information

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The distribution function is \(1 - \exp\left(-\frac{3t^{2}}{2} - 2t\right)\), \(t \ge 0\).

The failure rate determines the distribution through the cumulative hazard \(H(t) = \int_{0}^{t} h(u)\, du\):

\[ \begin{aligned} H(t) &= \int_{0}^{t} (3u + 2)\, du = \frac{3t^{2}}{2} + 2t \\ F(t) &= 1 - e^{-H(t)} = 1 - \exp\left(-\frac{3t^{2}}{2} - 2t\right) \end{aligned} \]

So the distribution can be obtained, which rules out the last option. The options with \((3t + 2)^{2}\) square the hazard instead of integrating it, and the one without '1 -' gives a survival-type function that is not even a valid distribution function, since it decreases.

Discuss in Forum

12.

Suppose the lifetime of a system follows the exponential distribution with failure rate 0.8. Then the probability that the system will survive longer than twice its mean time to failure is closest to :

A.0.331
B.0.261
C.0.135
D.0.101

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The probability is closest to 0.135.

With failure rate \(\lambda = 0.8\), the mean time to failure is \(1/\lambda = 1.25\), and twice that is 2.5. The survival function of the exponential gives

\[ \begin{aligned} P(T \gt 2/\lambda) &= e^{-\lambda \cdot 2/\lambda} \\ &= e^{-2} \approx 0.135 \end{aligned} \]

The rate cancels: for any exponential lifetime, the chance of surviving beyond twice the mean is \(e^{-2}\), and beyond the mean itself is \(e^{-1} \approx 0.368\).

None of 0.331, 0.261 or 0.101 equals \(e^{-2}\); the trap is to evaluate \(e^{-0.8t}\) at a time other than 2.5, for example the mean 1.25 itself, which gives \(e^{-1} \approx 0.368\).

Discuss in Forum

13.

Suppose the lifetimes of two systems, system 1 and system 2, follow exponential distributions with failure rates 2 and ½ respectively. Let the probability that system 'i' fails prior to its mean time to failure be denoted by p_i, i = 1, 2. Which of the following is true ?

A.\(p_{1} \gt p_{2}\)
B.\(p_{1} = p_{2}\)
C.\(p_{1} \lt p_{2}\)
D.\(p_{1} \cdot p_{2} = 1\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

p₁ = p₂.

For an exponential lifetime with failure rate \(\lambda\), the mean time to failure is \(1/\lambda\), and

\[ \begin{aligned} P\left(T \lt \frac{1}{\lambda}\right) &= 1 - e^{-\lambda \cdot 1/\lambda} \\ &= 1 - e^{-1} \\ &\approx 0.632 \end{aligned} \]

whatever the rate. So both systems have the same probability, about 0.632, of failing before their own mean life; the rates 2 and ½ change the mean but not this probability. Their product is about 0.4, not 1.

Discuss in Forum

14.

A system consists of 4 components, 3 of which form a parallel subsystem connected in series with the 4th component. The components work independently of each other. If the reliability of each component is 0.8, then the system reliability is :

A.0.8381
B.0.7132
C.0.4096
D.0.7936

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The system reliability is 0.7936.

The parallel block of three components fails only if all three fail, and it is in series with the fourth component:

\[ \begin{aligned} R_{\text{parallel}} &= 1 - (0.2)^{3} = 0.992 \\ R &= 0.992 \times 0.8 = 0.7936 \end{aligned} \]

0.4096 is \(0.8^{4}\), all four components in series. The other values do not follow from this structure.

Discuss in Forum

15.

Suppose only the first 6 failure times of 18 items kept on test are observed. Suppose the failure times are independent and follow an exponential distribution with mean \(\theta\). If the data are :

1.5, 2, 2.5, 3, 4, 5

the maximum likelihood estimate of \(\theta\) :

A.9
B.cannot be obtained
C.is 3
D.is 13

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The maximum likelihood estimate is 13.

This is type II censoring: r = 6 failures are observed out of n = 18, and the other 12 items are still working at the sixth failure time, 5. The MLE of the exponential mean is the total time on test divided by the number of failures:

\[ \begin{aligned} T &= (1.5 + 2 + 2.5 + 3 + 4 + 5) + 12 \times 5 \\ &= 18 + 60 = 78 \\ \hat{\theta} &= \frac{78}{6} = 13 \end{aligned} \]

Averaging only the six observed times gives 3, which ignores the 12 surviving items. Censored data still give an MLE, so 'cannot be obtained' is wrong.

Discuss in Forum

16.

Consider a series system with 3 components. Suppose the component lifetimes are independent but with failure rates \(\lambda_{1}\), \(\lambda_{2}\) and \(\lambda_{3}\). Then the failure rate of the system lifetime :

A.is \(\lambda_{1} \lambda_{2} \lambda_{3}\)
B.cannot be obtained from the given information
C.\(\min(\lambda_{1}, \lambda_{2}, \lambda_{3})\)
D.\(\lambda_{1} + \lambda_{2} + \lambda_{3}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The failure rate of the series system is \(\lambda_{1} + \lambda_{2} + \lambda_{3}\).

A series system works only while every component works, so its reliability is the product of the component reliabilities. With independent components,

\[ \begin{aligned} R(t) &= R_{1}(t)R_{2}(t)R_{3}(t) \\ h(t) &= -\frac{d}{dt}\ln R(t) = h_{1}(t) + h_{2}(t) + h_{3}(t) \end{aligned} \]

so failure rates add. For constant rates the system lifetime is exponential with rate \(\lambda_{1} + \lambda_{2} + \lambda_{3}\).

The product of the rates has the wrong units, the minimum would describe the most reliable component alone, and the information given is enough.

Discuss in Forum

17.

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a lifetime distribution F with parameter \(\theta\). Let f denote the corresponding probability density function and let \(T_{0}\) be a fixed time. We observe \(X_{i} = x_{i}\) only if \(x_{i} \le T_{0}\), i = 1, 2, ....., n. Let \(n_{u}\) denote the number of observed lifetimes and \(x_{(1)} \lt x_{(2)} \lt \ldots\) be the ordered observations. Which of the following is the most appropriate 'likelihood' for estimating \(\theta\) ?

A.\(\prod_{i=1}^{n} f(x_{(i)})\)
B.\(\prod_{i=1}^{n_{u}} f(x_{(i)})\)
C.\(\prod_{i=1}^{n_{u}} f(x_{(i)}) / (f(T_{0}))^{n-n_{u}}\)
D.\(\prod_{i=1}^{n_{u}} f(x_{(i)})\,(1 - F(T_{0}))^{n-n_{u}}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The appropriate likelihood is \(\prod_{i=1}^{n_{u}} f(x_{(i)})\,(1 - F(T_{0}))^{n - n_{u}}\).

This is Type I censoring at the fixed time \(T_{0}\). Each observed failure contributes its density \(f(x_{(i)})\). Each of the \(n - n_{u}\) items still working at \(T_{0}\) contributes only the information that its lifetime exceeds \(T_{0}\), with probability \(1 - F(T_{0})\), the survival function.

Option A uses all n densities as if every lifetime were observed. Option B ignores the censored items, which biases the estimate towards short lifetimes. Option C divides by the density at \(T_{0}\), which is not a probability of surviving past \(T_{0}\).

Reliability and Survival Analysis Questions and Answers

Topics

Other Subjects

Tags: Reliability and Survival Analysis Questions and Answers

About Us

Reliability and Survival Analysis Questions and Answers

Topics

Other Subjects

Tags: Reliability and Survival Analysis Questions and Answers

About Us

Stay informed about all important Dates & Notifications