Weibull Survival Probabilities and Percentiles

Survival and Failure Probabilities for a Weibull Lifetime

A lifetime model is useful only when its probabilities can be connected to practical questions: What is the chance a component lasts at least a specified time? What is the chance it fails during a particular interval? If it has already survived to an inspection, how likely is it to fail before the next inspection? For a Weibull lifetime, each question can be answered from the survival function.

The Weibull shape parameter controls how the lifetime distribution changes over time, while the scale parameter sets the time scale. Here we use the shape-scale parameterization: shape $k>0$, scale $\lambda>0$, and lifetime $T$ measured in the same time units as $\lambda$. The probabilities below apply to a nonnegative continuous lifetime.

Definition: If $T$ has a Weibull distribution with shape $k$ and scale $\lambda$, write $T\sim\operatorname{Weibull}(k,\lambda)$. For $t\geq 0$, its cumulative distribution function and survival function are $$ F_T(t)=P(T\leq t)=1-\exp\left[-\left(\frac{t}{\lambda}\right)^k\right], \qquad S_T(t)=P(T>t)=\exp\left[-\left(\frac{t}{\lambda}\right)^k\right]. $$ For $t<0$, $F_T(t)=0$ and $S_T(t)=1$. The scale $\lambda$ has units of time; the shape $k$ is dimensionless.

The two functions describe complementary events: $F_T(t)$ is the probability that failure has occurred by time $t$, while $S_T(t)$ is the probability that the lifetime exceeds $t$. Since this is a continuous distribution, the probability of failure at exactly one specified time is zero. Thus it does not matter for these calculations whether an interval endpoint is included or excluded.

Calculating Survival and Interval Failure

For a specified time $t$, substitute the time and the model parameters into $S_T(t)$. The probability of failure by that time is its complement, $F_T(t)=1-S_T(t)$. For two times $a<b$, the probability of failure between them is the difference between the probability of surviving beyond $a$ and the probability of surviving beyond $b$:

P(a<T\leq b)=F_T(b)-F_T(a)=S_T(a)-S_T(b).

This difference is an unconditional probability: it is calculated for a unit observed from time zero, without conditioning on its status at time $a$. If the question instead says the unit has survived to time $a$, the probability must be divided by $S_T(a)$, as derived below.

Worked Example: Failure Between Two Inspection Times

A hypothetical sensor has lifetime $T\sim\operatorname{Weibull}(2,1000)$, where time is measured in operating hours. Calculate (1) the probability it survives beyond 600 hours, (2) the probability it fails between 600 and 900 hours, and (3) the probability it fails in that interval given that it survived to 600 hours. Interest rate: not applicable; no financial discounting is used. Mortality assumption: not applicable; $T$ is a mechanical component lifetime, not a human lifetime.

Method. Use the survival function for both inspection times, take their difference for the unconditional interval probability, and divide by survival to 600 hours for the conditional probability.

At 600 hours, the scaled time is $600/1000=0.6$, so $$ S_T(600)=\exp[-(0.6)^2]=e^{-0.36}=0.6976763. $$ The sensor's probability of surviving beyond 600 hours is $0.6977$, to four decimal places.

At 900 hours, $900/1000=0.9$, giving $$ S_T(900)=\exp[-(0.9)^2]=e^{-0.81}=0.4448581. $$ The unconditional probability of failure between 600 and 900 hours is therefore $$ P(600<T\leq900)=0.6976763-0.4448581=0.2528182, $$ or $0.2528$ to four decimal places.

For the conditional probability, divide the interval probability by the probability of reaching 600 hours: $$ P(600<T\leq900\mid T>600) =\frac{0.2528182}{0.6976763} =0.362372\ldots. $$ Thus, given survival to 600 hours, the probability of failure before 900 hours is $0.3624$, to four decimal places.

Check. The conditional probability is greater than the unconditional interval probability because the conditioning restricts attention to the units that reached 600 hours. The conditional result is also less than 1, as required. A common mistake is to report $S_T(900)=0.4449$ as the conditional probability; that is the probability of surviving past 900 hours from time zero, not the probability of failing between 600 and 900 hours given survival to 600.

Conditional Failure After Surviving to a Given Time

For $t\geq 0$ and $s>0$, consider a unit known to have survived to time $t$. Failure during the next $s$ time units means $t<T\leq t+s$. The conditioning event $T>t$ has probability $S_T(t)$, so the conditional probability is the interval probability divided by $S_T(t)$.

\begin{aligned} P(t<T\leq t+s\mid T>t) &=\frac{P(t<T\leq t+s)}{P(T>t)}\\ &=\frac{S_T(t)-S_T(t+s)}{S_T(t)}\\ &=1-\frac{S_T(t+s)}{S_T(t)}\\ &=1-\exp\left\{-\left[ \left(\frac{t+s}{\lambda}\right)^k -\left(\frac{t}{\lambda}\right)^k \right]\right\}. \end{aligned}

The final line follows by substituting the Weibull survival function and cancelling the common survival factor. This is a useful way to see why the conditional calculation differs from the unconditional one: it uses the change in the exponent between $t$ and $t+s$, rather than the exponent at $t+s$ alone. The formula requires $S_T(t)>0$, which holds for every finite $t$ under this Weibull model.

Identify the probability being asked for.
Distinguish survival beyond a time, failure by a time, failure over an interval, and failure over an interval conditional on prior survival.

Check the time scale.
Use the same time units for $t$ and $\lambda$. Convert months to years, or cycles to thousands of cycles, before forming $t/\lambda$.

Substitute into the appropriate formula.
Use $S_T(t)$ for survival, $1-S_T(t)$ for failure by $t$, $S_T(a)-S_T(b)$ for unconditional interval failure, or the conditional formula when survival to $a$ is given.

Check the result.
Probabilities must lie between 0 and 1. For $a<b$, $S_T(a)\geq S_T(b)$, so $S_T(a)-S_T(b)$ cannot be negative.

Lifetime Percentiles

A lifetime percentile specifies a time by the cumulative proportion of lifetimes at or below it. For $0<r<1$, the $r$-th percentile $t_r$ is defined by $F_T(t_r)=r$. Equivalently, the survival probability at this time is $1-r$. Solving the Weibull CDF for $t_r$ gives:

Percentile formula: For $T\sim\operatorname{Weibull}(k,\lambda)$ and $0<r<1$, $$ F_T(t_r)=r \quad\Longrightarrow\quad t_r=\lambda\left[-\ln(1-r)\right]^{1/k}. $$ The result has the same time units as $\lambda$. In particular, the median is $t_{0.5}=\lambda(\ln 2)^{1/k}$.

To derive the formula, start with the defining percentile equation and isolate the exponential term:

\begin{aligned} r&=1-\exp\left[-\left(\frac{t_r}{\lambda}\right)^k\right]\\ 1-r&=\exp\left[-\left(\frac{t_r}{\lambda}\right)^k\right]\\ -\ln(1-r)&=\left(\frac{t_r}{\lambda}\right)^k\\ t_r&=\lambda\left[-\ln(1-r)\right]^{1/k}. \end{aligned}

The logarithm is taken of $1-r$, not $r$, because $r$ is a cumulative failure probability and $1-r$ is the survival probability. A percentile calculation is therefore an inverse-probability calculation: first translate the percentile into the corresponding tail probability, then solve for time.

Worked Example: Finding the Ninetieth Lifetime Percentile

A hypothetical pump bearing has lifetime $T\sim\operatorname{Weibull}(2,5)$, with time measured in years. Find its ninetieth lifetime percentile. Interest rate: not applicable; no financial discounting is used. Mortality assumption: not applicable; $T$ is a bearing lifetime, not a human lifetime.

Method. Set $r=0.90$ in the percentile formula. The ninetieth percentile is the time by which 90% of bearings have failed, so its survival probability is $1-0.90=0.10$.

Substitute the shape $k=2$, scale $\lambda=5$ years, and percentile fraction $r=0.90$: $$ t_{0.90}=5\left[-\ln(1-0.90)\right]^{1/2} =5\sqrt{-\ln(0.10)}. $$ Since $-\ln(0.10)=\ln(10)=2.302585$, $$ t_{0.90}=5\sqrt{2.302585} =5(1.517427) =7.587135\text{ years}. $$ Rounded to two decimal places, the ninetieth percentile is $7.59$ years.

Check. At the unrounded result, $(7.587135/5)^2$ is approximately $2.302585$, so the survival probability is $e^{-2.302585}=0.10$, and the failure probability is $0.90$. A common mistake is to substitute $0.90$ for the survival probability and calculate $5\sqrt{-\ln(0.90)}$. That produces a low percentile, because it corresponds to only 10% failure by that time, not 90%.

Putting Conditional Probabilities and Percentiles Together

A question may describe an item that has survived to an inspection age and ask about a later interval. The elapsed time and the additional time must be kept distinct. If an item has survived to $t$, then “fails within the next $s$ months” refers to the interval from $t$ to $t+s$, not from zero to $s$. The Weibull model does not in general have the memoryless property; treating the remaining lifetime as a fresh lifetime with the original distribution can therefore give the wrong probability.

Worked Example: Failure in the Next Year After an Inspection

A hypothetical test instrument has lifetime $T\sim\operatorname{Weibull}(0.5,16)$, with time measured in months. Given that it is operating at month 4, calculate the probability it fails by month 16. Also calculate the unconditional probability of failure between months 4 and 16. Interest rate: not applicable; no financial discounting is used. Mortality assumption: not applicable; $T$ is an instrument lifetime, not a human lifetime.

Method. Evaluate the Weibull survival function at both calendar times. For the conditional probability, divide the interval probability by survival to month 4.

At month 4, $4/16=0.25$, and $(0.25)^{0.5}=0.5$. Thus, $$ S_T(4)=e^{-0.5}=0.6065307. $$ At month 16, $16/16=1$, so $$ S_T(16)=e^{-1}=0.3678794. $$ The unconditional probability of failure between months 4 and 16 is $$ P(4<T\leq16)=S_T(4)-S_T(16) =0.6065307-0.3678794 =0.2386513. $$ To condition on operating at month 4, divide this value by $S_T(4)$: $$ P(4<T\leq16\mid T>4) =\frac{0.2386513}{0.6065307} =0.393469\ldots. $$ Therefore, the probability of failure by month 16 given operation at month 4 is $0.3935$, to four decimal places. The unconditional interval probability is $0.2387$, to four decimal places.

Check. The conditional probability can also be evaluated as $1-S_T(16)/S_T(4)=1-e^{-0.5}=0.3935$. The unconditional probability is smaller because it includes all units at time zero, including those that failed before month 4. Using $1-S_T(12)$ would answer a different question: failure by month 12 from the start, not failure within 12 months after the month-4 inspection.

Common Exam Traps

Reversing the percentile tail. The $r$-th percentile has $F_T(t_r)=r$ and $S_T(t_r)=1-r$. For the 90th percentile, use survival $0.10$, not $0.90$.
Using the endpoint survival alone for an interval. The unconditional probability of failure from $a$ to $b$ is $S_T(a)-S_T(b)$, not $S_T(b)$.
Forgetting the condition in a conditional question. Given survival to $a$, divide the interval probability by $S_T(a)$; do not report the unconditional interval probability.
Mixing time units. The ratio $t/\lambda$ must be dimensionless. If the shape-scale model uses $\lambda=2$ years, a time of 6 months must be converted to $0.5$ years before substitution.
Confusing scale and shape. The exponent is $k$, while the scale $\lambda$ appears in the denominator inside $t/\lambda$. Swapping their roles changes the model.

Key takeaway: The Weibull survival function is $S_T(t)=\exp[-(t/\lambda)^k]$. Use it directly for survival, take differences for unconditional interval failure, divide by survival to the starting time for conditional interval failure, and set $F_T(t_r)=r$ to obtain a percentile. These calculations are aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each hypothetical lifetime model, identify whether the requested probability is unconditional or conditional before calculating. Round probabilities to four decimal places and times to two decimal places where needed.

A filter has lifetime $T\sim\operatorname{Weibull}(1,8)$, measured in days. Find the probability it survives beyond 5 days and the probability it fails by day 5. (Answer: survival $0.5353$; failure $0.4647$.)
A relay has lifetime $T\sim\operatorname{Weibull}(2,6)$, measured in months. Find the unconditional probability it fails between months 3 and 9. (Answer: $e^{-0.25}-e^{-2.25}=0.6734$.)
A valve has lifetime $T\sim\operatorname{Weibull}(1.5,10)$, measured in years. Given that it has survived 5 years, find the probability it fails by year 8. (Answer: $1-\exp[-(0.8^{1.5}-0.5^{1.5})]=0.3037$.)
A battery has lifetime $T\sim\operatorname{Weibull}(3,2)$, measured in years. Find its 95th lifetime percentile. (Answer: $2[-\ln(0.05)]^{1/3}=2.88$ years.)
A component has lifetime $T\sim\operatorname{Weibull}(2,1000)$, measured in operating cycles. Given that it has survived 600 cycles, find how many additional cycles are needed for the conditional probability of surviving that long to be $0.90$. (Answer: $1000\sqrt{0.36-\ln(0.90)}-600=82.17$ additional cycles.)

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