QCE Specialist Mathematics - Unit 4 - Applications of integral calculus
Exponential Random Variables | QCE Specialist Mathematics
Learn QCE Specialist exponential random variables, density functions, means, quantiles and probability modelling.
Updated 2026-05-18 - 4 min read
QCAA official coverage - Specialist Mathematics 2025 v1.4
Exact syllabus points covered
- Understand and use the probability density function $f(t)=\lambda e^{-\lambda t}$ for $t\ge0$ of the exponential random variable with parameter $\lambda>0$, with mean $\frac1\lambda$ and standard deviation $\frac1\lambda$.
- Model and solve problems that involve exponential random variables and associated probabilities and quantiles, with and without technology.
The exponential distribution is a continuous model for waiting time. It is often used when the question describes the time until an event happens, such as the time until a machine fails or the time between arrivals.
If $T$ is exponential with rate parameter $\lambda>0$, then its probability density function is:
$ f(t)=\lambda e^{-\lambda t},\quad t\ge0. $
For $t<0$, the density is $0$, because a waiting time cannot be negative.
Why it is a valid density
A probability density must be non-negative and must have total area $1$.
Since $\lambda>0$ and $e^{-\lambda t}>0$, the density is non-negative. Also:
$ \int_0^\infty \lambda e^{-\lambda t}\,dt=1. $
That is why the coefficient $\lambda$ is not optional. It makes the total area under the curve equal $1$.
The probability of an interval is area under the density:
$ P(a<T<b)=\int_a^b \lambda e^{-\lambda t}\,dt. $
For the exponential distribution, this simplifies to:
$ P(a<T<b)=e^{-\lambda a}-e^{-\lambda b}. $
The cumulative distribution function is piecewise:
$ F(t)=P(T\le t)= \begin{cases} 0, & t<0,\\ 1-e^{-\lambda t}, & t\ge0. \end{cases} $
Writing the $t<0$ part is not just formality. It shows that the model is a waiting-time model and has no probability before time zero.
Mean and standard deviation
For an exponential random variable,
$ E(T)=\frac1\lambda $
and
$ \operatorname{Var}(T)=\frac1{\lambda^2} $
so
$ \operatorname{SD}(T)=\frac1\lambda. $
So if the average waiting time is $8$ minutes, then:
$ \lambda=\frac18 $
when time is measured in minutes.
Probabilities and quantiles
The cumulative distribution function is:
$ P(T\le t)=1-e^{-\lambda t},\quad t\ge0. $
So:
$ P(T>t)=e^{-\lambda t}. $
To find a quantile, set the cumulative probability equal to the required value and solve for $t$.
For a $p$-quantile $q_p$:
$ 1-e^{-\lambda q_p}=p. $
So:
$ q_p=-\frac{\ln(1-p)}{\lambda}. $
The median is the special case $p=0.5$:
$ q_{0.5}=\frac{\ln2}{\lambda}. $
If a question gives the median instead of the mean, use the CDF. For example, if the median lifetime is $4$ hours, then:
$ P(T\le4)=\frac12. $
So:
$ 1-e^{-4\lambda}=\frac12. $
This gives:
$ e^{-4\lambda}=\frac12,\quad \lambda=\frac{\ln2}{4}. $
The survival function is often faster than the CDF:
$ P(T>t)=e^{-\lambda t}. $
Then an interval probability can be found by subtraction:
$ P(a<T<b)=P(T>a)-P(T>b). $
Modelling assumptions
The exponential distribution is strongest when it is reasonable to model the waiting time as having a constant rate. In practical questions, that means the chance of the event happening in the next small time interval should not depend strongly on how long you have already waited.
This is why the model is common for idealised waiting-time and lifetime contexts, but it is not automatically correct for every lifetime problem. A car part that becomes more likely to fail as it ages may need a different model.
The exponential distribution is memoryless:
$ P(T>s+t\mid T>s)=P(T>t). $
In plain language, once the process has already lasted $s$ units of time, the model treats the remaining waiting time as if the clock has just restarted. That is useful for idealised arrival processes, but it can be unrealistic for ageing equipment or biological lifetimes.
Worked context example
Suppose floods at a site are modelled by an exponential waiting time $T$ with mean $10$ years. Then:
$ \lambda=\frac1{10}=0.1 $
per year.
The probability of waiting more than $12$ years is:
$ P(T>12)=e^{-0.1(12)}=e^{-1.2}. $
The probability of at least one flood within the next $6$ months uses $t=0.5$ years:
$ P(T\le0.5)=1-e^{-0.1(0.5)}=1-e^{-0.05}. $
To find the time by which there is a $90\%$ chance a flood has occurred:
$ 1-e^{-0.1t}=0.9. $
So:
$ e^{-0.1t}=0.1,\quad t=-\frac{\ln(0.1)}{0.1}. $
This is the same quantile formula, but writing the CDF equation first usually makes the context easier to follow.
Always name the time unit in the final sentence. A rate measured per year produces answers in years; a rate measured per minute produces answers in minutes.