The exponential probability distribution
Master the exponential probability distribution for Year 12 Specialist Mathematics in Queensland (QCAA). This is the continuous model for the time until a random event, with probability density function \(\lambda e^{-\lambda x}\) and probabilities read as areas under a decaying curve.
You will learn to find the mean and standard deviation \(\dfrac{1}{\lambda}\), compute tail, interval and "less than" probabilities, and solve for medians and quantiles — an application of integral calculus that models real lifetimes and waiting times in Unit 4.
Theory
The exponential probability distribution models the waiting time until a random event — a component failing, a call arriving — in Year 12 Specialist Mathematics (QCAA, Queensland). Its probability density function is \(f(x)=\lambda e^{-\lambda x}\) for \(x\ge 0\), the mean and standard deviation both equal \(\dfrac{1}{\lambda}\), and tail probabilities come straight from \(P(X>a)=e^{-\lambda a}\). This page shows how to find probabilities, means and quantiles.
A continuous random variable \(X\) is exponentially distributed with parameter \(\lambda>0\) if its probability density function is \(f(x)=\lambda e^{-\lambda x}\) for \(x\ge 0\) (and \(0\) for \(x<0\)). It is the standard model for the time until a random event occurs, such as a lifetime or a waiting time.
Like every pdf, the total area under it is \(1\): \(\displaystyle\int_0^{\infty}\lambda e^{-\lambda x}\,dx=1\). Probabilities are areas under the curve, found by integrating between the limits. Because the antiderivative of \(\lambda e^{-\lambda x}\) is \(-e^{-\lambda x}\), the two most useful results are \(P(X>a)=e^{-\lambda a}\) and \(P(X The cumulative distribution function is \(F(x)=P(X\le x)=1-e^{-\lambda x}\). A probability between two values is a difference of tails, \(P(a\le X\le b)=e^{-\lambda a}-e^{-\lambda b}\). The mean and the standard deviation are both \(\dfrac{1}{\lambda}\); a large \(\lambda\) means events happen quickly and the curve decays fast. A quantile \(x_p\) (the value with probability \(p\) below it) is found by solving \(1-e^{-\lambda x}=p\); the median is the case \(p=\dfrac{1}{2}\).
For an exponential random variable \(X\) with parameter \(\lambda>0\), the probability density function and the tail / head probabilities are:
A probability between two values is a difference of tails:
The mean and standard deviation are equal, and a quantile comes from the CDF:
Solving an exponential distribution problem
- Find \(\lambda\): read it from \(f(x)=\lambda e^{-\lambda x}\) directly, or, if you are given an average, use \(\lambda=\dfrac{1}{\text{mean}}\).
- Choose the probability form: "more than" uses \(P(X>a)=e^{-\lambda a}\); "less than" uses \(P(X
- Substitute and evaluate: put in the numbers, leaving an exact \(e^{-k}\) if an exact answer is asked, or rounding to the required decimal places.
- For a quantile or median: set the CDF equal to \(p\), i.e. \(1-e^{-\lambda x}=p\), then take logs to solve for \(x\); the median is the case \(p=\dfrac{1}{2}\).
Read off \(\lambda\) from the pdf, then use \(E(X)=\operatorname{sd}(X)=\dfrac{1}{\lambda}\):
| \(\lambda\) | \(=\) | \(0.4\) |
| \(E(X)\) | \(=\) | \(\dfrac{1}{\lambda}\) |
| \(=\) | \(\dfrac{1}{0.4}\) | |
| \(=\) | \(2.5\) |
The standard deviation of an exponential distribution equals the mean:
| \(\operatorname{sd}(X)\) | \(=\) | \(\dfrac{1}{\lambda}\) |
| \(=\) | \(2.5\) |
The mean is \(2.5\) hours and the standard deviation is \(2.5\) hours.
Use the tail result \(P(X>a)=\displaystyle\int_a^{\infty}\lambda e^{-\lambda x}\,dx=e^{-\lambda a}\):
| \(P(X>3)\) | \(=\) | \(\int_3^{\infty} 0.5\,e^{-0.5x}\,dx\) |
| \(=\) | \(\Big[-e^{-0.5x}\Big]_3^{\infty}\) | |
| \(=\) | \(0-\left(-e^{-1.5}\right)\) | |
| \(=\) | \(e^{-1.5}\) |
Evaluate to four decimal places:
| \(e^{-1.5}\) | \(=\) | \(0.2231\) |
\(P(X>3)=e^{-1.5}\approx 0.2231\).
First convert the mean to \(\lambda=\dfrac{1}{\text{mean}}\):
| \(\lambda\) | \(=\) | \(\dfrac{1}{5}\) |
| \(=\) | \(0.2\) |
Use \(P(a\le X\le b)=e^{-\lambda a}-e^{-\lambda b}\):
| \(P(2\le X\le 6)\) | \(=\) | \(e^{-0.2\times 2}-e^{-0.2\times 6}\) |
| \(=\) | \(e^{-0.4}-e^{-1.2}\) | |
| \(=\) | \(0.6703-0.3012\) | |
| \(=\) | \(0.3691\) |
\(P(2\le X\le 6)\approx 0.3691\).
Convert the mean to \(\lambda\), then solve \(1-e^{-\lambda m}=\dfrac{1}{2}\) for the median \(m\):
| \(\lambda\) | \(=\) | \(\dfrac{1}{8}\) |
| \(1-e^{-m/8}\) | \(=\) | \(\dfrac{1}{2}\) |
| \(e^{-m/8}\) | \(=\) | \(\dfrac{1}{2}\) |
Take natural logs of both sides and solve:
| \(-\dfrac{m}{8}\) | \(=\) | \(\ln\dfrac{1}{2}=-\ln 2\) |
| \(m\) | \(=\) | \(8\ln 2\) |
| \(=\) | \(5.55\) |
The median lifetime is \(8\ln 2\approx 5.55\) hours — below the mean of \(8\) hours, as always for an exponential distribution.
Common pitfalls
Frequently asked questions
What is the exponential probability density function?
It is \(f(x)=\lambda e^{-\lambda x}\) for \(x\ge 0\), where \(\lambda>0\) is the parameter. It models the waiting time until a random event, such as a component failing.
What are the mean and standard deviation of an exponential distribution?
Both equal \(\dfrac{1}{\lambda}\). So if \(\lambda=0.25\) the mean and standard deviation are both \(4\).
How do you find \(P(X>a)\) for an exponential distribution?
Integrate the pdf from \(a\) to infinity, which gives \(P(X>a)=e^{-\lambda a}\). For \(P(X
How do you find the median of an exponential distribution?
Set the CDF equal to \(\dfrac{1}{2}\): solve \(1-e^{-\lambda m}=\dfrac{1}{2}\), which gives the median \(m=\dfrac{1}{\lambda}\ln 2\).
How is \(\lambda\) related to the mean?
They are reciprocals: \(\lambda=\dfrac{1}{\text{mean}}\). If a problem says events happen on average every \(5\) minutes, then \(\lambda=\dfrac{1}{5}=0.2\).
What is the cumulative distribution function of an exponential distribution?
It is \(F(x)=P(X\le x)=1-e^{-\lambda x}\) for \(x\ge 0\). It gives the probability of a value at or below \(x\) and is used to find quantiles.