Exponential Distribution Formula: PDF, CDF, Mean and Examples

The exponential distribution describes the waiting time until an event when the event happens at a constant average rate. Its density is f(x) = λe^−λx for x ≥ 0, its mean is 1/λ, and its defining feature is a constant hazard rate. The part most guides skip is the parameterization: Excel and R take the rate λ, while SciPy and the NIST Handbook take the scale 1/λ. Mixing them up is the commonest way to get a wrong answer.

What is the exponential distribution?

The exponential distribution is a continuous probability distribution for non-negative values, most often a time until something happens. The NIST/SEMATECH e-Handbook describes it as used to model data with a constant failure rate and as primarily applied in reliability work. It is also the continuous counterpart of the waiting-time idea behind the Poisson distribution: if events arrive at a constant rate, the gaps between them are exponentially distributed.

The key property is that the hazard, the instantaneous chance of the event given it has not yet happened, never changes. NIST states that the exponential is the only distribution with a constant failure rate. A unit that is “as good as new” at every moment is the mental picture.

Exponential distribution formula: every symbol explained

Rate form (λ)

The NIST reliability chapter writes the model with the rate λ:

• Density (PDF): f(t) = λe^−λt
• Cumulative distribution (CDF): F(t) = 1 − e^−λt
• Survival (reliability) function: R(t) = e^−λt
• Hazard rate: h(t) = λ
• Mean (the mean time to failure): 1/λ

Here t is elapsed time (or any non-negative quantity), λ > 0 is the rate in events per unit of time, e is Euler’s number (about 2.71828), F(t) is the probability the event has happened by time t, and R(t) = 1 − F(t) is the probability it has not.

Scale form (β)

The NIST Engineering Statistics Handbook uses a scale parameter β and an optional location μ: f(x) = (1/β)e^{−(x−μ)/β} for x ≥ μ, with β > 0. It notes that the scale parameter is often called λ, which equals 1/β. The standard exponential has μ = 0 and β = 1. With μ = 0, the CDF is 1 − e^−x/β, the survival function is e^−x/β, and the hazard is the constant 1/β.

Summary measures

Measure	Value (location μ = 0)	Source
Mean	1/λ (β)	NIST
Median	(ln 2)/λ ≈ 0.693 × mean	NIST (β ln 2)
Mode	0	NIST (μ)
Standard deviation	1/λ	NIST
Variance	1/λ2 (the square of the standard deviation)	derived from NIST
Coefficient of variation	1	NIST
Skewness / kurtosis	2 / 9	NIST

Worked example with the arithmetic shown

Take the parameters NIST uses in its reliability chapter: λ = 0.01 per hour, evaluated at t = 100 hours. Think of a component with a constant failure rate of 0.01 per hour (an illustrative setting, not data).

1. Mean time to failure = 1/λ = 1/0.01 = 100 hours.
2. λt = 0.01 × 100 = 1, so e⁻¹ = 0.3679.
3. Density at 100 hours: f(100) = 0.01 × 0.3679 = 0.003679, which rounds to NIST’s 0.0037.
4. CDF: F(100) = 1 − 0.3679 = 0.6321. About 63.2% of units fail by the mean time to failure.
5. Survival: R(100) = 0.3679, and R(150) = e^−1.5 = 0.2231.
6. Median: (ln 2)/0.01 = 0.6931/0.01 = 69.31 hours. Half the units fail before 69.3 hours, well below the 100-hour mean.

Exponential distribution examples

Two settings cover most uses. In reliability, t is time to failure and λ is the failure rate, as in the example above. In waiting-time problems, t is the gap between events in a Poisson process. If calls reach a line at an average of 3 per hour (λ = 3 per hour), the chance the next call arrives within 10 minutes (1/6 hour) is 1 − e^−3/6 = 1 − e^−0.5 = 1 − 0.6065 = 0.3935, or 39.35%. Note that 10 minutes must be converted to hours so it matches the unit of λ. See the Poisson distribution guide for the count version of the same process.

The memoryless property

Because the survival function is S(x) = e^−λx, the conditional survival after time s follows by division (a derivation from the NIST survival function):

P(X > s + t | X > s) = S(s + t) / S(s) = e^−λ(s+t) / e^−λs = e^−λt = P(X > t)

In the worked example, a unit that has already survived 100 hours has the same chance of surviving another 50 hours as a new unit: e^−0.5 = 0.6065. This equals R(150)/R(100) = 0.2231/0.3679 = 0.6065. Memorylessness is the same fact as a constant hazard, seen from the survival side.

Interpretation: reading the numbers

Quantity	How to read it
Rate λ	Events per unit time. Larger λ means shorter waits; the hazard equals λ at every time.
Mean 1/λ	The average wait, in the same time unit as t.
Median ≈ 0.693 × mean	The median sits below the mean, so a typical wait is shorter than the average wait.
Coefficient of variation = 1	The standard deviation equals the mean. Data with a much smaller or larger spread are a poor fit.
Skewness = 2	The distribution is right-skewed with a long upper tail.
F(1/λ) = 0.6321	About 63.2% of events occur by the time equal to the mean.

Assumptions and common mistakes

• Constant hazard. The model assumes the event rate does not change with age or elapsed time. If the hazard rises (wear-out) or falls (early failures), the exponential does not fit; a Weibull or another parametric survival model is the usual alternative. See survival analysis with Kaplan-Meier, Cox and parametric models and accelerated failure time models.
• Non-negative values. The variable is a time or amount that cannot be negative (x ≥ 0 when the location is 0).
• Mean is not the median. Quoting the mean as “typical” overstates the typical wait by about 44% (1/0.693).
• Rate versus scale. The same distribution is entered differently in each tool (next table).
• Units. λ and t must use the same time unit.

Tool	Parameter you supply	For mean 100 (λ = 0.01)
Excel EXPON.DIST	lambda (rate)	lambda = 0.01
R dexp / pexp	rate (mean = 1/rate)	rate = 0.01
SciPy expon	scale = 1/lambda	scale = 100
NIST Engineering Statistics Handbook	β (scale)	β = 100

How to report it (APA example)

State the parameterization, the unit and the estimate, so a reader cannot confuse rate with scale. An APA-style results sentence might read:

Time to failure was modeled with an exponential distribution with rate λ = 0.01 failures per hour (mean time to failure = 100 hours). The probability of failure by 100 hours was F(100) = .63.

Also say whether λ was fixed from a standard or estimated from data, and report an interval if it was estimated.

How to calculate it in Excel, R or Python

All three snippets compute the density and CDF at t = 100 for λ = 0.01.

Excel

=EXPON.DIST(100, 0.01, FALSE)   → 0.003679 (PDF)
=EXPON.DIST(100, 0.01, TRUE)    → 0.632121 (CDF)

Microsoft documents EXPON.DIST(x, lambda, cumulative): TRUE returns the CDF and FALSE the PDF, and x < 0 or lambda ≤ 0 returns #NUM!. Its own example, x = 0.2 and lambda = 10, gives a CDF of 0.86466472 and a PDF of 1.35335283.

R

dexp(100, rate = 0.01)   # 0.003678794
pexp(100, rate = 0.01)   # 0.6321206
qexp(0.5, rate = 0.01)   # 69.31472 (median)
rexp(5, rate = 0.01)     # five random draws

Python (SciPy)

from scipy.stats import expon
lam = 0.01
d = expon(scale=1/lam)
d.pdf(100)   # 0.003678794
d.cdf(100)   # 0.6321206
d.sf(150)    # 0.2231302
d.ppf(0.5)   # 69.31472

SciPy’s standardized density is exp(−x) and the documentation states that this corresponds to scale = 1/lambda, which is why the code passes scale=1/lam.

Frequently asked questions

What is the exponential distribution?

It is a continuous distribution for non-negative waiting times with a constant hazard rate λ and mean 1/λ. Its CDF is 1 − e^−λx.

What is the exponential distribution formula?

The density is f(x) = λe^−λx for x ≥ 0, or equivalently (1/β)e^−x/β with β = 1/λ.

What are examples of the exponential distribution?

Time to failure of a component with a constant failure rate, and the time between events in a Poisson process, such as the gap between successive arrivals at a constant average rate.

Is the mean or the median larger?

The mean. The median is (ln 2)/λ, about 69.3% of the mean.

When should I not use it?

When the hazard changes over time. Compare candidate survival models before assuming a constant rate; the survival analysis guide and the hazard ratio guide cover the next steps, and the normal distribution guide covers a symmetric alternative for data that are not waiting times.

Sources

• NIST/SEMATECH e-Handbook of Statistical Methods, “Exponential Distribution” (Engineering Statistics Handbook, section 1.3.6.6.7): https://www.itl.nist.gov/div898/handbook/eda/section3/eda3667.htm. Checked 2026-10-29.
• NIST/SEMATECH e-Handbook of Statistical Methods, “Exponential” reliability model: https://www.itl.nist.gov/div898/handbook/apr/section1/apr161.htm. Checked 2026-10-29.
• R Documentation, “The Exponential Distribution” (stats package): https://stat.ethz.ch/R-manual/R-devel/library/stats/html/Exponential.html. Checked 2026-10-29.
• SciPy documentation, scipy.stats.expon: https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.expon.html. Checked 2026-10-29.
• Microsoft Support, “EXPON.DIST function”: https://support.microsoft.com/en-us/office/expon-dist-function-4c12ae24-e563-4155-bf3e-8b78b6ae140e. Checked 2026-10-29.