Exponential Distribution Calculator

Find exponential distribution probabilities, CDF, PDF, mean, and median for any rate parameter and value x.

Rate parameter λ (events per unit time)2

120

Value x (time or distance)1

010

Rate parameter λ2

120

Lower bound a0.5

010

Upper bound b2

010

P(X ≤ x): Cumulative

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P(X > x): Survival

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PDF f(x) = λe^−λx

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Mean (1/λ)

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Variance (1/λ²)

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Std Dev (1/λ)

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Median (ln 2 / λ)

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P(a ≤ X ≤ b)

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P(X < a): Below Range

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P(X > b): Above Range

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Mean (1/λ)

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Std Dev (1/λ)

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📉 What is the Exponential Distribution?

The exponential distribution is a continuous probability distribution that models the time between events in a Poisson process: a process where events occur independently at a constant average rate lambda. If you know that on average 3 customers arrive per hour at a coffee shop, the waiting time until the next customer follows an exponential distribution with lambda = 3. The distribution is fully specified by a single parameter, the rate lambda (events per unit time), and it is defined for all non-negative real values x.

The exponential distribution appears throughout applied probability, reliability engineering, queueing theory, and survival analysis. Specific real-world examples include: the time between radioactive decays (where lambda is the decay constant), the service time of a call-center agent (when service rates are approximately constant), the lifetime of electronic components that fail at a constant hazard rate (exponential failure model), the time between arrivals at an emergency room, and the distance between potholes on a road if potholes occur at a constant average density.

One of the most important properties of the exponential distribution is that it is the only continuous memoryless distribution. Memorylessness means that the probability of waiting at least t more time units is independent of how long you have already been waiting. This makes the exponential distribution a natural model whenever the process has no "aging" or "wear" component: a machine does not become more likely to fail just because it has been running longer. When aging or wear is present, the Weibull distribution is a more appropriate model.

The relationship between the exponential distribution and the Poisson distribution is fundamental. If the number of events in a time interval follows a Poisson distribution with mean lambda times t, then the waiting time until the first event follows an exponential distribution with rate lambda. This duality lets you switch between counting events (Poisson) and timing events (exponential) depending on which question you are asking about the same underlying process.

📐 Formula

f(x) = λe^−λx | P(X ≤ x) = 1 − e^−λx

λ = rate parameter (average events per unit time; λ greater than 0)

x = time (or distance) value; x greater than or equal to 0

f(x) = probability density function (PDF): height of the density curve at x

P(X ≤ x) = cumulative distribution function (CDF): probability the event occurs by time x

P(X > x) = survival function = e^−λx

P(a ≤ X ≤ b) = e^−λa − e^−λb

Mean: μ = 1 ÷ λ

Variance: σ² = 1 ÷ λ²

Median: ln(2) ÷ λ ≈ 0.6931 ÷ λ

Example: λ = 2, x = 1: P(X ≤ 1) = 1 − e⁻² ≈ 1 − 0.1353 = 86.47%

📖 How to Use This Calculator

Steps

Choose a mode. Select "Calculate Probability" to find P(X at most x) and P(X greater than x) for a specific value, or "Between Values" to find P(a at most X at most b) over a range.

Enter lambda. Set the rate parameter to the average number of events per unit time. For example, if customers arrive at a rate of 5 per hour, enter 5.

Enter x (or a and b). In Probability mode enter the specific time value x. In Between Values mode enter the lower bound a and upper bound b (b must be greater than a).

Read the results. The calculator shows CDF, survival probability, PDF value, mean, variance, standard deviation, and median in one view.

💡 Example Calculations

Example 1: Customer Arrivals at a Coffee Shop (lambda = 3, x = 0.5 hours)

Customers arrive at an average rate of 3 per hour. What is the probability the next customer arrives within 30 minutes (0.5 hours)?

Set lambda = 3 (customers per hour) and x = 0.5 (hours). Use the CDF formula.

P(X at most 0.5) = 1 - e^(-3 times 0.5) = 1 - e^(-1.5) = 1 - 0.2231 = 0.7769.

The mean is 1/3 hours = 20 minutes. The probability the next customer arrives within 30 minutes is about 77.69%.

P(X at most 0.5) = 77.69% | Mean = 0.333 hours (20 min)

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Example 2: Equipment Failure Rate (lambda = 0.2, x = 10 years)

A machine fails at an average rate of once every 5 years (lambda = 0.2 per year). What is the probability it lasts more than 10 years without failure?

Set lambda = 0.2 (failures per year) and x = 10 (years). We want P(X greater than 10) = e^(-lambda times x).

P(X greater than 10) = e^(-0.2 times 10) = e^(-2) = 0.1353, about 13.53%.

The mean is 1/0.2 = 5 years. There is only a 13.53% chance the machine survives 10 years, which is twice the mean lifetime.

P(X greater than 10) = 13.53% | Mean = 5 years

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Example 3: Bus Arrival Between 5 and 15 Minutes (lambda = 0.1 per minute)

A bus arrives every 10 minutes on average (lambda = 0.1 per minute). What is the probability the next bus arrives between 5 and 15 minutes from now?

Set lambda = 0.1, a = 5, b = 15. Use P(a at most X at most b) = e^(-lambda times a) - e^(-lambda times b).

e^(-0.1 times 5) - e^(-0.1 times 15) = e^(-0.5) - e^(-1.5) = 0.6065 - 0.2231 = 0.3834.

There is a 38.34% chance the bus arrives between 5 and 15 minutes. The mean wait is 1/0.1 = 10 minutes.

P(5 at most X at most 15) = 38.34% | Mean = 10 minutes

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❓ Frequently Asked Questions

What is the exponential distribution and when is it used?+

The exponential distribution models the waiting time until the first event in a Poisson process, where events occur independently at a constant rate lambda. It is used in reliability engineering (time to failure), queueing theory (service times), telecommunications (call durations), and physics (radioactive decay). The key requirement is that the event rate is constant and events are independent.

What is the exponential distribution CDF formula?+

The CDF is P(X at most x) = 1 - e^(-lambda times x) for x greater than or equal to 0. This gives the probability that the first event occurs by time x. For example, with lambda = 2 and x = 1, P(X at most 1) = 1 - e^(-2) = 1 - 0.1353 = 0.8647, or about 86.47%.

What is the mean and standard deviation of the exponential distribution?+

Both the mean and the standard deviation equal 1/lambda. This is a unique property: the coefficient of variation (std dev divided by mean) is always 1, regardless of lambda. For lambda = 0.5, mean = 2 and std dev = 2. For lambda = 4, mean = 0.25 and std dev = 0.25.

Why is the median of the exponential distribution less than the mean?+

The median is ln(2)/lambda, about 0.693/lambda, while the mean is 1/lambda. The ratio median/mean = ln(2) = 0.693, so the median is always about 69.3% of the mean. This gap exists because the exponential distribution is right-skewed: a long tail of rare large values pulls the mean above the median.

What does memoryless mean for the exponential distribution?+

Memoryless (or Markov) means that, given you have already waited s time units without an event, the additional waiting time still follows the same Exp(lambda) distribution. Formally, P(X greater than s + t | X greater than s) = P(X greater than t) = e^(-lambda times t). This property is unique to the exponential distribution among all continuous distributions.

What is the connection between the exponential and Poisson distributions?+

They describe the same Poisson process from two perspectives. If events arrive at rate lambda per unit time, then the number of events in any interval of length t is Poisson(lambda times t), and the waiting time between consecutive events is Exp(lambda). Switching between the two distributions lets you answer "how many events?" (Poisson) versus "how long until the next event?" (exponential).

How do I convert between rate lambda and mean waiting time?+

Mean = 1/lambda and lambda = 1/mean. If the average waiting time is 4 minutes, then lambda = 0.25 per minute. If the rate is 5 events per hour, then the mean waiting time is 1/5 = 0.2 hours = 12 minutes. Always use consistent units: if lambda is per minute, x must also be in minutes.

What is the survival function of the exponential distribution?+

The survival function (also called the reliability function) is P(X greater than x) = e^(-lambda times x). It gives the probability that the event (such as a failure) has not yet occurred by time x. For lambda = 1 and x = 2, P(X greater than 2) = e^(-2) = 0.1353, about 13.53%.

How do I calculate P(a less than X less than b) for the exponential distribution?+

P(a at most X at most b) = CDF(b) - CDF(a) = e^(-lambda times a) - e^(-lambda times b). Alternatively, this equals the integral of f(x) from a to b. Use the Between Values mode of this calculator to compute it directly by entering lambda, the lower bound a, and the upper bound b.

When does the exponential distribution fail as a model?+

The exponential model is inappropriate when the event rate is not constant. Mechanical components that wear out over time have an increasing failure rate, which is better modelled by a Weibull distribution. Human lifetimes have a bathtub-shaped hazard rate. Light bulbs (close to exponential in the middle of life) and biological aging both violate the constant-rate assumption at the extremes of their lifetime.

What is the mode of the exponential distribution?+

The mode is 0 for all values of lambda greater than 0. The PDF f(x) = lambda times e^(-lambda times x) is a strictly decreasing function that starts at f(0) = lambda and falls toward 0 as x increases. The most probable value is always at the very start of the distribution, meaning very short waiting times are always the most likely.

🔗 Related Calculators

What is the exponential distribution formula?

The PDF is f(x) = lambda times e^(-lambda times x) for x greater than or equal to 0, where lambda is the rate parameter (events per unit time). The CDF is P(X less than or equal to x) = 1 - e^(-lambda times x). The survival function is P(X greater than x) = e^(-lambda times x).

What is the mean of the exponential distribution?

The mean (expected value) is mu = 1/lambda. If events arrive at a rate of 2 per hour, the average waiting time is 1/2 = 0.5 hours. The variance is 1/lambda^2 and the standard deviation is also 1/lambda, so mean equals standard deviation.

What is the median of the exponential distribution?

The median is ln(2)/lambda, approximately 0.693/lambda. For lambda = 1, the median is about 0.693. The median is always less than the mean (0.693 times mean), reflecting the right-skewed shape of the distribution.

What does the rate parameter lambda mean?

Lambda is the rate of events per unit time (or per unit length, distance, etc.). For example, lambda = 3 means on average 3 events occur per hour. The mean waiting time between events is 1/lambda = 1/3 of an hour = 20 minutes.

What is the memoryless property of the exponential distribution?

Memoryless means the distribution forgets its history. If you have already waited s time units with no event, the probability of waiting at least t more time units is the same as if you had just started. Formally, P(X greater than s+t | X greater than s) = P(X greater than t) = e^(-lambda times t).

How is the exponential distribution related to the Poisson distribution?

If events occur in a Poisson process with rate lambda (events per unit time), the time between consecutive events follows an exponential distribution with the same rate lambda. The number of events in a fixed interval follows a Poisson distribution. These two models are two sides of the same process.

What is the difference between the exponential and geometric distribution?

Both are memoryless. The geometric distribution is discrete and counts trials until the first success. The exponential distribution is continuous and measures time until the first event. The geometric is the discrete analogue; the exponential is the continuous analogue of the same waiting-time concept.

When is the exponential distribution appropriate to use?

Use the exponential distribution when: events occur independently at a constant average rate, you are modelling the time (or distance) until the next event, and the memoryless property is a reasonable assumption. Common applications include radioactive decay, customer service times, equipment failure rates, and phone call durations.

📌 Quick Tips

💡The exponential distribution is memoryless: P(X greater than s + t | X greater than s) = P(X greater than t). Past waiting time gives no information about future waiting time.

💡The mean and standard deviation of the exponential distribution are both equal to 1/lambda. A high standard deviation relative to the mean is a signature property.

💡The median is ln(2)/lambda, which is always less than the mean 1/lambda. The ratio median/mean equals ln(2) = 0.693 regardless of lambda.

💡If events follow a Poisson process with rate lambda, the time between consecutive events follows Exp(lambda). These two distributions are inseparably linked.

💡Use the Between Values mode to find the probability of waiting between a and b time units. For example, P(1 less than X less than 3) with lambda = 0.5 gives the chance of a wait of 1 to 3 time units.