# Exponential Distribution Calculator with Examples

## Exponential Distribution Calculator

Exponential Distribution Calculator is used to find the probability density and cumulative probabilities for Exponential distribution with parameter $\theta$.

Exponential Distribution Calculator
Parameter $\theta$:
Value of A
Value of B
Results
Probability X less than A: P(X < A)
Probability X greater than B: P(X > B)
Probability X is between A and B: P(A < X < B)
Mean = $1/\theta$
Variance = $1/\theta^2$
Standard deviation = $1/\theta$

## How to calculate probabilities of Laplace Distribution?

Step 1 - Enter the parameter $\theta$

Step 2 - Enter the value of $A$

Step 3 - Enter the value of $B$

Step 4 - Click on "Calculate" button to get Exponential distribution probabilities

Step 5 - Gives the output of $P(X < A)$ for Exponential distribution

Step 6 - Gives the output of $P(X > B)$ for exponential distribution

Step 7 - Gives the output of $P(A < X < B)$ for Exponential distribution

Step 8 - Gives the output of mean, variance and standard ddeviation for Exponential distribution

## Definition of Exponential distribution

A continuous random variable $X$ is said to have an exponential distribution with parameter $\theta$ if its p.d.f. is given by

 \begin{align*} f(x)&= \begin{cases} \theta e^{-\theta x}, & x>0;\theta>0 \\ 0, & Otherwise. \end{cases} \end{align*}

In notation, it can be written as $X\sim \exp(\theta)$.

## Distribution Function of exponential distribution

The distribution function of exponential distribution is $F(x) = 1-e^{-\theta x}$.

## Mean and Variance of Exponential Distribution

Let $X\sim \exp(\theta)$. Then the mean and variance of $X$ are $\frac{1}{\theta}$ and $\frac{1}{\theta^2}$ respectively.

## Exponential Distribution Example 1

The time (in hours) required to repair a machine is an exponential distributed random variable
with paramter $\lambda =1/2$. What is

a. the probability that a repair time exceeds 4 hours,

b. the probability that a repair time takes at most 3 hours,

c. the probability that a repair time takes between 2 to 4 hours,

d. the conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours?

#### Solution

Let $X$ denote the time (in hours) required to repair a machine. Given that $X$ is exponentially distributed with $\lambda = 1/2$.

The pdf of $X$ is

 \begin{aligned} f(x) &= \lambda e^{-\lambda x},\; x>0\\ &= \frac{1}{2}e^{-x/2},\; x>0 \end{aligned}

The distribution function of $X$ is

 \begin{aligned} F(x) &= P(X\leq x) = 1- e^{-x/2}. \end{aligned}

a. The probability that a repair time exceeds 4 hours is

 \begin{aligned} P(X> 4) &= 1- P(X\leq 4)\\ & = 1- F(4)\\ & = 1- \big[1- e^{-4/2}\big]\\ &= e^{-2}\\ & = 0.1353 \end{aligned}

b. The probability that a repair time takes at most 3 hours is

 \begin{aligned} P(X\leq 3) &= F(3)\\ &=1- e^{-3/2}\\ &= 1-e^{-1.5}\\ & = 0.7769 \end{aligned}

c. The probability that a repair time takes between 2 to 4 hours is

 \begin{aligned} P(2< X< 4) &= F(4)-F(2)\\ &=\big[1- e^{-4/2}\big]-\big[1- e^{-2/2}\big]\\ &= e^{-1}-e^{-2}\\ & = 0.3679-0.1353\\ & = 0.2326 \end{aligned}

d. The conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours is

 \begin{aligned} P(X \geq 10|X>9) &= \frac{P(X\geq 10)}{P(X>9)}\\ & = \frac{1- P(X<10)}{1-P(X<9)}\\ & = \frac{1- F(10)}{1-F(9)}\\ &= \frac{1-(1-e^{-10/2})}{1-(1-e^{-9/2})}\\ & = \frac{e^{-10/2}}{e^{-9/2}}\\ &=0.6065 \end{aligned}

OR

Using memoryless property of exponential distribution,

$$P(X>s+t|X>s)=P(X>t)$$

 \begin{aligned} P(X \geq 10|X>9) &= P(X> 9+1|X> 9)\\ &= P(X> 1)\\ &=1- P(X\leq 1)\\ &= 1- F(1)\\ &= 1-(1-e^{-1/2})\\ &= e^{-1/2}\\ &=0.6065 \end{aligned}

## Exponential Distribution Example 2

The time to failure $X$ of a machine has exponential distribution with probability density function

$f(x) = 0.01e^{-0.01 x}, x>0$.

Find

a. distribution function of $X$,

b. the probability that the machine fails between 100 and 200 hours,

c. the probability that the machine fails before 100 hours,

d. the value of $x$ such that $P(X> x)=0.5$.

#### Solution

Let $X$ denote the time (in hours) to failure of a machine machine. Given that $X$ is exponentially distributed with $\lambda = 0.01$.

The pdf of $X$ is

 \begin{aligned} f(x) &= \lambda e^{-\lambda x},\; x>0\\ &= 0.01e^{-0.01x},\; x>0 \end{aligned}

a. The distribution function of $X$ is

 \begin{aligned} F(x) &= P(X\leq x) = 1- e^{-0.01x}. \end{aligned}

b. The probability that the machine fails between $100$ and $200$ hours is

 \begin{aligned} P(100< X< 200) &= F(200)-F(100)\\ &=\big[1- e^{-200\times0.01}\big]-\big[1- e^{-100\times0.01}\big]\\ &= e^{-1}-e^{-2}\\ & = 0.3679-0.1353\\ & = 0.2326 \end{aligned}

c. The probability that a repair time takes at most $100$ hours is

 \begin{aligned} P(X\leq 100) &= F(100)\\ &=1- e^{-100\times0.01}\\ &= 1-e^{-1}\\ & = 0.6321 \end{aligned}

d. The value of $x$ such that $P(X>x)=0.5$ is

 \begin{aligned} & P(X> x) = 0.5\\ \Rightarrow & P(X\leq x)= 0.5\\ \Rightarrow & F(x)= 0.5\\ \Rightarrow & 1- e^{-0.01x}= 0.5\\ \Rightarrow & e^{-0.01x}= 0.5\\ \Rightarrow & -0.01x= \ln 0.5\\ \Rightarrow & -0.01x= -0.693\\ \Rightarrow & x= 69.3 \end{aligned}

## Conclusion

In this tutorial, you learned about how to calculate probabilities of Exponential distribution. You also learned about how to solve numerical problems based on Exponential distribution.

To read more about the step by step tutorial on Exponential distribution refer the link Exponential Distribution.

This tutorial will help you to understand Exponential distribution and you will learn how to derive mean, variance, moment generating function of Exponential distribution and other properties of Exponential distribution.

To learn more about other probability distributions, please refer to the following tutorial:

Probability distributions

Let me know in the comments if you have any questions on Exponential Distribution Examples and your thought on this article.