Exponential Distribution Calculator

Exponential Distribution Calculator

Use this calculator to find the probability density and cumulative probabilities for Exponential distribution with parameter $\theta$.

How ito use Exponential Probability Density Function Calculator?

Step 1 - Enter the Parameter $\theta$

Step 2 - Enter the Value of A and Value of B

Step 3 - Click on Calculate button to calculate exponential probability

Step 4 - Calculates Probability X less than A: P(X < A)

Step 5 - Calculates Probability X greater than B: P(X > B)

Step 6 - Calculates Probability X is between A and B: P(A < X < B)

Step 7 - Calculates Mean = $1/\theta$

Step 8 - Calculates Variance = $1/\theta^2$

Step 9 - Calculates Standard deviation = $1/\theta$

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

$$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \theta e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Distribution Function of Exponential Distribution

The distribution function of an exponential random variable is

$$ \begin{equation*} F(x)=\left\{ \begin{array}{ll} 1- e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Mean of Exponential Distribution

The mean of an exponential random variable is $E(X) = \dfrac{1}{\theta}$.

Variance of Exponential Distribution

The variance of an exponential random variable is $V(X) = \dfrac{1}{\theta^2}$.

Lets understand how to solve numerical problems based on exponential distribution.

Example - 1 Exponential Distribution Calculator

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 4 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,

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

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

Example - 2 Exponential Probability Distribution Calculator

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

Hope you like above article on Exponential Distribution Calculator helpful.

Read more about other Statistics Calculator on below links

• Descriptive Statistics Calculators
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