Uniform Distribution Calculator computes probability density and cumulative probabilities associated with continuous uniform distribution. This distribution is characterized by a constant probability density function over a specified interval.
The continuous uniform distribution is the simplest probability distribution where all the values belonging to its support have the same probability density. It is also known as rectangular distribution.
Use this Uniform distribution calculator to calculate probability density,probability X less than x and probability X greater than x using minimum value of alpha, maximum value of beta,value of x.
Uniform Distribution Calculator with Steps
Continuous Uniform Distribution Calculator  

Minimum Value $a$:  
Maximum Value $b$  
Value of x  
Uniform Probability Distribution Results  
Probability density : f(x)  
Probability X less than x: P(X < x)  
Probability X greater than x: P(X > x)  
Stepbystep procedure to use continuous uniform distribution calculator:

Enter the value of a (alpha) and b (beta) in the input field

Enter random number x to evaluate probability which lies between limits of distribution

Click on “Calculate” button to calculate uniform probability distribution

Calculate Probability Density,Probability X less than x and Probability X greater than x
Uniform Distribution Formula
A continuous random variable $X$ is said to have uniform distribution with parameter $\alpha$ and $\beta$ if its probability density function is given by
$$f(x; \alpha,\beta) = \dfrac{1}{\beta\alpha}; \alpha< x< \beta $$
where:
 f(x;\alpha,\beta) is the probability density function
 \alpha is the lower bound of the distribution
 \beta is the upper bound of the distribution
 (\beta  \alpha) represents the range of the distribution
Distribution function of continuous uniform distribution is
$$F(x) = \dfrac{x\alpha}{\beta\alpha}; \alpha< x< \beta $$
The mean of uniform distribution is $E(X) = \dfrac{\alpha+\beta}{2}$.
The variance of uniform distribution is $V(X) = \dfrac{(\beta  \alpha)^2}{2}$.
The variables in uniform distribution are called as uniform random variable.
Continuous Uniform Distribution Examples
Below are the solved examples using Continuous Uniform Distribution probability Calculator to calculate probability density,mean of uniform distribution,variance of uniform distribution.
Example 1 : Find Standard Deviation of a Uniform Distribution
The waiting time at a bus stop is uniformly distributed between 1 and 10 minute.
a. What is the probability density function?
b. What is the probability that the rider waits 8 minutes or less?
c. What is the expected waiting time?
d. What is standard deviation of waiting time?
Solution
Let $X$ denote the waiting time at a bust stop. The waiting time at a bus stop is uniformly distributed between 1 and 10 minute. That is $X\sim U(1,10)$.
using Continuous Uniform Distribution formula calculate probability density, mean of uniform distribution and variance of distribution.
a. The uniform probability density function of $X$ is
$$ \begin{aligned} f(x) & = \frac{1}{101},\; 1\leq x \leq 10\\ & = \frac{1}{9},\; 1\leq x \leq 10. \end{aligned} $$
The uniform probability density function calculated as : 0.1111
b. The probability that the rider waits 8 minutes or less is
$$ \begin{aligned} P(X\leq 8) & = \int_1^8 f(x) \; dx\\ & = \frac{1}{9}\int_1^8 \; dx\\ & = \frac{1}{9} \big[x \big]_1^8\\ &= \frac{1}{9}\big[ 81\big]\\ &= \frac{7}{9}\\ &= 0.7778. \end{aligned} $$
lower cumulative distribution : 0.7778
c. The expected (mean of uniform distribution) wait time is $E(X) =\dfrac{\alpha+\beta}{2} =\dfrac{1+10}{2} = 5.5$
d. The variance of waiting time is $V(X) =\dfrac{(\beta\alpha)^2}{10} =\dfrac{(101)^2}{10} = 8.1$.
Example 2: Find Mean and Standard Deviation of Uniform Distribution
Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 2,500 pounds to 4,500 pounds.
a. What is the mean and standard deviation of weight of a randomly chosen vehicle?
b. What is the probability that a vehicle will weigh less than 3,000 pounds?
c. More than 3,900 pounds?
d. Between 3,000 and 3,800 pounds?
Solution
Let the random variable $X$ denote the weight of randomly chosen American passenger car. It is given that $X\sim U(2500, 4500)$. That is $\alpha=2500$ and $\beta=4500$
using Continuous Uniform Distribution formula calculate probability density, mean of uniform distribution and variance of distribution.
The probability density function of $X$ is
$$ \begin{aligned} f(x)&=\frac{1}{4500 2500},\quad2500 \leq x\leq 4500\\ &=\frac{1}{2000},\quad 2500 \leq x\leq 4500 \end{aligned} $$
and the distribution function of $X$ is
$$ \begin{aligned} F(x)&=\frac{x2500}{4500 2500},\quad 2500 \leq x\leq 4500\\ &=\frac{x2500}{2000},\quad 2500 \leq x\leq 4500. \end{aligned} $$
a. The mean weight of a randomly chosen vehicle is
$$ \begin{aligned} E(X) &=\dfrac{\alpha+\beta}{2}\\ &=\dfrac{2500+4500}{2} =3500 \end{aligned} $$
The standard deviation of weight of randomly chosen vehicle is
$$ \begin{aligned} sd(X) &= \sqrt{V(X)}\\ &=\sqrt{\dfrac{(\beta\alpha)^2}{12}}\\ &=\sqrt{\dfrac{(45002500)^2}{12}}\\ &=577.35 \end{aligned} $$
b. The probability that a vehicle will weigh less than $3000$ pounds is
$$ \begin{aligned} P(X< 3000) &=F(3000)\\ &=\dfrac{3000  2500}{2000}\\ &=\dfrac{500}{2000}\\ &=0.25 \end{aligned} $$
c. The probability that a vehicle will weigh more than $3900$ pounds is
$$ \begin{aligned} P(X>3900) &=1P(X\leq 3900)\\ &=1F(3900)\\ &=1\dfrac{3900  2500}{2000}\\ &=1\dfrac{1400}{2000}\\ &=10.7\\ &=0.3\\ \end{aligned} $$
d. The probability that a vehicle will weight between $3000$ and $3800$ pounds is
$$ \begin{aligned} P(3000< X<3800) &= F(3800)  F(3000)\\ &=\frac{38002500}{2000} \frac{30002500}{2000}\\ &= \frac{1300}{2000}\frac{500}{2000}\\ &= 0.650.25\\ &= 0.4. \end{aligned} $$
Example 3: Find the Uniform Distribution Variance and Mean
Assume that voltages in a circuit follows a continuous uniform distribution between 6 volts and 12 volts.
a. What is the mean and variance of voltage in a circuit? b. What is the distribution function of voltage in a circuit? c. If a voltage is randomly selected, find the probability that the given voltage is less than 11 volts. d. If a voltage is randomly selected, find the probability that the given voltage is more than 9 volts. e. If a voltage is randomly selected, find the probability that the given voltage is between 9 volts and 11 volt.
Solution
Let the random variable $X$ denote the voltage in a circuit. It is given that $X\sim U(6, 12)$. That is $\alpha=6$ and $\beta=12$
The probability density function of $X$ is
$$ \begin{aligned} f(x)&=\frac{1}{12 6},\quad6 \leq x\leq 12\\ &=\frac{1}{6},\quad 6 \leq x\leq 12 \end{aligned} $$
a. The mean (of uniform distribution) voltage in a circuit is
$$ \begin{aligned} E(X) &=\dfrac{\alpha+\beta}{2}\\ &=\dfrac{6+12}{2}\\ &=9 \end{aligned} $$
The standard deviation of uniform distribution of voltage in a circuit is
$$ \begin{aligned} sd(X) &= \sqrt{V(X)}\\ &=\sqrt{\dfrac{(\beta\alpha)^2}{12}}\\ &=\sqrt{\dfrac{(126)^2}{12}}\\ &=1.73 \end{aligned} $$
b. The distribution function of $X$ is
$$ \begin{aligned} F(x)&=\frac{x6}{12 6},\quad 6 \leq x\leq 12\\ &=\frac{x6}{6},\quad 6 \leq x\leq 12. \end{aligned} $$
b. The probability that given voltage is less than $11$ volts is
$$ \begin{aligned} P(X < 11) &=F(11)\\ &=\dfrac{11  6}{6}\\ &=\dfrac{5}{6}\\ &=0.8333 \end{aligned} $$
c. The probability that given voltage is more than $9$ volts is
$$ \begin{aligned} P(X > 9) &=1P(X\leq 9)\\ &=1F(9)\\ &=1\dfrac{9  6}{6}\\ &=1\dfrac{3}{6}\\ &=10.5\\ &=0.5\\ \end{aligned} $$
d. The probability that voltage is between $9$ and $11$ volts is
$$ \begin{aligned} P(9 < X < 11) &= F(11)  F(9)\\ &=\frac{116}{6} \frac{96}{6}\\ &= \frac{5}{6}\frac{3}{6}\\ &= 0.83330.5\\ &= 0.3333. \end{aligned} $$
Example 4: Find Probability Density Function of Uniform Distribution
The daily amount of coffee, in liters, dispensed by a machine located in an airport lobby is a random variable $X$ having a continuous uniform distribution with $A = 7$ and $B = 10$. Find the probability that on a given day the amount, of coffee dispensed by this machine will be
a. at most 8.8 liters; b. more than 7.4 liters but less than 9.5 liters; c. at least 8.5 liters.
Solution
Let the random variable $X$ represent the daily amount of coffee dispensed by a machine. It is given that $X\sim U(7, 10)$. That is $\alpha=7$ and $\beta=10$
The probability density function of $X$ is
$$ \begin{aligned} f(x)&=\frac{1}{10 7},\quad7 \leq x\leq 10\\ &=\frac{1}{3},\quad 7 \leq x\leq 10 \end{aligned} $$
The distribution function of $X$ is
$$ \begin{aligned} F(x)&=\frac{x7}{10 7},\quad 7 \leq x\leq 10\\ &=\frac{x7}{3},\quad 7 \leq x\leq 10. \end{aligned} $$
a. The probability that on a given day the amount of coffee dispensed by the machine will be at most $8.8$ liters is
$$ \begin{aligned} P(X < 8.8) &=F(8.8)\\ &=\dfrac{8.8  7}{3}\\ &=\dfrac{1.8}{3}\\ &=0.6 \end{aligned} $$
b. Let us find the probability that on a given day the amount of coffee dispensed by the machine will be more than $7.4$ liters but less than $9.5$ liters.
$$ \begin{aligned} P(7.4 < X < 9.5) &= F(9.5)  F(7.4)\\ &=\frac{9.57}{3} \frac{7.47}{3}\\ &= \frac{2.5}{3}\frac{0.4}{3}\\ &= 0.83330.1333\\ &= 0.7. \end{aligned} $$
c. Let us determine the probability that on a given day the amount of coffee dispensed by the machine will be at least $8.5$ liters.
$$ \begin{aligned} P(X > 8.5) &=1P(X\leq 8.5)\\ &=1F(8.5)\\ &=1\dfrac{8.5  7}{3}\\ &=1\dfrac{1.5}{3}\\ &=10.5\\ &=0.5\\ \end{aligned} $$
Example 5: Find the Function of Uniform Distribution Random Variable
A bus arrives every 10 minutes at a bus stop. It is assumed that the waiting time for a particular individual is a random variable with a continuous uniform distribution.
a. What is the probability that the individual waits more than 7 minutes? b. What is the probability that the individual waits between 2 and 7 minutes?
Solution
Let the random variable $X$ represent the waiting time for a particular individual. It is given that $X\sim U(0, 10)$. That is $\alpha=0$ and $\beta=10$
The probability density function of $X$ is
$$ \begin{aligned} f(x)&=\frac{1}{10 0},\quad0 \leq x\leq 10\\ &=\frac{1}{10},\quad 0 \leq x\leq 10 \end{aligned} $$
The distribution function of $X$ is
$$ \begin{aligned} F(x)&=\frac{x0}{10 0},\quad 0 \leq x\leq 10\\ &=\frac{x}{10},\quad 0 \leq x\leq 10. \end{aligned} $$
a. Let us determine the probability that an individual waits more than $7$ minutes.
$$ \begin{aligned} P(X > 7) &=1P(X\leq 7)\\ &=1F(7)\\ &=1\dfrac{7  0}{10}\\ &=1\dfrac{7}{10}\\ &=10.7\\ &=0.3\\ \end{aligned} $$
b. Let us find the probability that an individual waits between $2$ and $7$ minutes.
$$ \begin{aligned} P(2 \leq X \leq 7) &= F(7)  F(2)\\ &=\frac{70}{10} \frac{20}{10}\\ &= \frac{7}{10}\frac{2}{10}\\ &= 0.70.2\\ &= 0.5. \end{aligned} $$
Conclusion
Hope you find Continous Uniform Distribution Calculator and step by step guide to solve uniform distribution examples helpful and educational.
Click on Theory button to understand conitnuous uniform distribution, mean, variance of uniform distribution,Raw Moments of uniform distribution with proof,M.G.F of uniform distribution with proof.
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