Continuous Uniform Distribution Calculator With Examples
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.
This tutorial will help you understand how to solve the numerical examples based on continuous uniform distribution.
Continuous Uniform Distribution Calculator
Use this calculator to find the probability density and cumulative probabilities for continuous Uniform distribution with parameter $a$ and $b$.
Uniform Distribution Calculator | |
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Minimum Value $a$: | |
Maximum Value $b$ | |
Value of x | |
Results | |
Probability density : f(x) | |
Probability X less than x: P(X < x) | |
Probability X greater than x: P(X > x) | |
How to find Continuous Uniform Distribution Probabilities?
Step 1 – Enter the minimum value $a$
Step 2 – Enter the maximum value $b$
Step 3 – Enter the value of $x$
Step 4 – Click on "Calculate" button to get Continuous Uniform distribution probabilities
Step 5 – Gives the output probability at $x$ for Continuous Uniform distribution
Step 6 – Gives the output cumulative probabilities for Continuous Uniform distribution
Definition of Uniform Distribution
A continuous random variable $X$ is said to have a Uniform distribution (or rectangular distribution) with parameters $\alpha$ and $\beta$ if its p.d.f. is given by
$$ \begin{align*} f(x)&= \begin{cases} \frac{1}{\beta - \alpha}, & \alpha \leq x\leq \beta \\ 0, & Otherwise. \end{cases} \end{align*} $$
Notation: $X\sim U(\alpha, \beta)$.
Distribution Function
The distribution function of uniform distribution $U(\alpha,\beta)$ is
$$ \begin{align*} F(x)&= \begin{cases} 0, & x<\alpha\\ \frac{x-\alpha}{\beta - \alpha}, & \alpha \leq x\leq \beta \\ 1, & x>\beta \end{cases} \end{align*} $$
Continuous Uniform Distribution Example 1
The waiting time at a bus stop is uniformly distributed between 1 and 12 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 12 minute. That is $X\sim U(1,12)$.
(a) The probability density function of $X$ is
$$ \begin{aligned} f(x) & = \frac{1}{12-1},\; 1\leq x \leq 12\\ & = \frac{1}{11},\; 1\leq x \leq 12. \end{aligned} $$
(b) The probability that the rider waits 8 minutes or less is
$$ \begin{aligned} P(X\leq 8) & = \int_1^8 f(x) \; dx\\ & = \frac{1}{11}\int_1^8 \; dx\\ & = \frac{1}{11} \big[x \big]_1^8\\ &= \frac{1}{11}\big[ 8-1\big]\\ &= \frac{7}{11}\\ &= 0.6364. \end{aligned} $$
(c) The expected wait time is $E(X) =\dfrac{\alpha+\beta}{2} =\dfrac{1+12}{2} =6.5$ minutes.
(d) The variance of waiting time is $V(X) =\dfrac{(\beta-\alpha)^2}{12} =\dfrac{(12-1)^2}{12} =10.08$.
Continuous Uniform Distribution Example 2
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. What is the probability that a vehicle will weigh more than 3,900 pounds?
d. What is the probability that a vehicle will weigh 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 $\beta=4500$
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
$$ \begin{aligned} F(x)&=\frac{x-2500}{4500- 2500},\quad 2500 \leq x\leq 4500\\ &=\frac{x-2500}{2000},\quad 2500 \leq x\leq 4500. \end{aligned} $$
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
$$ \begin{aligned} sd(X) &= \sqrt{V(X)}\\ &=\sqrt{\dfrac{(\beta-\alpha)^2}{12}}\\ &=\sqrt{\dfrac{(4500-2500)^2}{12}}\\ &=577.35 \end{aligned} $$
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
$$ \begin{aligned} P(X > 3900) &=1-P(X\leq 3900)\\ &=1-F(3900)\\ &=1-\dfrac{3900 - 2500}{2000}\\ &=1-\dfrac{1400}{2000}\\ &=1-0.7\\ &=0.3\\ \end{aligned} $$
d. The probability that a vehicle will weight between $3000$ and $3800$ pounds is
$$ \begin{aligned} P(3000 < X < 3800) &= F(3800) - F(3000)\\ &=\frac{3800-2500}{2000}- \frac{3000-2500}{2000}\\ &= \frac{1300}{2000}-\frac{500}{2000}\\ &= 0.65-0.25\\ &= 0.4. \end{aligned} $$
Continuous Uniform Distribution Example 3
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 voltage in a circuit is
$$ \begin{aligned} E(X) &=\dfrac{\alpha+\beta}{2}\\ &=\dfrac{6+12}{2}\\ &=9 \end{aligned} $$
The standard deviation of voltage in a circuit is
$$ \begin{aligned} sd(X) &= \sqrt{V(X)}\\ &=\sqrt{\dfrac{(\beta-\alpha)^2}{12}}\\ &=\sqrt{\dfrac{(12-6)^2}{12}}\\ &=1.73 \end{aligned} $$
b. The distribution function of $X$ is
$$ \begin{aligned} F(x)&=\frac{x-6}{12- 6},\quad 6 \leq x\leq 12\\ &=\frac{x-6}{6},\quad 6 \leq x\leq 12. \end{aligned} $$
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
$$ \begin{aligned} P(X > 9) &=1-P(X\leq 9)\\ &=1-F(9)\\ &=1-\dfrac{9 - 6}{6}\\ &=1-\dfrac{3}{6}\\ &=1-0.5\\ &=0.5\\ \end{aligned} $$
d. The probability that voltage is between $9$ and $11$ volts is
$$ \begin{aligned} P(9 < X < 11) &= F(11) - F(9)\\ &=\frac{11-6}{6}- \frac{9-6}{6}\\ &= \frac{5}{6}-\frac{3}{6}\\ &= 0.8333-0.5\\ &= 0.3333. \end{aligned} $$
Continuous Uniform Distribution Example 4
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
$$ \begin{aligned} F(x)&=\frac{x-7}{10- 7},\quad 7 \leq x\leq 10\\ &=\frac{x-7}{3},\quad 7 \leq x\leq 10. \end{aligned} $$
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.
$$ \begin{aligned} P(7.4 < X < 9.5) &= F(9.5) - F(7.4)\\ &=\frac{9.5-7}{3}- \frac{7.4-7}{3}\\ &= \frac{2.5}{3}-\frac{0.4}{3}\\ &= 0.8333-0.1333\\ &= 0.7. \end{aligned} $$
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.
$$ \begin{aligned} P(X > 8.5) &=1-P(X\leq 8.5)\\ &=1-F(8.5)\\ &=1-\dfrac{8.5 - 7}{3}\\ &=1-\dfrac{1.5}{3}\\ &=1-0.5\\ &=0.5\\ \end{aligned} $$
Continuous Uniform Distribution Example 5
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
$$ \begin{aligned} F(x)&=\frac{x-0}{10- 0},\quad 0 \leq x\leq 10\\ &=\frac{x}{10},\quad 0 \leq x\leq 10. \end{aligned} $$
a. Let us determine the probability that an individual waits more than $7$ minutes.
$$ \begin{aligned} P(X > 7) &=1-P(X\leq 7)\\ &=1-F(7)\\ &=1-\dfrac{7 - 0}{10}\\ &=1-\dfrac{7}{10}\\ &=1-0.7\\ &=0.3\\ \end{aligned} $$
b. Let us find the probability that an individual waits between $2$ and $7$ minutes.
$$ \begin{aligned} P(2 \leq X \leq 7) &= F(7) - F(2)\\ &=\frac{7-0}{10}- \frac{2-0}{10}\\ &= \frac{7}{10}-\frac{2}{10}\\ &= 0.7-0.2\\ &= 0.5. \end{aligned} $$
Conclusion
In this tutorial, you learned about how to calculate mean, variance and probabilities of Continuous Uniform distribution. You also learned about how to solve numerical problems based on Continuous Uniform distribution.
To read more about the step by step tutorial on Continuous Uniform distribution refer the link Continuous Uniform Distribution. This tutorial will help you to understand Continuous Uniform distribution and you will learn how to derive mean of Continuous Uniform distribution, variance of Continuous Uniform distribution, moment generating function and other properties of Continuous Uniform distribution.
To learn more about other probability distributions, please refer to the following tutorial:
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