## Continuous Uniform Distribution Calculator

Use **Continuous Uniform Distribution Calculator** to find the probability density and cumulative probabilities for continuous Uniform distribution with parameter $a$ and $b$.

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 Probability Distribution Calculator

Continuous Uniform Distribution Calculator | |
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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) | |

## How to use Continuous Uniform Distribution Calculator?

### Step-by-step procedure to use continuous uniform distribution calculator:

Step 1: Enter the value of a (alpha) and b (beta) in the input field

Step 2: Enter random number x to evaluate probability which lies between limits of distribution

Step 3: Click on “Calculate” button to calculate uniform probability distribution

Step 4: Calculate Probability Density,Probability X less than x and Probability X greater than x

## Uniform Distribution Definition

A continuous random variable $X$ is said to have uniform distribution with parameter $\alpha$ and $\beta$ if its p.d.f. is given by

$$f(x; \alpha,\beta) = \dfrac{1}{\beta-\alpha}; \alpha< x< \beta $$

## Distribution function of Uniform Distribution

Distribution function of continuous uniform distribution is

$$F(x) = \dfrac{x-\alpha}{\beta-\alpha}; \alpha< x< \beta $$

## Mean of Uniform Distribution

The mean of uniform distribution is $E(X) = \dfrac{\alpha+\beta}{2}$.

## Variance of Uniform Distribution

The variance of uniform distribution is $V(X) = \dfrac{(\beta - \alpha)^2}{2}$.

## Variables

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 - Continuous Uniform Distribution Mean and Standard Deviation Calculation

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

`$$ \begin{aligned} f(x) & = \frac{1}{10-1},\; 1\leq x \leq 10\\ & = \frac{1}{9},\; 1\leq x \leq 10. \end{aligned} $$`

The *uniform probability density function* calculated as : 0.1111

b. The probability that the rider waits 8 minutes or less is

`$$ \begin{aligned} P(X\leq 8) & = \int_1^8 f(x) \; dx\\ & = \frac{1}{9}\int_1^8 \; dx\\ & = \frac{1}{9} \big[x \big]_1^8\\ &= \frac{1}{9}\big[ 8-1\big]\\ &= \frac{7}{9}\\ &= 0.7778. \end{aligned} $$`

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 $V(X) =\dfrac{(\beta-\alpha)^2}{10} =\dfrac{(10-1)^2}{10} = 8.1$.

## Example 2 - Uniform Probability Distribution Calculator

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

`$$ \begin{aligned} F(x)&=\frac{x-2500}{4500- 2500},\quad 2500 \leq x\leq 4500\\ &=\frac{x-2500}{2000},\quad 2500 \leq x\leq 4500. \end{aligned} $$`

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

`$$ \begin{aligned} P(3000< X<3800) &= F(3800) - F(3000)\\ &=\frac{3800-2500}{2000}- \frac{3000-2500}{2000}\\ &= \frac{1300}{2000}-\frac{500}{2000}\\ &= 0.65-0.25\\ &= 0.4. \end{aligned} $$`

## Example 3 - Uniform Distribution Probability Calculator

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

`$$ \begin{aligned} sd(X) &= \sqrt{V(X)}\\ &=\sqrt{\dfrac{(\beta-\alpha)^2}{12}}\\ &=\sqrt{\dfrac{(12-6)^2}{12}}\\ &=1.73 \end{aligned} $$`

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

`$$ \begin{aligned} P(9 < X < 11) &= F(11) - F(9)\\ &=\frac{11-6}{6}- \frac{9-6}{6}\\ &= \frac{5}{6}-\frac{3}{6}\\ &= 0.8333-0.5\\ &= 0.3333. \end{aligned} $$`

## Example 4 - Continuous Uniform Probability Distribution Calculator

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.

`$$ \begin{aligned} P(X > 8.5) &=1-P(X\leq 8.5)\\ &=1-F(8.5)\\ &=1-\dfrac{8.5 - 7}{3}\\ &=1-\dfrac{1.5}{3}\\ &=1-0.5\\ &=0.5\\ \end{aligned} $$`

## Example 5 - Uniform Distribution Probability Calculator

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

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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