## 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.

## Calculator

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) | |

## 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 Calculator** to calculate probability density,mean of uniform distribution,variance of uniform distribution.

## Continuous Uniform Distribution Example 1

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

## Uniform Probability Distribution Calculator 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. 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

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

## Uniform Distribution Probability Calculator 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 (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

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

## Continuous Uniform Distribution Calculator 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.

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

## Uniform Distribution Probability Calculator 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

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 and mean deviation about mean of Uniform distribution.