# 5.2: The Uniform Distribution

The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints.

The mathematical statement of the uniform distribution is

$$f(x) = \frac{1}{b-a}$$ for $$a \leq x \leq b$$

where $$a =$$ the lowest value of $$x$$ and $$b =$$ the highest value of $$x$$.

Formulas for the theoretical mean and standard deviation are

$$\mu=\frac{a+b}{2}$$ and $$\sigma=\sqrt{\frac{(b-a)^{2}}{12}}$$

Exercise $$\PageIndex{1}$$

The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of $$a$$ and $$b$$. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.

 1 12 4 10 4 14 11 7 11 4 13 2 4 6 3 10 0 12 6 9 10 5 13 4 10 14 12 11 6 10 11 0 11 13 2

Table 5.1

Example $$\PageIndex{2}$$

The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.

a. What is the probability that a person waits fewer than 12.5 minutes?

a. Let $$X$$ = the number of minutes a person must wait for a bus. $$a = 0$$ and $$b = 15$$. $$X \sim U(0, 15)$$. Write the probability density function. $$f(x) = \frac{1}{15-0}=\frac{1}{15}$$ for $$0 \leq x \leq 15$$.

Find $$P(x < 12.5)$$. Draw a graph.

$P(x<k)=\text { (base) (height) }=(12.5-0)\left(\frac{1}{15}\right)=0.8333\nonumber$

The probability a person waits less than 12.5 minutes is 0.8333.

b. On the average, how long must a person wait? Find the mean, $$\mu$$, and the standard deviation, $$\sigma$$.

b. $$\mu=\frac{a+b}{2}=\frac{15+0}{2}=7.5$$. On the average, a person must wait 7.5 minutes.

$$\sigma=\sqrt{\frac{(b-a)^{2}}{12}}=\sqrt{\frac{(15-\theta)^{2}}{12}}=4.3$$. The Standard deviation is 4.3 minutes.

c. Ninety percent of the time, the time a person must wait falls below what value?

Note

This asks for the 90th percentile.

c. Find the 90th percentile. Draw a graph. Let $$k =$$ the 90th percentile.

$$P(x<k)> \(0.90=(k)\left(\frac{1}{15}\right)$$

$$k=(0.90)(15)=13.5$$

The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.

Exercise $$\PageIndex{2}$$

The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

1. Find $$a$$ and $$b$$ and describe what they represent.
2. Write the distribution.
3. Find the mean and the standard deviation.
4. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?