Session:5 Continuous Random Variables

5.2 The Uniform Distribution

Introductory Business Statistics | Leadership Development – Micro-Learning Session

Rice University 2020 | Michael Laverty, Colorado State University Global Chris Littel, North Carolina State University| https://openstax.org/details/books/introductory-business-statistics

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) = 1b−a

$\frac{1}{�-�}$

for a ≤ x ≤ b

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

Formulas for the theoretical mean and standard deviation are

μ=a+b2

$�=\frac{�+�}{2}$

and σ=(b−a)212−−−−−√

$�=\sqrt{\frac{{(�-�)}^2}{12}}$

EXAMPLE 5.2

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

Problem

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

Solution

a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0, 15). Write the probability density function. f (x) = 115 − 0

$\frac{1}{15-0}$ = 115

$\frac{1}{15}$ for 0 ≤ x ≤ 15.

Find P (x < 12.5). Draw a graph.

P(x<k)=(base)(height)=(12.5−0)(115)=0.8333

$$�(�<�)=(\text{base})(\text{height})=(12.5–0)\left(\frac{1}{15}\right)=0.8333$$ 

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

Figure 5.11

Problem

b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.

Solution

b. μ = a + b2

$\frac{�+�}{2}$ = 15 + 02

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

σ = (b−a)212−−−−−√=(15−0)212−−−−−−√

$\sqrt{\frac{(�–�{)}^2}{12}}=\sqrt{\frac{(\mathrm{15}–0{)}^2}{12}}$ = 4.3. The Standard deviation is 4.3 minutes.

Problem

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

NOTE

This asks for the 90^th percentile.

Solution

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

P(x<k)=(base)(height)=(k−0)(115)

$�(�<�)=(\text{base})(\text{height})=(�-0)(\frac{1}{15})$

0.90=(k)(115)

$0.90=\left(�\right)\left(\frac{1}{15}\right)$

k=(0.90)(15)=13.5

$�=(0.90)(15)=13.5$

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

Figure 5.12