Business Statistics:
A Decision-Making Approach
8th Edition
Chapter 6
Introduction to Continuous
Probability Distributions
6-1
Chapter Goals
After completing this chapter, you should be
able to:

Convert values from any normal distribution to a
standardized z-score

Find probabilities using a normal distribution table

Apply the normal distribution to business problems

Recognize when to apply the uniform and
exponential distributions (not covered)
6-2
Continuous Probability Distributions

A continuous probability distribution differs from a
discrete probability distribution in several ways.



A continuous random variable can take on an infinite
number of values. That is, the probability for a specific
variable is always zero.
As a result, a continuous probability distribution cannot be
expressed in tabular form.
Instead, an equation or formula is used to describe a
continuous probability distribution.
6-3
Probability Density Function

The equation for describing a continuous probability
distribution is called a probability density function
and noted as “f(x).”


Sometimes refer to as Density function
pdf or PDF
1
 ( x   ) 2 / 2 2
f ( x) 
2
e
6-4
Probability Density Function Example

The shaded area in the graph
represents the probability that the
random variable X is less than or
equal to a.



Cumulative probability
Always to the left
The probability that the random
variable X is exactly equal to a
would be zero.
6-5
Types of Continuous Distributions

Three types



Normal
Uniform (so easy…by yourself)
Exponential (not cover)
6-6
The Normal Distribution
‘Bell Shaped’
 Symmetrical
 Mean=Median=Mode

Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+  to  
f(x)
σ
x
μ
Mean
Median
Mode
6-7
Position and Shape of the Normal Curve
6-8
Basis for the empirical rules



68.26% of the area under the curve falls within +/- 1 Stdev.
95.44% of the area under the curve falls within +/- 2 Stdev.
99.73% of the area under the curve falls within +/- 3 Stdev.
9
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve
is symmetric, so half is above the mean, half is below
f(x)
P(   x  μ)  0.5
0.5
P(μ  x   )  0.5
0.5
μ
x
P(   x   )  1.0
6-10
Normal Probability Distribution

The normal distribution refers to a family
of continuous probability distributions described by the
following equation.
1
 ( x   ) 2 / 2 2
f ( x) 
e
2
Where:
 = mean
 = standard deviation
 = 3.14159
e = 2.71828
6-11
The Standard Normal Distribution



Also known as the “z” distribution
Mean is defined to be 0
Standard Deviation is 1
f(z)
1
0
z
Values above the mean have positive z-values
Values below the mean have negative z-values
6-12
Translation to the Standard
Normal Distribution

Translate from x to the standard normal (the “z”
distribution) by subtracting the mean of x and dividing
by its standard deviation:
x μ
z
σ
z is the number of standard deviations units that
x is away from the population mean
6-13
Example 1

If x is distributed normally with mean of 100
and standard deviation of 50, the z value for
x = 250 is
x  μ 250  100
z

 3.0
σ
50

This says that x = 250 is three standard
deviations (3 increments of 50 units) above
the mean of 100.
6-14
Comparing x and z units
μ = 100
σ = 50
100
0
250
3.0
x
z
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (x) or in standardized units (z)
6-15
Why standardizing?


To measure variable with different means and/or
standard deviations on a single scale.
Several sections of BA 301 by different instructors.



Distributions of scores might be different due to differences in
teaching methods and grading procedures….
By calculating Z value for each student, the distributions of the
Z-value would be approximately the same in each section.
That is, it is very easy to interpret the result using the Z-value
since it is standardized.
6-16
Example 2

Statistically (based standard normal distribution), are
they really or fundamentally different matters?

SAT mean 1020 with standard deviation of 160. You scored
1260. What is your percentile?

Average adult male height is 70 inches with standard deviation
of 2 inches. You are 73 inches tall. What is your height
percentile?
6-17
Z-scores for Two Problems
x  μ 1260  1020
z

 1.5
σ
160
x μ
73  70
z

 1.5
σ
2
Both are 1.5 Std Dev above the mean….
Download the Normal Dist. using Excel file from the class website
6-18
Four Excel Functions

NORMDIST(x, mean, standard_dev, cumulative)


Gives the probability that a number falls at or below a given
value of a normal distribution
NORMSDIST(z)

Translates the number of standard deviations (z) into
cumulative probabilities
6-19
Four Excel Functions


The distribution of heights of American women aged 18 to
24 is approximately normally distributed with a mean of
65.5 inches (166.37 cm) and a standard deviation of 2.5
inches (6.35 cm).
What percentage of these women is taller than 5' 8", that
is, 68 inches (172.72 cm)?




NORMDIST(68, 65.5, 2.5, TRUE) = 84.13%
STANDARDIZE(68, 65.5, 2.5) = 1, NORMSDIST(1) = 84.13%
Cumulative probability (to the left on the normal curve)
Thus, 1- 84.13% = 15.87% or NORMSDIST(-1) = 15.87%
6-20
Four Excel Functions

NORMINV(probability, mean, standard_dev)


Calculates the x variable given a probability
NORMSINV(probability)

Given the probability that a variable is within a certain distance
of the mean, it finds the z value
6-21
Four Excel Functions

How tall would a woman need to be if she
wanted to be among the tallest 75% of
American women (find 25% shorter instead as
the right figure shows)?


NORMINV(0.25, 65.5, 2.5) = 63.81 inches
Find the z values that mark the boundary that is
25% less than the mean and 25% (75%) more
than the mean.


NORMSINV(0.25) = - 0.674
NORMSINV(0.75) = 0.674
6-22
Formula Table
We have
X
We want
X
Z
p
z
Z
p
X=μ+zσ
Norminv(p,μ,σ)
x

Normsinv(p)
z  standardiz e( x,  ,  )
Normdist(x,μ,σ,1)
Normsdist(z)
6-23
The Standard Normal Table


The Standard Normal table in the textbook
(Appendix D)
Gives the probability from the mean (zero)
up to a desired value for z
0.4772
Example:
P(0 < z < 2.00) = 0.4772
0
2.00
z
6-24
The Standard Normal Table
(continued)



The Standard Normal Table gives the
probability between the mean and a certain z
value
The z value ALWAYS refers to the area
between some value (-z or +z) and the mean
Since the distribution is symmetrical, the
Standard Normal Table only displays
probabilities for ½ of the full distribution
6-25
The Standard Normal Table
(continued)
The column gives the value of
z to the second decimal point
z
The row shows
the value of z
to the first
decimal point
0.00
0.01
0.02
…
0.1
0.2
.
.
.
2.0
.4772
P(0 < z < 2.00)2.0
= 0.4772
The value within the
table gives the
probability from z = 0
up to the desired z
value
6-26
General Procedure for
Finding Probabilities
1. Determine  and 
2. Define the event of interest
e.g., P(x > x1)
3. Convert to standard normal
x μ
z
σ
4. Use the table to find the probability
6-27
z Table Example

Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
Calculate z-values:
x μ 8 8
z

0
σ
5
x  μ 8.6  8
z

 0.12
σ
5
8 8.6
x
0 0.12
Z
P(8 < x < 8.6)
= P(0 < z < 0.12)
6-28
z Table Example
(continued)

Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
=8
=5
8 8.6
P(8 < x < 8.6)
=0
=1
x
0 0.12
z
P(0 < z < 0.12)
6-29
Solution: Finding P(0 < z < 0.12)
Standard Normal Probability
Table (Portion)
z
.00
.01
P(8 < x < 8.6)
= P(0 < z < 0.12)
.02
0.0478
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
Z
0.3 .1179 .1217 .1255
0.00
0.12
6-30
Finding Normal Probabilities
Suppose x is normal with mean 8.0
and standard deviation 5.0.
 Now Find P(x < 8.6)


The probability of obtaining a value less than 8.6
P = 0.5
Z
8.0
8.6
6-31
Finding Normal Probabilities
(continued)
Suppose x is normal with mean 8.0
and standard deviation 5.0.
 Now Find P(x < 8.6)

P(x < 8.6)
0.0478
0.5000
= P(z < 0.12)
= P(z < 0) + P(0 < z < 0.12)
= 0.5000 + 0.0478 = 0.5478
Z
0.00
0.12
6-32
Upper Tail Probabilities
Suppose x is normal with mean 8.0
and standard deviation 5.0.
 Now Find P(x > 8.6)

Z
8.0
8.6
6-33
Upper Tail Probabilities
(continued)

Now Find P(x > 8.6)…
P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)
= 0.5000 - 0.0478 = 0.4522
0.5000
Z
0
0.12
0.0478
0.4522
Z
0
0.12
6-34
Lower Tail Probabilities
Suppose x is normal with mean 8.0
and standard deviation 5.0.
 Now Find P(7.4 < x < 8)

Z
8.0
7.4
6-35
Lower Tail Probabilities
(continued)
Now Find P(7.4 < x < 8)…the probability
between 7.4 and the mean of 8
The Normal distribution is
symmetric, so we use the
same table even if z-values
are negative:
0.0478
P(7.4 < x < 8)
= P(-0.12 < z < 0)
Z
= 0.0478
8.0
7.4
6-36
Empirical Rules
What can we say about the distribution of values
around the mean if the distribution is normal?
f(x)
μ ± 1σ covers about 68%
of x’s
σ
σ
Recall
Tchebyshev
from Chpt. 3
μ1σ
μ
μ+1σ
x
68.26%
6-37
The Empirical Rule
(continued)
μ ± 2σ covers about 95% of x’s
μ ± 3σ covers about 99.7% of x’s
2σ
3σ
2σ
μ
95.44%
x
3σ
μ
x
99.72%
6-38
Importance of the Rule

If a value is about 2 or more standard
deviations away from the mean in a normal
distribution, then it is far from the mean

The chance that a value that far or farther
away from the mean is highly unlikely, given
that particular mean and standard deviation
6-39
Normal Probabilities in PHStat

We can use Excel and PHStat to quickly
generate probabilities for any normal
distribution

We will find P(8 < x < 8.6) when x is
normally distributed with mean 8 and
standard deviation 5
6-40
PHStat Dialogue Box
Select desired options
and enter values
6-41
PHStat Output
6-42
The Uniform Distribution
The uniform distribution is a probability distribution
that has equal probabilities for all possible
outcomes of the random variable

Referred to as the distribution of “little information”

Probability is the same for ANY interval of the same width

Useful when you have limited information about how the
data “behaves” (e.g., is it skewed left?)
6-43
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
f(x) =
1
ba
if a  x  b
0
otherwise
where
f(x) = value of the density function at any x value
a = lower limit of the interval of interest
b = upper limit of the interval of interest
6-44
The Mean and Standard Deviation
for the Uniform Distribution
The mean (expected value) is:
a+b
E(x)  μ 
2
The standard deviation is
(b  a)2
σ
12
where
a = lower limit of the interval from a to b
b = upper limit of the interval from a to b
6-45
Steps for Using the
Uniform Distribution
1. Define the density function
2. Define the event of interest
3. Calculate the required probability
f(x)
x
6-46
Uniform Distribution
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6
f(x)
.25
2
6
x
6-47
Uniform Distribution
Example: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
2+6
E(x)  μ 
4
2
(b  a)
(6  2)
σ

 1.1547
12
12
2
2
6-48
The Exponential Distribution

Used to measure the time that elapses
between two occurrences of an event (the
time between arrivals)


Examples:
 Time between trucks arriving at a dock
 Time between transactions at an ATM Machine
 Time between phone calls to the main operator
Recall l = mean for Poisson (see Chpt. 5)
6-49
The Exponential Distribution

The probability that an arrival time is equal to or
less than some specified time a is
P(0  x  a)  1  e
 λa
where 1/l is the mean time between events and e = 2.7183
NOTE: If the number of occurrences per time period is Poisson
with mean l, then the time between occurrences is exponential
with mean time 1/ l and the standard deviation also is 1/l
6-50
Exponential Distribution
(continued)

Shape of the exponential distribution
f(x)
l = 3.0
(mean = .333)
l = 1.0
(mean = 1.0)
l= 0.5
(mean = 2.0)
x
6-51
Example
Example: Customers arrive at the claims counter at
the rate of 15 per hour (Poisson distributed). What
is the probability that the arrival time between
consecutive customers is less than five minutes?
 Time between arrivals is exponentially distributed
with mean time between arrivals of 4 minutes (15
per 60 minutes, on average)

1/l = 4.0, so l = .25

P(x < 5) = 1 - e-la = 1 – e-(.25)(5) = 0.7135
There is a 71.35% chance that the arrival time between
consecutive customers is less than 5 minutes
6-52
Using PHStat
6-53
Chapter Summary

Reviewed key continuous distributions



normal
uniform
exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems
6-54
All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system, or transmitted, in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the publisher.
Printed in the United States of America.
6-55