Lind, Marchal, Wathen (2012), Statistical Techniques in Business and Economics, 15th Edition

1. Continuous
Probability
Distributions
Chapter 7
McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved.

2. Learning Objectives
LO1 List the characteristics of the uniform distribution.
LO2 Compute probabilities by using the uniform distribution.
LO3 List the characteristics of the normal probability distribution.
LO4 Convert a normal distribution to the standard normal distribution.
LO5 Find the probability that an observation on a normally distributed
random variable is between two values.
LO6 Find probabilities using the Empirical Rule.
LO7 Approximate the binomial distribution using the normal
distribution.
LO8 Describe the characteristics and compute probabilities using the
exponential distribution.
7-2

3. The Uniform Distribution
The uniform probability
distribution is perhaps
the simplest
distribution for a
continuous random
variable.
This distribution is
rectangular in shape
and is defined by
minimum and maximum
values.
LO1 List the characteristics of
the uniform distribution.
7-3

4. The Uniform Distribution – Mean and
Standard Deviation
LO1
7-4

5. The Uniform Distribution - Example
Southwest Arizona State University provides bus service to
students while they are on campus. A bus arrives at the North
Main Street and College Drive stop every 30 minutes between 6
A.M. and 11 P.M. during weekdays. Students arrive at the bus stop
at random times. The time that a student waits is uniformly
distributed from 0 to 30 minutes.
1. Draw a graph of this distribution.
2. Show that the area of this uniform distribution is 1.00.
3. How long will a student “typically” have to wait for a bus? In other
words what is the mean waiting time? What is the standard
deviation of the waiting times?
4. What is the probability a student will wait more than 25 minutes
5. What is the probability a student will wait between 10 and 20
minutes?
LO2 Compute probabilities by
using the uniform distribution.
7-5

6. The Uniform Distribution - Example
1. Graph of this distribution.
LO2
7-6

7. The Uniform Distribution - Example
2. Show that the area of this distribution is 1.00
LO2
7-7

8. The Uniform Distribution - Example
3. How long will a student
“typically” have to wait for a
bus? In other words what is
the mean waiting time?
What is the standard
deviation of the waiting
times?
LO2
7-8

9. The Uniform Distribution - Example
4. What is the
probability a
student will wait
more than 25
minutes? 0.1667
)
5
(
)
0
30
(
1
ase)
(height)(b
30)
Time
Wait
25
(






P
LO2
7-9

10. The Uniform Distribution - Example
5. What is the
probability a
student will wait
between 10 and 20
minutes? 0.3333
)
10
(
)
0
30
(
1
ase)
(height)(b
20)
Time
Wait
10
(






P
LO2
7-10

11. Characteristics of a Normal
Probability Distribution
1. It is bell-shaped and has a single peak at the center of the
distribution.
2. It is symmetrical about the mean
3. It is asymptotic: The curve gets closer and closer to the X-
axis but never actually touches it. To put it another way, the
tails of the curve extend indefinitely in both directions.
4. The location of a normal distribution is determined by the
mean,, the dispersion or spread of the distribution is
determined by the standard deviation,σ .
5. The arithmetic mean, median, and mode are equal
6. The total area under the curve is 1.00; half the area under
the normal curve is to the right of this center point, the mean,
and the other half to the left of it.
LO3 List the characteristics of the
normal probability distribution.
7-11

12. The Normal Distribution - Graphically
LO3
7-12

13. The Family of Normal Distribution
Different Means and
Standard Deviations
Equal Means and Different
Standard Deviations
Different Means and Equal Standard Deviations
LO3
7-13

14. The Standard Normal Probability
Distribution
 The standard normal distribution is a normal
distribution with a mean of 0 and a standard
deviation of 1.
 It is also called the z distribution.
 A z-value is the signed distance between a
selected value, designated X, and the population
mean , divided by the population standard
deviation, σ.
 The formula is:
LO4 Convert a normal distribution to the
standard normal distribution.
7-14

15. Areas Under the Normal Curve
LO4
7-15

16. The Normal Distribution – Example
The weekly incomes of
shift foremen in the
glass industry follow the
normal probability
distribution with a mean
of $1,000 and a
standard deviation of
$100.
What is the z value for
the income, let’s call it X,
of a foreman who earns
$1,100 per week? For a
foreman who earns
$900 per week?
LO5 Find the probability that an observation on a normally
distributed random variable is between two values.
7-16

17. Normal Distribution – Finding Probabilities
In an earlier example
we reported that the
mean weekly income
of a shift foreman in
the glass industry is
normally distributed
with a mean of $1,000
and a standard
deviation of $100.
What is the likelihood
of selecting a foreman
whose weekly income
is between $1,000
and $1,100?
LO5
7-17

18. Normal Distribution – Finding Probabilities
LO5
7-18

19. Finding Areas for Z Using Excel
The Excel function
=NORMDIST(x,Mean,Standard_dev,Cumu)
=NORMDIST(1100,1000,100,true)
generates area (probability) from
Z=1 and below
LO5
7-19

20. Normal Distribution – Finding Probabilities
(Example 2)
Refer to the information
regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Between $790 and $1,000?
LO5
7-20

21. Normal Distribution – Finding Probabilities
(Example 3)
Refer to the information
regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Less than $790?
LO5
7-21

22. Normal Distribution – Finding Probabilities
(Example 4)
Refer to the information
regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Between $840 and $1,200?
LO5
7-22

23. Normal Distribution – Finding Probabilities
(Example 5)
Refer to the information
regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Between $1,150 and $1,250
LO5
7-23

24. Using Z in Finding X Given Area - Example
Layton Tire and Rubber
Company wishes to set a
minimum mileage guarantee on
its new MX100 tire. Tests
reveal the mean mileage is
67,900 with a standard
deviation of 2,050 miles and
that the distribution of miles
follows the normal probability
distribution. Layton wants to set
the minimum guaranteed
mileage so that no more than 4
percent of the tires will have to
be replaced.
What minimum guaranteed
mileage should Layton
announce?
24
LO5
7-24

25. Using Z in Finding X Given Area - Example
64,312
x
)
1.75(2,050
-
67,900
x
67,900
-
x
)
1.75(2,050
-
x
for
solving
then
,
2,050
67,900
-
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1.75
-
:
equation
the
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ng
substituti
Then
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of
alue
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a
gives
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0.4599,
is
0.4600
to
closest
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the
B.1,
Appendix
Using
0.0400
-
0.5000
by
found
0.4600,
is
x
and
67,900
between
area
The
n
informatio
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the
using
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of
value
The
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,
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050
2
900
67
x
x


LO5
7-25

26. Using Z in Finding X Given Area - Excel
LO5
7-26

27. The Empirical Rule
 About 68 percent of
the area under the
normal curve is within
one standard
deviation of the
mean.
 About 95 percent is
within two standard
deviations of the
mean.
 Practically all is
within three standard
deviations of the
mean.
LO6 Find probabilities using the
Empirical Rule.
7-27

28. The Empirical Rule - Example
As part of its quality
assurance program, the
Autolite Battery Company
conducts tests on battery
life. For a particular D-cell
alkaline battery, the mean
life is 19 hours. The useful
life of the battery follows a
normal distribution with a
standard deviation of 1.2
hours.
Answer the following questions.
1. About 68 percent of the
batteries failed between
what two values?
2. About 95 percent of the
batteries failed between
what two values?
3. Virtually all of the batteries
failed between what two
values?
LO6
7-28

29. Normal Approximation to the Binomial
 The normal distribution (a continuous distribution)
yields a good approximation of the binomial
distribution (a discrete distribution) for large values
of n.
 The normal probability distribution is generally a good
approximation to the binomial probability distribution
when n and n(1- ) are both greater than 5.
LO7 Approximate the binomial distribution
using the normal distribution.
7-29

30. Normal Approximation to the Binomial
Using the normal distribution (a continuous distribution) as a substitute
for a binomial distribution (a discrete distribution) for large values of n
seems reasonable because, as n increases, a binomial distribution gets
closer and closer to a normal distribution.
LO7
7-30

31. Continuity Correction Factor
The value .5 subtracted or added, depending on the
problem, to a selected value when a binomial probability
distribution (a discrete probability distribution) is being
approximated by a continuous probability distribution (the
normal distribution).
LO7
7-31

32. How to Apply the Correction Factor
Only one of four cases may arise:
1. For the probability at least X occurs, use the area above (X -.5).
2. For the probability that more than X occurs, use the area above
(X+.5).
3. For the probability that X or fewer occurs, use the area below (X -
.5).
4. For the probability that fewer than X occurs, use the area below
(X+.5).
LO7
7-32

33. Suppose the
management of the
Santoni Pizza Restaurant
found that 70 percent of
its new customers return
for another meal. For a
week in which 80 new
(first-time) customers
dined at Santoni’s, what
is the probability that
60 or more will return
for another meal?
Normal Approximation to the Binomial -
Example
LO7
7-33

34. Normal Approximation to the Binomial - Example
P(X ≥ 60) = 0.063+0.048+ … + 0.001) = 0.197
Binomial distribution solution:
LO7
7-34

35. Step 1. Find the
mean and the
variance of a binomial
distribution and find
the z corresponding
to an X of 59.5 (x-.5,
the correction factor)
Step 2: Determine
the area from 59.5
and beyond
Normal Approximation to the Binomial -
Example
LO7
7-35

36. The Family of Exponential Distributions
Characteristics and Uses:
1. Positively skewed, similar to
the Poisson distribution (for
discrete variables).
2. Not symmetric like the
uniform and normal
distributions.
3. Described by only one
parameter, which we identify
as λ, often referred to as the
“rate” of occurrence
parameter.
4. As λ decreases, the shape of
the distribution becomes “less
skewed.”
LO8 Describe the characteristics and compute
probabilities using the exponential distribution.
The exponential distribution usually describes
inter-arrival situations such as:
• The service times in a system.
• The time between “hits” on a web site.
• The lifetime of an electrical component.
• The time until the next phone call arrives in a
customer service center
7-36

37. Exponential Distribution - Example
Orders for prescriptions arrive at a
pharmacy management website
according to an exponential
probability distribution at a mean of
one every twenty seconds.
Find the probability the next order
arrives in:
1) in less than 5 seconds,
2) in more than 40 seconds,
3) or between 5 and 40 seconds.
LO8
7-37

38. 1353
0
8647
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1
1
40
1
40
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Arrival
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Arrival
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2212
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Arrival
P
LO8
7-38

39. Exponential Distribution - Example
Compton Computers wishes to set a
minimum lifetime guarantee on it new
power supply unit. Quality testing
shows the time to failure follows an
exponential distribution with a mean of
4000 hours. Note that 4000 hours is a
mean and not a rate. Therefore, we
must compute λ as 1/4000 or 0.00025
failures per hour.
Compton wants a warranty period such
that only five percent of the power
supply units fail during that period.
What value should they set for the
warranty period?
LO8
7-39

40. Use formula (7–7) . In this case, the rate parameter is 4,000 hours and
we want the area, as shown in the diagram, to be .05.
Now, we need to solve this equation for x.
Obtain the natural log of both sides of the equation:
X = 205.17. Hence, Compton can set the warranty period at 205 hours
and expect about 5 percent of the power supply units to be returned.
LO8
)
(
,
)
(
.
)
Time
Arrival
(
x
x
e
e
x
P
000
4
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1
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7-40