CHAPTER 7
Continuous Probability Distributions
to accompany
Introduction to Business Statistics
fourth edition,by Ronald M,Weiers
Presentation by Priscilla Chaffe-Stengel
Donald N,Stengel
? 2002 The Wadsworth Group
Chapter 7 - Learning Objectives
? Differentiate between the normal and
the exponential distributions.
? Use the standard normal distribution
and z-scores to determine probabilities
associated with the normal distribution.
? Use the normal distribution to
approximate the binomial distribution.
? Use the exponential distribution to
determine related probabilities.
? 2002 The Wadsworth Group
Chapter 7 - Key Terms
?Probability density function
?Probability distributions
–Standard normal distribution
?Mean,variance,applications
–Exponential distribution
?Mean,variance,applications
?Normal approximation to the
binomial distribution
? 2002 The Wadsworth Group
Chapter 7 - Key Concept
?The area under a probability
density function between
two bounds,a and b,is the
probability that a value will
occur within the bounded
interval between a and b.
? 2002 The Wadsworth Group
The Normal Distribution
? An important family of continuous
distributions
? Bell-shaped,symmetric,and asymptotic
? To specify a particular distribution in this
family,two parameters must be given:
– Mean
– Standard deviation
? 2002 The Wadsworth Group
Areas under the Normal Curve
Use the standard normal table to find:
? The z-score such that the area from the midpoint to z is 0.20.
In the interior of the standard normal
table,look up a value close to 0.20.
The closest value is 0.1985,which
occurs at
z = 0.52.
? 2002 The Wadsworth Group
20%
of the area
z
Areas under the Normal Curve
Use the standard normal table to find:
? The probability associated with z,P(0 ? z ? 1.32).
Locate the row whose header is 1.3,Proceed
along that row to the column whose
header is,02,There you find the
value,4066,which is the amount of
area capture between the mean
and a z of 1.32.
Answer,0.4066
? 2002 The Wadsworth Group
z = 1,32
Areas under the Normal Curve
Use the standard normal table to find:
? The probability associated with z,P(–1.10 ? z ? 1.32).
Find the amount of area between
the mean and z = 1.32 and add it
to the amount of area between
the mean and z = 1.10*.
0.3643 + 0.4066 = 0.7709
? 2002 The Wadsworth Group
z = 1.32 z =?,10
Area 1 Area 2
Areas under the Normal Curve -
Dealing with Negative Z’s
?Note - Because the normal curve is
symmetric,the amount of area
between the mean and z = –1.10 is
the same as the amount of area
between the mean and z = +1.10.
? 2002 The Wadsworth Group
Areas under the Normal Curve
Use the standard normal table to find:
? The probability associated with z,P(1.00 ? z ? 1.32).
Find the amount of area between
the mean and z = 1.00 and subtract
it from the amount of area
between the mean and z = 1.32.
0.4066 – 0.3413 = 0.0653
? 2002 The Wadsworth Group
z = 1,32
z = 1,00
Standardizing Individual Data
Values on a Normal Curve
? The standardized z-score is how far above or
below the individual value is compared to the
population mean in units of standard
deviation.
–,How far above or below”= data value – mean
–,In units of standard deviation”= divide by s
? Standardized individual value
z ? da t a v a l ue ? m e a n
s t a nd a r d d e v i a t i on
? x ?s
? 2002 The Wadsworth Group
Standard Normal Distribution:
An Example
? It has been reported that the average hotel check-in time,
from curbside to delivery of bags into the room,is 12.1
minutes,Mary has just left the cab that brought her to her
hotel,Assuming a normal distribution with a standard
deviation of 2.0 minutes,what is the probability that the time
required for Mary and her bags to get to the room will be:
a) greater than 14.1 minutes?
b) less than 8.1 minutes?
c) between 10.1 and 14.1 minutes?
d) between 10.1 and 16.1 minutes?
? 2002 The Wadsworth Group
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? a) Greater than 14.1 minutes
P(x > 14.1) = P(z > 1.00)
=,5 –,3413 = 0.1587
z ? x–?s ? 14.1–12.12.0 ? 1.00
? 2002 The Wadsworth Group
z = 1,00
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? b) Less than 8.1 minutes
P(x < 8.1) = P(z < –2.00)
=,5 –,4772 = 0.0228
z ? x–?s ? 8.1–12.12.0 ? –2.00
? 2002 The Wadsworth Group
z =?,00
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? c) Between 10.1 and 14.1 minutes
P(10.1 < x < 14.1)
= P(–1.00 < z < 1.00)
= 0.3413 + 0.3413 = 0.6826
zlower ? x–?s ? 10.1–12.12.0 ? –1.00
zupper ? x–?s ? 14.1–12.12.0 ? 1.00
? 2002 The Wadsworth Group
z = 1,00 z =?,00
Are a 1 Ar ea 2
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? d) Between 10.1 and 16.1 minutes
P(10.1 < x < 16.1)
= P(–1.00 < z < 2.00)
= 0.3413 + 0.4772 = 0.8185
z
l o we r
? x – ?s ? 10, 1 – 12, 1
2, 0
? 1, 00
z u p p e r ? x – ?s ? 16, 1 – 12, 1
2, 0
? 2, 00
? 2002 The Wadsworth Group
z = 2.00 z =?,00
Ar ea 1 A r ea 2
Example,Using Microsoft Excel
? Problem,What is the probability that the time required for
Mary and her bags to get to the room will be:
a) greater than 14.1 minutes?
In a cell on an Excel worksheet,type
=1-NORMDIST(14.1,12.1,2,true)
and you will see the answer,0.1587
b) less than 8.1 minutes?
In a cell on an Excel worksheet,type
=NORMDIST(8.1,12.1,2,true)
and you will see the answer,0.0228
? 2002 The Wadsworth Group
Example,Using Microsoft Excel
? Problem,What is the probability that the time required for
Mary and her bags to get to the room will be:
c) between 10.1 and 14.1 minutes?
In a cell on an Excel worksheet,type all on one line
=NORMDIST(14.1,12.1,2,true)-
NORMDIST(10.1,12.1,2,true)
and you will see the answer,0.6826
d) between 10.1 and 16.1 minutes?
In a cell on an Excel worksheet,type all on one line
=NORMDIST(16.1,12.1,2,true)-
NORMDIST(10.1,12.1,2,true)
and you will see the answer,0.8185
? 2002 The Wadsworth Group
The Exponential Distribution
where l = mean and standard deviation
e = 2.71828,a constant
? Probability:
? Application,Every day,drivers arrive at a
tollbooth,If the Poisson distribution were applied
to this process,what would be an appropriate
random variable? What would be the exponential
distribution counterpart to this random variable?
f ( x ) ? ?e – l x
P ( x ? k ) ? e – l k
? 2002 The Wadsworth Group
Continuous Probability Distributions
to accompany
Introduction to Business Statistics
fourth edition,by Ronald M,Weiers
Presentation by Priscilla Chaffe-Stengel
Donald N,Stengel
? 2002 The Wadsworth Group
Chapter 7 - Learning Objectives
? Differentiate between the normal and
the exponential distributions.
? Use the standard normal distribution
and z-scores to determine probabilities
associated with the normal distribution.
? Use the normal distribution to
approximate the binomial distribution.
? Use the exponential distribution to
determine related probabilities.
? 2002 The Wadsworth Group
Chapter 7 - Key Terms
?Probability density function
?Probability distributions
–Standard normal distribution
?Mean,variance,applications
–Exponential distribution
?Mean,variance,applications
?Normal approximation to the
binomial distribution
? 2002 The Wadsworth Group
Chapter 7 - Key Concept
?The area under a probability
density function between
two bounds,a and b,is the
probability that a value will
occur within the bounded
interval between a and b.
? 2002 The Wadsworth Group
The Normal Distribution
? An important family of continuous
distributions
? Bell-shaped,symmetric,and asymptotic
? To specify a particular distribution in this
family,two parameters must be given:
– Mean
– Standard deviation
? 2002 The Wadsworth Group
Areas under the Normal Curve
Use the standard normal table to find:
? The z-score such that the area from the midpoint to z is 0.20.
In the interior of the standard normal
table,look up a value close to 0.20.
The closest value is 0.1985,which
occurs at
z = 0.52.
? 2002 The Wadsworth Group
20%
of the area
z
Areas under the Normal Curve
Use the standard normal table to find:
? The probability associated with z,P(0 ? z ? 1.32).
Locate the row whose header is 1.3,Proceed
along that row to the column whose
header is,02,There you find the
value,4066,which is the amount of
area capture between the mean
and a z of 1.32.
Answer,0.4066
? 2002 The Wadsworth Group
z = 1,32
Areas under the Normal Curve
Use the standard normal table to find:
? The probability associated with z,P(–1.10 ? z ? 1.32).
Find the amount of area between
the mean and z = 1.32 and add it
to the amount of area between
the mean and z = 1.10*.
0.3643 + 0.4066 = 0.7709
? 2002 The Wadsworth Group
z = 1.32 z =?,10
Area 1 Area 2
Areas under the Normal Curve -
Dealing with Negative Z’s
?Note - Because the normal curve is
symmetric,the amount of area
between the mean and z = –1.10 is
the same as the amount of area
between the mean and z = +1.10.
? 2002 The Wadsworth Group
Areas under the Normal Curve
Use the standard normal table to find:
? The probability associated with z,P(1.00 ? z ? 1.32).
Find the amount of area between
the mean and z = 1.00 and subtract
it from the amount of area
between the mean and z = 1.32.
0.4066 – 0.3413 = 0.0653
? 2002 The Wadsworth Group
z = 1,32
z = 1,00
Standardizing Individual Data
Values on a Normal Curve
? The standardized z-score is how far above or
below the individual value is compared to the
population mean in units of standard
deviation.
–,How far above or below”= data value – mean
–,In units of standard deviation”= divide by s
? Standardized individual value
z ? da t a v a l ue ? m e a n
s t a nd a r d d e v i a t i on
? x ?s
? 2002 The Wadsworth Group
Standard Normal Distribution:
An Example
? It has been reported that the average hotel check-in time,
from curbside to delivery of bags into the room,is 12.1
minutes,Mary has just left the cab that brought her to her
hotel,Assuming a normal distribution with a standard
deviation of 2.0 minutes,what is the probability that the time
required for Mary and her bags to get to the room will be:
a) greater than 14.1 minutes?
b) less than 8.1 minutes?
c) between 10.1 and 14.1 minutes?
d) between 10.1 and 16.1 minutes?
? 2002 The Wadsworth Group
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? a) Greater than 14.1 minutes
P(x > 14.1) = P(z > 1.00)
=,5 –,3413 = 0.1587
z ? x–?s ? 14.1–12.12.0 ? 1.00
? 2002 The Wadsworth Group
z = 1,00
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? b) Less than 8.1 minutes
P(x < 8.1) = P(z < –2.00)
=,5 –,4772 = 0.0228
z ? x–?s ? 8.1–12.12.0 ? –2.00
? 2002 The Wadsworth Group
z =?,00
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? c) Between 10.1 and 14.1 minutes
P(10.1 < x < 14.1)
= P(–1.00 < z < 1.00)
= 0.3413 + 0.3413 = 0.6826
zlower ? x–?s ? 10.1–12.12.0 ? –1.00
zupper ? x–?s ? 14.1–12.12.0 ? 1.00
? 2002 The Wadsworth Group
z = 1,00 z =?,00
Are a 1 Ar ea 2
An Example,cont.
Given in the problem:
μ= 12.1 minutes,s = 2.0 minutes
? d) Between 10.1 and 16.1 minutes
P(10.1 < x < 16.1)
= P(–1.00 < z < 2.00)
= 0.3413 + 0.4772 = 0.8185
z
l o we r
? x – ?s ? 10, 1 – 12, 1
2, 0
? 1, 00
z u p p e r ? x – ?s ? 16, 1 – 12, 1
2, 0
? 2, 00
? 2002 The Wadsworth Group
z = 2.00 z =?,00
Ar ea 1 A r ea 2
Example,Using Microsoft Excel
? Problem,What is the probability that the time required for
Mary and her bags to get to the room will be:
a) greater than 14.1 minutes?
In a cell on an Excel worksheet,type
=1-NORMDIST(14.1,12.1,2,true)
and you will see the answer,0.1587
b) less than 8.1 minutes?
In a cell on an Excel worksheet,type
=NORMDIST(8.1,12.1,2,true)
and you will see the answer,0.0228
? 2002 The Wadsworth Group
Example,Using Microsoft Excel
? Problem,What is the probability that the time required for
Mary and her bags to get to the room will be:
c) between 10.1 and 14.1 minutes?
In a cell on an Excel worksheet,type all on one line
=NORMDIST(14.1,12.1,2,true)-
NORMDIST(10.1,12.1,2,true)
and you will see the answer,0.6826
d) between 10.1 and 16.1 minutes?
In a cell on an Excel worksheet,type all on one line
=NORMDIST(16.1,12.1,2,true)-
NORMDIST(10.1,12.1,2,true)
and you will see the answer,0.8185
? 2002 The Wadsworth Group
The Exponential Distribution
where l = mean and standard deviation
e = 2.71828,a constant
? Probability:
? Application,Every day,drivers arrive at a
tollbooth,If the Poisson distribution were applied
to this process,what would be an appropriate
random variable? What would be the exponential
distribution counterpart to this random variable?
f ( x ) ? ?e – l x
P ( x ? k ) ? e – l k
? 2002 The Wadsworth Group
