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Chapter
Seven
McGraw-Hill/Irwin
© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.
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Chapter Seven
Continuous Probability Distributions
GOALS
When you have completed this chapter, you will be able to:
ONE
Understand the difference between discrete and continuous
distributions.
TWO
Compute the mean and the standard deviation for a uniform
distribution.
THREE
Compute probabilities using the uniform distribution.
FOUR
List the characteristics of the normal probability distribution.
Goals
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Chapter Seven
continued
Continuous Probability Distributions
GOALS
When you have completed this chapter, you will be able to:
FIVE
Define and calculate z values.
SIX
Determine the probability an observation will lie between two points
using the standard normal distribution.
SEVEN
Determine the probability an observation will be above or below a
given value using the standard normal distribution.
EIGHT
Use the normal distribution to approximate the binomial probability
distribution.
Goals
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A Discrete distribution
is based on random
variables which can
assume only clearly
separated values.
A Continuous
distribution usually
results from measuring
something.
Discrete distributions
studied include:
o Binomial
o Hypergeometric
o Poisson.
Continuous distributions
include:
o Uniform
o Normal
o Others
Discrete and continuous
distributions
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The Uniform distribution
Is rectangular in shape
Is defined by minimum and
maximum values
Has a mean computed as
follows:
a standard deviation
computed as follows:
f(x)
x
m= a+b
2
Has
s=
where a and b are
the minimum and
maximum values
(b-a)2
12
The uniform distribution
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Calculates its height as
P(x) =
1
if a < x < b and 0 elsewhere
(b-a)
Calculates its area as
1
Area = height* base =(b-a) *(b-a)
The uniform distribution
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Suppose the time
that you wait on the
telephone for a live
representative of
your phone company
to discuss your
problem with you is
uniformly distributed
between 5 and 25
minutes.
What is the mean wait time?
m= a+b
2
= 5+25
= 15
2
What is the standard
deviation of the wait time?
(b-a)2
s=
12
2
(25-5)
=
= 5.77
12
Example 1
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What is the probability of waiting more than ten minutes?
The area from 10
to 25 minutes is
15 minutes.
Thus:
P(10 < wait time < 25) = height*base
=
1
(25-5) *15 = .75
What is the probability of waiting between 15 and 20
minutes?
The area from 15 P(15 < wait time < 20) = height*base
to 20 minutes is =
1
*5 = .25
5 minutes. Thus:
(25-5)
Example 2 continued
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The Normal probability distribution
is bell-shaped and has a single peak at the
center of the distribution.
Is
symmetrical about the mean.
is
asymptotic.
That is the curve gets closer and
closer to the X-axis but never actually touches it.
m, to determine its location and
its standard deviation, s, to determine its
Has
its mean,
dispersion.
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r
a
l
i
t r
b
u
i o
n
:
m
=
0
,
s2
=
1
Characteristics of a Normal Distribution
0
. 4
Normal
curve is
symmetrical
. 3
0
. 2
0
. 1
f ( x
0
Theoretically,
curve extends to
infinity
. 0
- 5
a
Mean, median, and
mode are equal
x
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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 distance between a selected
value, designated X, and the population mean m,
divided by the population standard deviation, s.
The formula is:
z 
X m
s
The Standard Normal
Probability Distribution
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z 
X m
s
= $2,200 - $2000
$200
= 1.00
The bi-monthly
starting salaries of
recent MBA
graduates follows
the normal
distribution with a
mean of $2,000 and
a standard deviation
of $200. What is
the z-value for a
salary of $2,200?
MBA
Example 2
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What is the
z-value for
$1,700?
z
X m
s
$1,700  $2,000

 1.50
$200
A z-value of 1 indicates that the value of
$2,200 is one standard deviation above the
mean of $2,000. A z-value of –1.50 indicates
that $1,700 is 1.5 standard deviation below the
mean of $2000.
EXAMPLE 2 continued
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About 68 percent of
the area under the
normal curve is within
one standard deviation
of the mean.
m + 1s
About 95 percent is within two standard
deviations of the mean.
m + 2s
Practically all is within three standard
deviations of the mean.
m + 3s
Areas Under the Normal
Curve
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The daily water usage per
person in New Providence,
New Jersey is normally
distributed with a mean of
20 gallons and a standard
deviation of 5 gallons.
About 68 percent of those
living in New Providence
will use how many gallons
of water?
About 68% of the daily
water usage will lie between
15 and 25 gallons (+ 1s ).
Example 3
295
RHODE
ISLAND
Providence
Scituate
Res
Warwick
95
Newport
z 
z 
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What is the probability that
a person from New
Providence selected at
random will use between 20
and 24 gallons per day?
X m
s
X m
s
24  20

 0.80
5
20  20

 0.00
5
EXAMPLE 4
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The area under a normal
curve between a z-value of
0 and a z-value of 0.80 is
0.2881.
See the following diagram
We conclude that 28.81
percent of the residents use
between 20 and 24 gallons
of water per day.
Example 4 continued
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What percent of
the population use
between 18 and 26
gallons per day?
z
z
X m
s
X m
s
26  20

 1.20
5
18  20

 0.40
5
EXAMPLE 4 continued
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The area
The area
associated with a
associated with a
z-value of –0.40 is
z-value of 1.20 is
.1554.
.3849.
Adding these
areas, the result is
.5403.
We conclude that 54.03 percent of the
residents use between 18 and 26 gallons
of water per day.
EXAMPLE 4 continued
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Professor Mann has
determined that the scores
in his statistics course are
approximately normally
distributed with a mean of
72 and a standard
deviation of 5. He
announces to the class that
the top 15 percent of the
scores will earn an A.
What is the lowest score a
student can earn and still
receive an A?
EXAMPLE 5
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To begin let X be the score that
separates an A from a B.
If 15 percent of the students score
more than X, then 35 percent must
score between the mean of 72 and X.
The z-value associated corresponding to 35
percent is about 1.04.
EXAMPLE 5 continued
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We let z equal 1.04 and
solve the standard normal
equation for X. The result
is the score that separates
students that earned an A
from those that earned a B.
Those with a
score of 77.2 or
more earn an A.
X  72
1.04 
5
X  72  1.04 (5)  72  5.2  77 .2
EXAMPLE 5 continued
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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 np and
n(1- p ) are both greater
than 5.
The Normal Approximation
to the Binomial
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Recall for the binomial experiment:
oThere are only two mutually exclusive
outcomes (success or failure) on each trial.
oA binomial distribution results from counting
the number of successes.
oEach trial is independent.
oThe probability is fixed from trial to trial, and
the number of trials n is also fixed.
The Normal Approximation
continued
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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).
Continuity Correction Factor
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How to Apply the Correction Factor:
For the
probability that
more than X
occur, use the
area above (X+.5).
For the
probability
that fewer
than X occur,
use the area
below (X-.5).
For the probability
at least X occur,
use the area above
(X-.5).
For the
probability
that X or
fewer occur,
use the area
below (X+.5).
Continuity Correction Factor
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A recent study by a
marketing
research firm
showed that 15%
of American
households owned
a video camera.
For a sample of
200 homes, how
many of the homes
would you expect
to have video
cameras?
m  np  (.15)(200)  30
This is the mean of a
binomial distribution.
EXAMPLE 6
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What is the variance?
s  np (1  p )  (30)(1.15)  255
.
2
What is the standard deviation?
s  25.5  5.0498
EXAMPLE 6 continued
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What is the
probability
that less
than 40
homes in
the sample
have video
cameras?
We use the correction
factor (X-.5) for fewer
than, so X-.5 is 39.5.
The value of z is 1.88.
z
X m
s
39 .5  30 .0

 1.88
5.0498
EXAMPLE 6 continued
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From Appendix
D the area between 0 and 1.88
on the z scale is .4699.
So
the area to the left of 1.88 is .5000 + .4699 =
.9699.
The
likelihood that less than 40 of the 200
homes have a video camera is about 97%.
EXAMPLE 6 continued
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