Transcript Slide 1

7
Chapter
Continuous Probability
Distributions
Describing a Continuous Distribution
Uniform Continuous Distribution
Normal Distribution
Standard Normal Distribution
Normal Approximation to the
Binomial
Normal Approximation to the
Poisson
Exponential Distribution
Triangular Distribution
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc.
Continuous Variables
Events as Intervals
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Discrete Variable – each value of X has its own
probability P(X).
Continuous Variable – events are intervals and
probabilities are areas underneath smooth curves.
A single point has no probability.
Describing a Continuous Distribution
PDFs and CDFs
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Probability Density Function
(PDF) – For a continuous
random variable,
the PDF is an
equation that shows
the height of the
curve f(x) at each
possible value of X
over the range of X.
Describing a Continuous Distribution
PDFs and CDFs
Continuous PDF’s:
• Denoted f(x)
• Must be nonnegative
• Total area under
curve = 1
• Mean, variance and
shape depend on
the PDF parameters
• Reveals the shape
of the distribution
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Normal PDF
Describing a Continuous Distribution
PDFs and CDFs
Continuous CDF’s:
• Denoted F(x)
• Shows P(X < x), the
cumulative proportion
of scores
• Useful for finding
probabilities
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Describing a Continuous Distribution
Probabilities as Areas
Continuous probability functions are smooth
curves.
• Unlike discrete
distributions, the
area at any
single point = 0.
• The entire area under
any PDF must be 1.
• Mean is the balance
point of the distribution.
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Describing a Continuous Distribution
Expected Value and Variance
7-7
Uniform Continuous Distribution
Characteristics of the Uniform Distribution
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If X is a random variable that is uniformly
distributed between a and b, its PDF has
constant height.
• Denoted U(a,b)
• Area =
base x height =
(b-a) x 1/(b-a) = 1
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Uniform Continuous Distribution
Characteristics of the Uniform Distribution
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Uniform Continuous Distribution
Characteristics of the Uniform Distribution
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The CDF increases linearly to 1.
CDF formula is
(x-a)/(b-a)
Uniform Continuous Distribution
Example: Anesthesia Effectiveness
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An oral surgeon injects a painkiller prior to
extracting a tooth. Given the varying
characteristics of patients, the dentist
views the time for anesthesia effectiveness
as a uniform random variable that takes
between 15 minutes and 30 minutes.
X is U(15, 30)
a = 15, b = 30, find the mean and standard
deviation.
Uniform Continuous Distribution
Example: Anesthesia Effectiveness
a + b 15 + 30
m=
=
= 22.5 minutes
2
2
(b – a)2 = (30 – 15)2 = 4.33 minutes
s=
12
12
Find the probability that the anesthetic
takes between 20 and 25 minutes.
P(c < X < d) = (d – c)/(b – a)
P(20 < X < 25) = (25 – 20)/(30 – 15)
= 5/15 = 0.3333 or 33.33%
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Uniform Continuous Distribution
Example: Anesthesia Effectiveness
P(20 < X < 25)
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Normal Distribution
Characteristics of the Normal Distribution
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7-14
Normal or Gaussian distribution was
named for German mathematician Karl
Gauss (1777 – 1855).
Defined by two parameters, m and s
Denoted N(m, s)
Domain is – < X < + 
Almost all area under the normal curve is
included in the range m – 3s < X < m + 3s
Normal Distribution
Characteristics of the Normal Distribution
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Normal Distribution
Characteristics of the Normal Distribution
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Normal PDF f(x) reaches a maximum at m
and has points of inflection at m + s
Bell-shaped curve
Figure 7.9
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Normal Distribution
Characteristics of the Normal Distribution
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Normal CDF
Figure 7.9
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Normal Distribution
Characteristics of the Normal Distribution
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All normal distributions have the same
shape but differ in the axis scales.
m = 42.70mm
s = 0.01mm
Diameters of golf balls
7-18
m = 70
s = 10
CPA Exam Scores
Normal Distribution
What is Normal?
A normal random variable should:
• Be measured on a continuous scale.
• Possess clear central tendency.
• Have only one peak (unimodal).
• Exhibit tapering tails.
• Be symmetric about the mean (equal tails).
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Standard Normal Distribution
Characteristics of the Standard Normal
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Since for every value of m and s, there is a
different normal distribution, we transform
a normal random variable to a standard
normal distribution with m = 0 and s = 1
using the formula:
x
–
m
z=
s
Denoted N(0,1)
Standard Normal Distribution
Characteristics of the Standard Normal
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Standard Normal Distribution
Characteristics of the Standard Normal
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Standard normal PDF f(x) reaches a
maximum at 0 and has points of inflection
at +1.
Shape is
unaffected by
the
transformation.
It is still a bellshaped curve.
Figure 7.11
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Standard Normal Distribution
Characteristics of the Standard Normal
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Standard normal CDF
Figure 7.11
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Standard Normal Distribution
Characteristics of the Standard Normal
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A common scale from -3 to +3 is used.
Entire area under the curve is unity.
The probability of an event P(z1 < Z < z2) is
a definite integral of f(z).
However, standard normal tables or Excel
functions can be used to find the desired
probabilities.
Standard Normal Distribution
Normal Areas from Appendix C-1
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Appendix C-1 allows you to find the area
under the curve from 0 to z.
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For example, find
P(0 < Z < 1.96):
Figure 7.12
Standard Normal Distribution
Table 7.4
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Standard Normal Distribution
Normal Areas from Appendix C-1
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Now find P(Z < 1.96):
.5000
.5000 - .4750 = .0250
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Standard Normal Distribution
Normal Areas from Appendix C-1
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Now find P(-1.96 < Z < 1.96).
Due to symmetry, P(-1.96 < Z) is the same
as P(Z < 1.96).
.9500
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So, P(-1.96 < Z < 1.96) = .4750 + .4750 =
.9500 or 95% of the area under the curve.
Standard Normal Distribution
Basis for the Empirical Rule
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Approximately 95% of the area under the
curve is between + 2s
Approximately 99.7% of the area under the
curve is between + 3s
68.26%
Figure 7.15
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Standard Normal Distribution
Normal Areas from Appendix C-2
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Appendix C-2 allows you to find the area
under the curve from the left of z (similar to
Excel).
For example,
.9500
P(Z < 1.96)
P(Z < -1.96)
Figure 7.15
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P(-1.96 < Z < 1.96)
Standard Normal Distribution
Table 7.5
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Standard Normal Distribution
Normal Areas from Appendices C-1 or C-2
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Appendices C-1 and C-2 yield identical results.
Use whichever table is easiest.
Finding z for a Given Area
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Appendices C-1 and C-2 be used to find the
z-value corresponding to a given probability.
For example, what z-value defines the top 1% of a
normal distribution?
This implies that 49% of the area lies between 0
and z.
Standard Normal Distribution
Finding z for a Given Area
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Look for an
area of .4900
in Appendix
C-1:
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Without
interpolation,
the closest we
can get is
z = 2.33
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Standard Normal Distribution
Finding z for a Given Area
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Some important Normal areas:
Table 7.7
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Standard Normal Distribution
Finding Normal Areas with Excel
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Standard Normal Distribution
Finding Normal Areas with Excel
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Standard Normal Distribution
Finding Normal Areas with Excel
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Standard Normal Distribution
Finding Normal Areas with Excel
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Standard Normal Distribution
Finding Areas by using Standardized Variables
• Suppose John took an economics exam and
scored 86 points. The class mean was 75
with a standard deviation of 7. What
percentile is John in (i.e., find P(X < 86)?
86
–
75
x
–
m
= 11/7 = 1.57
zJohn =
=
7
s
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So John’s score is 1.57 standard deviations
about the mean.
Standard Normal Distribution
Finding Areas by using Standardized Variables
• P(X < 86) = P(Z < 1.57) = .9418
(from Appendix C-2)
• So, John is approximately in the 94th
percentile.
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Figure 7.18
Standard Normal Distribution
Inverse Normal
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You can manipulate the transformation
formula to find the normal percentile values
(e.g., 5th, 10th, 25th, etc.): x = m + zs
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Here are some common percentiles
Standard Normal Distribution
Using Excel Without Standardizing
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Excel’s NORMDIST and NORMINV function allow
you to evaluate areas without standardizing.
For example, let m = 2.040 cm and s = .001 cm, what
is the probability that a given steel bearing will have
a diameter between 2.039 and 2.042cm?
In other words, P(2.039 < X < 2.042)
Excel only gives left tail areas, so break the formula
into two, find P(X < 2.039) and
P(X < 2.042), then subtract them to find the desired
probability:
Standard Normal Distribution
Using Excel Without Standardizing
P(X < 2.042) = .9773
P(X < 2.039) = .1587
P(2.039 < X < 2.042) = .9773 - .1587 = .8186 or 81.9%
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Normal Approximation to the
Binomial
When is Approximation Needed?
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Binomial probabilities are difficult to
calculate when n is large.
Use a normal approximation to the binomial.
As n becomes large, the binomial bars
become more continuous and smooth.
Normal Approximation to the
Binomial
When is Approximation Needed?
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Rule of thumb: when np > 10 and n(1-p) >
10, then it is appropriate to use the normal
approximation to the binomial.
In this case, the binomial mean and
standard deviation will be equal to the
normal m and s, respectively.
m = np
s = np(1-p)
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Normal Approximation to the
Binomial
Example Coin Flips
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If we were to flip a coin n = 32 times and
p = .50, are the requirements for a normal
approximation to the binomial met?
Are np > 10 and n(1-p) > 10?
np = 32 x .50 = 16
n(1-p) = 32 x (1 - .50) = 16
So, a normal approximation can be used.
When translating a discrete scale into a
continuous scale, care must be taken about
individual points.
Normal Approximation to the
Binomial
Example Coin Flips
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For example, find the probability of more
than 17 heads in 32 flips of a fair coin.
This can be written
as P(X > 18).
However, “more
than 17” actually
falls between 17
and 18 on a discrete
scale.
Figure 7.26
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Normal Approximation to the
Binomial
Example Coin Flips
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Since the cutoff point for “more than 17” is
halfway between 17 and 18, we add 0.5 to
the lower limit and find P(X > 17.5).
This addition to X is called the Continuity
Correction.
At this point, the problem can be completed
as any normal distribution problem.
Normal Approximation to the
Binomial
Continuity Correction
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The table below shows some events and
their cutoff point for the normal
approximation.
Normal Approximation to the Poisson
When is Approximation Needed?
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The normal approximation to the Poisson
works best when l is large (e.g., when l
exceeds the values in Appendix B).
Set the normal m and s equal to the Poisson
mean and standard deviation.
m=l
s= l
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Normal Approximation to the Poisson
Example Utility Bills
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On Wednesday between 10A.M. and noon
customer billing inquiries arrive at a mean
rate of 42 inquiries per hour at Consumers
Energy. What is the probability of receiving
more than 50 calls?
l = 42 which is too big to use the Poisson
table.
Use the normal approximation with
m = l = 42
s = l = 42 = 6.48074
Normal Approximation to the Poisson
Example Utility Bills
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To find P(X > 50) calls, use the continuitycorrected cutoff point halfway between 50
and 51 (i.e., X = 50.5).
At this point, the problem can be completed
as any normal distribution problem.
Exponential Distribution
Characteristics of the Exponential Distribution
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If events per unit of time follow a Poisson
distribution, the waiting time until the next
event follows the Exponential distribution.
Waiting time until the next event is a
continuous variable.
Exponential Distribution
Characteristics of the Exponential Distribution
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Exponential Distribution
Characteristics of the Exponential Distribution
Probability of waiting more than x
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Probability of waiting less than x
Exponential Distribution
Example Customer Waiting Time
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Between 2P.M. and 4P.M. on Wednesday, patient
insurance inquiries arrive at Blue Choice insurance
at a mean rate of 2.2 calls per minute.
What is the probability of waiting more than 30
seconds (i.e., 0.50 minutes) for the next call?
Set l = 2.2 events/min and x = 0.50 min
P(X > 0.50) = e–lx = e–(2.2)(0.5) = .3329
or 33.29% chance of waiting more than 30 seconds
for the next call.
Exponential Distribution
Example Customer Waiting Time
P(X > 0.50)
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P(X < 0.50)
Exponential Distribution
Inverse Exponential
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If the mean arrival rate is 2.2 calls per
minute, we want the 90th percentile for
waiting time (the top 10% of waiting time).
Find the x-value
that defines the
upper 10%.
Exponential Distribution
Inverse Exponential
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P(X < x) = .90 or P(X > x) = .10
So, e–lx = .10
-lx = ln(.10)
= -2.302585
x = 2.302585/l
= 2.302585/2.2
= 1.0466 min.
90% of the calls will arrive within 1.0466
minutes (62.8 seconds).
Exponential Distribution
Inverse Exponential
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Quartiles for Exponential with l = 2.2
Table 7.11
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Exponential Distribution
Mean Time Between Events
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Exponential waiting times are described as
Mean time between events (MTBE) = 1/l
1/MTBE = l = mean events per unit of time
In a hospital, if an event is patient arrivals in
an ER, and the MTBE is 20 minutes, then
l = 1/20 = 0.05 arrivals per minute (or
3/hour).
Exponential Distribution
Using Excel
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In Excel, use =EXPONDIST(x,l,1) to return
the left-tail area P(X < x).
Relation Between Exponential and Poisson
Table 7.12
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Triangular Distribution
Characteristics of the Triangular Distribution
• A simple distribution that can be symmetric
or skewed.
• Ranges from a to b and has a mode or
“peak” at c
• Denoted T(a,b,c)
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Triangular Distribution
Characteristics of the Triangular Distribution
Table 7.13
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Triangular Distribution
Special Cases: Symmetric Triangular
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You can easily generate random triangular data
T(0, 1, 2) in Excel by summing RAND()+RAND().
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The triangular distribution T(−2.45, 0, +2.45) closely
resembles a standard normal distribution N(0, 1).
Continuous Distributions Compared
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Applied Statistics in
Business & Economics
End of Chapter 7
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