Download Chapter 5: Some Discrete Probability Distributions

6.1
Chapter 6: Some Continuous Probability Distributions
Take Sample
Sample
Inference
Population
Again, PDFs are population quantities which gives us
information about the distribution of items in the population.
There are many PDFs where are used to understand
probabilities associated with random variables. There are a
few PDFs which are used for multiple real-life situations.
These PDFs are described next. From this chapter, it is
important to learn the following:
 What are these PDFs which can be used for multiple
situations
 When can these PDFs be used
 The means and variances for random variables with
these PDFs
All PDFs in this chapter will be for continuous random
variables.
 2005 Christopher R. Bilder
6.2
6.1: Continuous Uniform Distribution
The simplest PDF for continuous random variables is
when the probability of observing a particular range of
values for X is the same for all equal length ranges!
Since the probabilities are the same, this PDF is called
the uniform PDF.
The Uniform PDF – Let X be a random variable on the
interval [A,B]. The uniform PDF is
 1

f(x;A,B)   B  A
 0
for A  x  B
otherwise
Notes:
o We examined this PDF at the beginning of Section 3.3!
o The parameters, A and B, control the location of the
PDF. In general, this is what a graph of the PDF looks
like.
f(x)
f(x;A,B)
1

BA
A
B
 2005 Christopher R. Bilder
x
6.3
o The area under the curve is 1. Since the PDF looks
like a rectangle, we can take baseheight =
(B-A)[1/(B-A)] to find the area is 1.
Example: Uniform distribution with A=1 and B=4
(uniform.xls)
Uniform PDF
0.35
0.3
f(x)
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
x
Areas underneath the curve correspond to probabilities.
For example, P(1<X<3) = 0.67.
How could I find this using calculus?
Note the blue lines on the x-axis should be extended to
the end of the plot.
 2005 Christopher R. Bilder
6.4
Theorem 6.1 – The mean and variance of a random variable
X with a uniform PDF are
2
A B
(B

A)
  E(X) 
and 2  Var(X) 
2
12
Proving these are homework!
 2005 Christopher R. Bilder
6.5
6.2: Normal Distribution
This is the main PDF that we will be using since it occurs
in many applications.
Normal PDF – Let X be a random variable with mean E(X)=
and Var(X)=2. The normal PDF is
f(x; , ) 
1
 2

e
( x  )2
2 2
for -  x  
Notes:
 The parameters,  and , control the location and scale
of the distribution, respectively. These are the
population mean and standard deviation! Thus, a nice
simplification with the normal PDF is that the mean and
standard deviation can be represented easily as
parameters in the function.
 In most realistic applications,  and  will not be known
and we will need to estimate them. How to do this will
be discussed in future chapters.
 The book denotes f(x;,) by n(x;,).
 Terminology: Suppose X is a random variable with a
normal PDF. One can shorten how this is said by
saying X is a normal random variable.
 In general, this is what a graph of the distribution looks
like.
 2005 Christopher R. Bilder
6.6
1
 2
f(x)
f(x) 
x

o The curve graphed are (x, f(x)) connected points.
o The PDF is centered at  (symmetric about ). Thus,
P(X>) = P(X<) = 0.5. The parameter  is often
called a location parameter since it gives the central
location of the PDF.
o The area under the curve is 1.
o The left and right sides of the curve extend out to -
and + without touching the x-axis (although it will get
very close). Note the plot above may be a little
misleading with respect to this. The left and right sides
of the PDFs are often called the “tails” of the PDF.
o  controls the scale of the PDF. The larger , the
more spread out the PDF (large variability). The
smaller , the less spread out the PDF (small
variability). Below are three normal PDFs
demonstrating this.
 2005 Christopher R. Bilder
6.7
Normal PDF Example
0.7
0.6
0.5
f(x)
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
x (MPG)
24.3 & 0.6
24.3 & 1.3
23.1 & 0.6
 A VERY IMPORTANT specific case of a normal PDF is
the standard normal PDF. This PDF has =0 and =1.
Therefore,
x2
1 2
f(x) 
e for -<x<
2
Typically, “Z” is used instead of “X” to denote a
standard normal random variable. This will be
discussed more later.
 2005 Christopher R. Bilder
6.8

 Showing  f(x)dx =1 is not as easy as it was in
Chapter 3. The proof involves making a transformation
to polar coordinates. Pages 104-5 of Casella and
Berger’s (1990) textbook shows the proof (this book is
used for STAT 882).
Example: Interactive normal PDFs (normal_dist.xls)
This file is constructed to help you visualize the normal
probability distribution. For example, below is the
normal PDF for =50 and =3.
 2005 Christopher R. Bilder
6.9
Experiment on your own using different values of  and
 to see changes in the distribution. Make sure you
understand the following:
 What happens when  is increased or decreased?
 What happens when  is increased or decreased?
 Where is the highest point on the distribution? What is
this highest point?
Also in the file are examples of how to use the
NORMDIST( ) and NORMINV( ) Excel functions which
be discussed in detail in Section 6.3.
Below is the proof showing that E(X) = . A similar proof can
be done to show Var(X) = 2 (see p. 146 of the book).
E(X) 
1

 xe
 2 

(x  )2
2 2
dx
 2005 Christopher R. Bilder
6.10
 2005 Christopher R. Bilder
6.11
6.3-6.4: Areas Under the Normal Curve and Applications
of the Normal Distribution
Example: Grand Am (grand_am_normal.xls)
Suppose that it is reasonable to assume a Grand Am’s
MPG has a normal PDF with a mean MPG of =24.3
and a standard deviation of =0.6. Let X denote the
MPG for one tank of gas. Answer the following
questions.
1) Find the probability that a randomly selected Grand
Am gets less than 23 MPG for one tank of gas.
We need to find P(X<23) = F(23). This is the area to
the left of the red line underneath the PDF.
Grand Am Normal PDF Example
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
x (MPG)
24.3 & 0.6
 2005 Christopher R. Bilder
27
28
29
30
6.12
This probability can be found by:
23


1
0.6 2

e
( x  24.3)2
2(0.6)2
dx .
Using Maple without evaluating at the limits of
integration, we get:
> assume(sigma>0);
> f:=1/(sqrt(2*Pi)*0.6)*exp((x-24.3)^2/(2*0.6^2));
f := .8333333335
2 e
2
( 1.388888889( x~  24.3 ) )

> int(f,x);
.4999999998 erf( 1.178511302 x~  28.63782464 )
Notice that a capital P is used in the Pi function. See
http://mathworld.wolfram.com/Erf.html for more
information on the erf() function.
Using Maple with the limits of integration, we get:
> int(f,x=-infinity..23);
.0151301397
where it uses numerical approximations for the last
integral.
 2005 Christopher R. Bilder
6.13
To make finding probabilities easier, many software
packages (and calculators) have special functions
which do the integration for X in some interval. In
Excel, the NORMDIST(x, , , TRUE) function finds
F(x) for a normal random variable with mean  and
standard deviation .
For this example, use
NORMDIST(23,24.3,0.6,TRUE)
This results in 0.0151.
Chris Malone’s Excel Instructions website contains
help for this function at
http://www.statsclass.com/excel/tables/prob_values.ht
ml#prob_n. The web page shows another way to use
the function through a window based format.
Side note: To find the probability in Maple using its
specialized functions, you can use the following code:
> with(stats);
[ anova, describe , fit , importdata, random, statevalf, statplots, transform ]
> statevalf[cdf,normald[24.3,0.6]](23);
.01513014001
2) Suppose  is increased to =1.3. What do you
expect to happen to P(X<23)?
 2005 Christopher R. Bilder
6.14
The Excel function is NORMDIST(23,24.3,1.3,TRUE)
Grand Am Normal PDF Example
0.7
0.6
0.5
f(x)
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
x (MPG)
24.3 & 1.3
3) Suppose =0.6 again, but  is decreased to =23.1.
What do you expect to happen to P(X<23)?
The Excel function is NORMDIST(23,23.1,0.6,TRUE)
 2005 Christopher R. Bilder
6.15
Grand Am Normal PDF Example
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
x (MPG)
23.1 & 0.6
Below is a nice comparative graph for the 3 examples
above.
Grand Am Normal PDF Example
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
x (MPG)
24.3 & 0.6
24.3 & 1.3
 2005 Christopher R. Bilder
23.1 & 0.6
30
6.16
4) Suppose =0.6 and =24.3 again. What is
P(23<X<25)?
The probability needs to be broken up since the
NORMDIST( ) function only finds probabilities in the
form of F(x).
P(23<X<25) = P(X<25) – P(X<23) = F(25) – F(23).
This can be found with the Excel functions:
NORMDIST(25,24.3,0.6,TRUE)-NORMDIST(23,24.3,0.6,TRUE)
The probability is 0.8632.
Grand Am Normal Probability Distribution Example for
 =24.3,  =0.6
0.7
0.6
f(X)
0.5
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
X
P(23<X<25)
5) Suppose =0.6 and =24.3 again. What is P(X>23)?
 2005 Christopher R. Bilder
6.17
6) Suppose =0.6 and =24.3 again. What is P(X<23 or
X>25)?
7) What MPG is at least required for a car to be in the
top 5% of all Grand Ams? Suppose =0.6 and
=24.3 again.
This problem requires going in the opposite direction.
We are now given a probability and need to find the
corresponding “x” that works for P(X>x)=0.05. In
terms of integration, we are trying to find x in the
equation below:

0.05  
x

1
0.6 2
(y  24.3)2
e
2(0.6)2
dy
Equivalently,
x
0.95  


1
0.6 2
e
(y  24.3)2
2(0.6)2
dy
 2005 Christopher R. Bilder
6.18
Notice the limits of integration used are in terms of y.
This is done to avoid confusion of integrating from
“x=x to ”.
The x value can be found by using Excel’s
NORMINV(area,, ) function where area=P(X<x).
Be careful! Notice that the area is for P(X<x),
not P(X>x).
The x value can be found with the Excel function:
NORMINV(0.95,24.3,0.6)
Therefore, P(X>25.29)=0.05.
See
http://www.statsclass.com/excel/tables/crit_values.html#
crit_n for more information about this function. Note that
 2005 Christopher R. Bilder
6.19
we will eventually use these types of values as “critical
points” in hypothesis testing.
Here are other ways to find the value of x in Maple:
> with(stats);
[ anova, describe , fit , importdata, random, statevalf, statplots, transform ]
> statevalf[icdf,normald[24.3,0.6]](0.95);
25.28691218
> f:=1/(sqrt(2*Pi)*0.6)*exp((y-mu)^2/(2*sigma^2));
f := .8333333335
2e
2

  1/2 ( y   ) 



2 




> solve(0.95 = eval(int(f, y =
-infinity..x), [mu=24.3, sigma=0.6], x);
25.28691217
Example: Grading (grade_bell.xls)
Suppose the set of test #2 grades in the class has a
normal distribution with =73% and =8%. Let X be a
student’s grade. Answer the following.
 2005 Christopher R. Bilder
6.20
1) What is the probability that a randomly chosen student
in the class received a grade of 90% or better?
Grading Normal PDF Example
0.14
0.12
f(x)
0.1
0.08
0.06
0.04
0.02
0
50
55
60
65
70
75
80
85
90
95
100
x (Grade)
Let X be a normal random 
variable
with =73% and
   =8%. Find P(X>90). Thus, we need to find

P(X  90)  
90
1
8 2

e
( x  73)2
2(8)2
dx
The Excel function is 1-NORMDIST(90,73,8,TRUE)
and the answer is 0.0168.
2) What percentage of students scored between a 70%
and 90%?
 2005 Christopher R. Bilder
6.21
Grading Normal PDF Example
0.14
0.12
f(x)
0.1
0.08
0.06
0.04
0.02
0
50
55
60
65
70
75
80
85
90
95
100
x (Grade)
The Excel function is NORMDIST(90,73,8,TRUE)  
NORMDIST(70,73,8,TRUE) and the answer is 0.6294.
3) Suppose that your instructor curves the test #2 grades
and that ONLY the top 10% of test scores receive A’s.
Would a student be better off with a test #2 grade of
81% (still with =73% and =8%) or a grade of 68%
on a different test #2 that has a normal distribution
with =62% and =3%?
 2005 Christopher R. Bilder
6.22
Grading Normal PDF Example
0.14
0.12
f(x)
0.1
0.08
0.06
0.04
0.02
0
50
55
60
65
70
75
80
85
90
95
100
x (Grade)
73 & 8
62 & 3
Find the top 10% of the scores for each situation.
For =73% and =8%, find x for P(X>x)=0.10.
The Excel function to find this is NORMINV(0.9,73,8)
and the answer is 83.25.
For =62% and =3%, find x for P(X>x)=0.10.
The Excel function to find this is NORMINV(0.9,62,3)
and the answer is 65.84.
 2005 Christopher R. Bilder
6.23
A student would prefer the second test since an A
would be received.
Rule of thumb for the number of standard deviations all data
lies from its mean:
In Chapter 4, we discussed that approximately all data
lies within 2 or 3. We also discussed the more
formal expression of this using Chebyshev’s Rule.
Examine what happens if our data comes from a normal
PDF. The end result is what is often called the Empirical
Rule.
Example: Standard normal distribution template
(stand_norm_prob.xls)
Let Z be a random variable with a standard normal PDF.
Thus, =0 and =1. All of these results apply for 0
and 1 also. Below are three screen captures that
show a standard normal PDF. The distributions show
the area between 1, 2, and 3 standard deviations of the
mean.
 2005 Christopher R. Bilder
6.24
Notice how large the probability is that Z is between 2 or
3 standard deviations from the mean!
 2005 Christopher R. Bilder
6.25
Reminder about P(X=x)=0
What is P(X=x)? It is 0 since X is a continuous random
variable. To see why this is true, consider this proof by
example.
Let Z be a standard normal random variable. The
following table of probabilities can then be constructed.
P(0.95<Z<1.05)
P(0.98<Z<1.02)
P(0.99<Z<1.01)
P(0.99<Z<1)
P(1<Z<1.01)

P(Z=1)
Probability
0.0242
0.0096
0.0049
0.0042
0.0025

0
Notice the probability gets smaller and smaller as the
interval gets smaller. Eventually, the probability will
become 0.
b
Remember that P(a  X  b)   f(x)dx for some PDF f(x)
a
where X is a continuous random variable. When a=b,
a
then P(a  X  a)   f(x)dx  0 .
a
Standard normal PDF
 2005 Christopher R. Bilder
6.26
Probabilities associated with the standard normal PDF
have been tabled.
Example: Standard normal distribution tables
(stand_norm_table.xls)
Before there were readily assessable software package
or calculators with functions for the normal PDF, people
used tables based on the standard normal PDF in order
to find probabilities associated with ANY normal PDF.
Table A.3 on p.670-1 of the book is one of these tables.
It provides F(z), the CDF for a standard normal random
variable Z. The reason why I am using Z here is
because this is the common practice when discussing
standard normal random variables.
Thus, Table A.3 gives probabilities such as the one
shown below.
 2005 Christopher R. Bilder
6.27
Below is an excerpt from the table contained in
stand_norm_table.xls.
-3.4
-3.3
-3.2
-3.1
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
0.00
0.0003
0.0005
0.0007
0.0010
0.0013
0.0019
0.0026
0.0035
0.0047
0.0062
0.0082
0.0107
0.0139
0.0179
0.0228
0.0287
0.01
0.0003
0.0005
0.0007
0.0009
0.0013
0.0018
0.0025
0.0034
0.0045
0.0060
0.0080
0.0104
0.0136
0.0174
0.0222
0.0281
0.02
0.0003
0.0005
0.0006
0.0009
0.0013
0.0018
0.0024
0.0033
0.0044
0.0059
0.0078
0.0102
0.0132
0.0170
0.0217
0.0274
0.03
0.0003
0.0004
0.0006
0.0009
0.0012
0.0017
0.0023
0.0032
0.0043
0.0057
0.0075
0.0099
0.0129
0.0166
0.0212
0.0268
0.04
0.0003
0.0004
0.0006
0.0008
0.0012
0.0016
0.0023
0.0031
0.0041
0.0055
0.0073
0.0096
0.0125
0.0162
0.0207
0.0262
0.05
0.0003
0.0004
0.0006
0.0008
0.0011
0.0016
0.0022
0.0030
0.0040
0.0054
0.0071
0.0094
0.0122
0.0158
0.0202
0.0256
0.06
0.0003
0.0004
0.0006
0.0008
0.0011
0.0015
0.0021
0.0029
0.0039
0.0052
0.0069
0.0091
0.0119
0.0154
0.0197
0.0250
0.07
0.0003
0.0004
0.0005
0.0008
0.0011
0.0015
0.0021
0.0028
0.0038
0.0051
0.0068
0.0089
0.0116
0.0150
0.0192
0.0244
0.08
0.0003
0.0004
0.0005
0.0007
0.0010
0.0014
0.0020
0.0027
0.0037
0.0049
0.0066
0.0087
0.0113
0.0146
0.0188
0.0239
0.09
0.0002
0.0003
0.0005
0.0007
0.0010
0.0014
0.0019
0.0026
0.0036
0.0048
0.0064
0.0084
0.0110
0.0143
0.0183
0.0233
This table uses the NORMDIST(z, 0, 1, TRUE) function
to find P(Z<z). For example,
 2005 Christopher R. Bilder
6.28
P(Z<-3.41) = 0.0003,
P(Z<-3.03) = 0.0012,
P(Z<-2.57) = 0.0051,

Why are we concerned with this table of standard normal
probabilities?
A simple transformation can be made from ANY normal
PDF to the standard normal PDF using the following
formula:
Z
X

where X is a normal random variable with mean  and
standard deviation  and Z is a standard normal random
variable with mean 0 and standard deviation 1.
Therefore, using this one table, we can find all normal
PDF probabilities WITHOUT Excel or other means.
Example: Grand Am (grand_am_normal.xls)
Suppose that it is reasonable to assume a Grand Am’s
MPG has a normal PDF with a mean MPG of =24.3
and a standard deviation of =0.6. Let X denote the
 2005 Christopher R. Bilder
6.29
MPG for one tank of gas. Answer the following
questions.
1) Find the probability that a randomly selected Grand
Am gets less than 23 MPG for one tank of gas.
We need to find P(X<23) = F(23). This is the area to
the left of the red line underneath the PDF.
Grand Am Normal PDF Example
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
x (MPG)
24.3 & 0.6
The function, NORMDIST(23,24.3,0.6,TRUE), can be
used in Excel to find the probability to be 0.0151.
Using the tables, P(X<23)
 X   23  24.3 

= P
 = P(Z<-2.1667)  P(Z<-2.17)

0.6


= 0.0150.
 2005 Christopher R. Bilder
6.30
2) Suppose  is increased to =1.3. What do you
expect to happen to P(X<23)?
The function, NORMDIST(23,24.3,1.3,TRUE), can be
used to find the probability to be 0.1587.
Grand Am Normal PDF Example
0.7
0.6
0.5
f(x)
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
x (MPG)
24.3 & 1.3
Using the tables, P(X<23)
 X   23  24.3 

= P
= P(Z<-1) = 0.1587.

1.3
 

3) Suppose =0.6 again, but  is decreased to =23.1.
What do you expect to happen to P(X<23)?
 2005 Christopher R. Bilder
30
6.31
The function, NORMDIST(23,23.1,0.6,TRUE), can be
used to find the probability to be 0.4338.
Using the tables, P(X<23)
 X   23  23.1
= P
= P(Z<-0.1667)  P(Z<-0.17)


0.6 
 
= 0.4325
4) Suppose =0.6 and =24.3 again. What is
P(23<X<25)?
The function,
NORMDIST(25,24.3,0.6,TRUE)-NORMDIST(23,24.3,0.6,TRUE)
can be used to find the probability to be 0.8632
Using the tables, P(23<X<25)
 23  24.3 X   25  24.3 


= P
0.6

0.6 

= P(-2.1667<Z<1.1667)
 P(-2.17<Z<1.17)
= P(Z<1.17) – P(Z<-2.17)
= 0.87900 – 0.01500
= 0.86400
5) Suppose =0.6 and =24.3 again. What is P(X>23)?
6) Suppose =0.6 and =24.3 again. What is P(X<23 or
X>25)?
 2005 Christopher R. Bilder
6.32
7) What is MPG required for a car to be in the top 5% of
all Grand Ams? Suppose =0.6 and =24.3 again.
The x value for P(X>x)=0.05 was found with the
Excel function =NORMINV(0.95,24.3,0.6). This
produced P(X>25.29)=0.05.
Using the tables, P(X>x)
 X   x  24.3 

 P  Z  z   0.05 .
=P 

0.6 
 
Note that P(Z<z)=0.95 produces z1.64.
 X   x  24.3 

 P  Z  1.64   0.05 and
Then P 

0.6 
 
x  24.3
 1.64 . Thus, z=25.284. Therefore,
0.6
P(X>25.284)0.05.
 2005 Christopher R. Bilder
6.33
Observing a sample from a population characterized by a
normal PDF
Suppose a population can be characterized by a normal
PDF. What characteristics would you expect for a
sample taken from that population?
Example: MPG (gen_norm.xls)
MPG example from before: X is a normal random
variable with =E(X)=24.3 and = Var(X) =0.6.
Suppose 1,000 different x’s are observed. In other
words, a sample of 1,000 is taken from the population
Questions:
1) What would you expect the average value of the
1,000 observed x’s to be approximately?
2) What range would you expect most of the x’s to fall
within?
Observed values of a normal random variable can also
be generated in the same way as what was done in the
Chapters 3 and 5. Excel also has a specific normal PDF
option in the Random Number Generation window. The
file, gen_norm.xls, gives an example of using the
window below. More directions are available at Chris
Malone’s Excel help website at
http://www.statsclass.com/excel/misc/norm_dist.html.
 2005 Christopher R. Bilder
6.34
In this case, 1 variable with 1,000 observed values are
generated. The mean =24.3 and standard deviation
=0.6 are used to coincide with the Grand Am example.
The seed number gives Excel a random place to start
when generating these observed values. I can use this
seed number again and generate the exact same data!
 2005 Christopher R. Bilder
6.35
Below are part of the results.
MPG
population sample
23.74819 mean
24.3
24.32106
23.48401 standard deviation
0.6
0.596285
24.59496
24.19059
24.32339
23.11923
classes
Bin
Frequency
24.47166
22.6
22.6
0
24.62146
22.8
22.8
3
24.72662
23
23
11
24.14857
23.2
23.2
23
23.62592
23.4
23.4
23
23.90335
23.6
23.6
53
24.57064
23.8
23.8
83
24.05448
24
24
110
23.71645
24.2
24.2
113
25.07131
24.4
24.4
121
24.32436
24.6
24.6
111
24.1024
24.8
24.8
136
24.62523
25
25
93
24.28823
25.2
25.2
55
24.7843
25.4
25.4
37
24.04441
25.6
25.6
16
24.54797
25.8
25.8
5
24.26204
26
26
4
23.68477
26.2
26.2
1
24.76348
26.4
26.4
1
24.05258
26.6
26.6
1
24.38633
26.8
26.8
0
23.26503
More
0
23.78286
24.09225
23.61932
23.32867
23.02444
23.91397
24.42597
24.42067

 2005 Christopher R. Bilder
6.36
Notes:
 Notice how close  and  are to the sample mean and
standard deviation. The sample standard deviation is
calculated as
n

 xi  x 
2
where x is the sample mean
n 1
and xi for i=1,…,n is the ith observed value. Explanation
for why this formula was used will be given in Chapter
8.
 Here is an example of how to simulate a sample from a
normal PDF using Maple:
i 1
> randomize(1514);
1514
> data:=stats[random, normald[24.3,
0.6]](100);
data := 25.36372908 , 24.96025314 , 24.44663243 , 25.27318122 , 23.94262355 ,
23.69829609 , 23.96992063 , 23.72400640 , 24.06923492 , 24.38832186 ,
24.54452405 , 23.47219191 , 24.51653894 , 24.22545826 , 24.58063212 ,
24.40056631 , 24.22519976 , 24.73647509 , 23.04956592 , 24.94875357 ,
24.02254401 , 24.35341391 , 24.67885308 , 24.81796173 , 23.60716054 ,
24.15571156 , 24.48549168 , 23.84686372 , 25.62993784 , 24.95907390 ,
24.13187013 , 24.40491872 , 25.04623787 , 23.81147131 , 23.04161664 ,
25.57549338 , 23.34059716 , 24.46719408 , 24.23062843 , 23.80346201 ,
25.20382342 , 23.72508178 , 23.35185260 , 23.99842442 , 24.55421301 ,
24.06936962 , 23.50756715 , 24.22223306 , 24.28139128 , 24.47253728 ,
 2005 Christopher R. Bilder
6.37
24.50969275
24.44507802
25.09714835
25.60826660
24.63990563
24.27232784
24.23063511
23.34198654
25.63674860
23.65460645
, 25.31179898
, 24.28610049
, 25.12040542
, 24.19990788
, 24.74360918
, 23.27915406
, 24.06653251
, 23.30621749
, 24.48445061
, 24.16122685
, 24.30883191
, 24.04085590
, 24.69746294
, 25.02525917
, 23.45013063
, 25.15368420
, 24.43778592
, 24.58547842
, 24.56049376
, 24.74861908
, 24.39745116
, 25.13232101
, 24.51272238
, 24.41097845
, 24.52780462
, 24.38724182
, 24.04812858
, 24.40825270
, 23.33174552
, 24.58956375
, 24.34240361
, 24.66322075
, 23.75350627
, 24.17714648
, 24.47851759
, 23.47378351
, 25.20231330
, 23.90859335
, 24.26972911
, 24.41964871
,
,
,
,
,
,
,
,
,
> evalf(stats[describe,mean]([data]),4);
24.30
> evalf(stats[describe,
standarddeviation]([data]),4);
.5871
 Page 6.35 of the notes shows one possible frequency
distribution for the sample. This gives information
about how often observed values fell into chosen
classes. In Excel, I originally entered in the values in
the “classes” column. Through performing a few steps,
Excel automatically generates a frequency distribution.
One needs to be VERY careful with interpreting what
Excel gives. Below is another representation of it:
classes
Frequency
22.6
0
>22.6 and 22.8
3
>22.8 and 23
11
>23 and 23.2
23
>23.2 and 23.4
23
>23.4 and 23.6
53
 2005 Christopher R. Bilder
6.38
classes
Frequency
>23.6 and 23.8
83
>23.8 and 24
110
>24 and 24.2
113
>24.2 and 24.4
121
>24.4 and 24.6
111
>24.6 and 24.8
136
>24.8 and 25
93
>25 and 25.2
55
>25.2 and 25.4
37
>25.4 and 25.6
16
>25.6 and 25.8
5
>25.8 and 26
4
>26 and 26.2
1
>26.2 and 26.4
1
>26.4 and 26.6
1
>26.6 and 26.8
0
>26.8
0
Thus, 136 sampled values are greater than 24.6 and
less than or equal to 24.8.
Why were these classes chosen? There are more
than one set of classes which can be used. Here
are some guidelines:
a) Find the minimum and maximum observed
values. You can use the MIN() and MAX()
functions in Excel to do this.
b) Choose classes which are of equal size.
c) Choose the classes between the minimum and
maximum values which make sense relative to
 2005 Christopher R. Bilder
6.39
the data set. You may need to choose a few
different ones until you think the frequency
distribution represents the data well.
d) Note that 1, 2, or 3 classes do not work!
 The frequency distribution is often plotted. This plot is
called a histogram. Below is the histogram created by
Excel.
Histogram of 1,000 MPG observed values
160
140
Frequency
120
100
80
60
40
20
25
.4
25
.8
26
.2
26
.6
M
or
e
25
23
.4
23
.8
24
.2
24
.6
23
22
.6
0
x = MPG
Does the histogram have a similar shape to the normal
PDF with =24.3 and =0.6? If so, a normal PDF
approximation to the distribution of MPG would be
appropriate.
 2005 Christopher R. Bilder
6.40
Grand Am Normal PDF Example
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
x (MPG)
24.3 & 0.6
 Below is an outline of the steps to find the frequency
distribution and histogram for this example. General
information about how to find a frequency distribution
and histogram are available at
http://www.statsclass.com/excel/graphs/histogram.html.
1)Find the minimum and maximum values.
min
22.71475
max
26.46672
min
max
=MIN(B10:B1009) =MAX(B10:B1009)
2)In an empty area in the spreadsheet, create a column of
classes.
classes
22.6
22.8
23
 2005 Christopher R. Bilder
6.41
23.2
23.4
23.6
23.8
24
24.2
24.4
24.6
24.8
25
25.2
25.4
25.6
25.8
26
26.2
26.4
26.6
26.8
3)Select TOOLS > DATA ANALYSIS from the main Excel
menu bar.
4)Select HISTOGRAM and OK from the DATA ANALYSIS
window.
 2005 Christopher R. Bilder
6.42
5)The HISTOGRAM window will then appear. In the
window, do the following:
a)Input the cell range of the 1,000 observed values in the
INPUT RANGE.
b)Input the cell range of the classes into the BIN RANGE.
c) Select an OUTPUT RANGE for the corresponding
frequency distribution to start at. I usually specify the
first cell to the right of my classes.
d)Select the CHART OUTPUT option to have a histogram
created.
e)Select OK to have the frequency distribution and the
histogram created!
Below is what my spreadsheet looks like immediately
after OK is selected.
 2005 Christopher R. Bilder
6.43
6)Edit the histogram so that it looks nicer:
Histogram of 1,000 MPG observed values
160
140
Frequency
120
100
80
60
40
20
25
25
.4
25
.8
26
.2
26
.6
M
or
e
23
23
.4
23
.8
24
.2
24
.6
22
.6
0
x = MPG
 Chris Malone has created a spreadsheet called,
data_summary.xls, which can be used when one wants
 2005 Christopher R. Bilder
6.44
to determine if a normal PDF approximation is
appropriate. Below is the spreadsheet result when
used with the 1,000 MPG observed values.
The curve drawn on the histogram is a normal PDF
with mean 24.3211 and standard deviation of
0.5963. Thus, the sample mean and standard
deviation are substituted in for the population mean
and standard deviation. You are not responsible for
knowing how this plot was created, but you will need
to be able to use the spreadsheet. There are also
other summary measures displayed (box plot and
dot plot) which may be discussed in future chapters.
 2005 Christopher R. Bilder
6.45
From the results in data_summary.xls, does a
normal PDF approximation for MPG seem
appropriate? Explain.
 Please see p. 18-19 of the book for more information
about frequency distributions and histograms.
Validity of the normal PDF assumption
All of the probabilities found using the normal PDF
ASSUME the normal PDF is the correct PDF for the
random variable. What if this assumption is incorrect?
The probabilities found using this assumption are
WRONG!
Example: Grand Am (grand_am.xls)
Suppose X really has an uniform distribution with A=22.3
and B=26.3. The P(X<23) is baseheight =
0.70.25=0.175. With the normal assumption of =24.3
and =0.6, the probability was found to be 0.0151
 2005 Christopher R. Bilder
6.46
Grand Am Normal and Uniform PDF Example
0.7
0.6
0.5
f(x)
0.4
0.3
0.2
0.1
0
20
21
22
23
24
25
26
27
28
29
30
x (MPG)
Normal mean=24.3 s.d.=0.6
Uniform A=22.3, B=26.3
How does one know when the normal PDF assumption is
valid?
Rarely, if ever, will it be 100% correct.
If a sample from the population is possible, construct a
histogram of the observed values and check to see if it
has the shape of a normal PDF. In addition, calculate
the sample mean and variance to see if they are close to
the population mean and variance (if they are known). If
the histogram does have a similar shape to a normal
 2005 Christopher R. Bilder
6.47
PDF and the sample and population mean and variance
are about the same (if the population values are known),
then the normal PDF assumption is a reasonable
approximation.
Suppose a histogram was constructed and the data did not
appear to come from a normal or other known PDF. What
can you do?
You can still use the normal PDF with the sample mean
provided the sample size is large enough. The central
limit theorem is used here in order to make a normal
PDF approximation. Chapter 8 talks about this in detail.
 2005 Christopher R. Bilder
6.48
6.5: Normal Approximation to the Binomial
Skip!
Theorem 6.2: Note that if X is a binomial random variable
with mean  = E(X) = np and variance Var(X) = 2 = np(1-p),
then the limiting form of the PDF for
X 
X  np
Z


np(1  p)
as n, is the standard normal PDF. Another way this can
be worded is X can be approximated by a normal random
variable with mean np and variance np(1-p).
Thus as the number of trials increases, Z increasingly
becomes more like a normal random variable.
This information will be used in Section 9.10.
 2005 Christopher R. Bilder
6.49
6.6: Gamma and Exponential Distributions
We have already been using the Gamma and
Exponential PDFs! These PDFs are often used in
survival and reliability analysis. For example, these
PDFs are used for modeling lifetimes of individuals or
manufactured products.
Definition 6.2: The gamma function is defined by

()   x 1e x dx
for >0.
0
Notes:
 When  is a positive integer, () = (-1)!; for
example, (3) = (3-1)! = 2! = 21 = 2
 Through integrating by parts, one can show () =
(-1)(-1)
 (1/2) = 
 In Maple, this is represented by the GAMMA() function
where GAMMA needs to be in capital letters. For
example,
> GAMMA(3);
2
 2005 Christopher R. Bilder
6.50
Gamma PDF: The continuous random variable X has a
gamma PDF, with parameters  and , if its PDF is given by
 1
x 1e x /  for x  0
 
f(x)    ()

0
otherwise

where >0 and >0.
Notes:
 In most realistic applications,  and  will not be known
and we will need to estimate them. How to do this will
be discussed in future chapters.
  controls the shape of the PDF since it mostly
influences the “peakedness” of the PDF.
  controls the scale of the PDF since most of its
influence is for the spread of the PDF.
 In Maple, this can be programmed in as
> assume(x>0);
> assume(alpha>0);
> assume(beta>0);
> about(x, alpha, beta);
Originally x, renamed x~:
is assumed to be: RealRange(Open(0),infinity)
Originally alpha, renamed alpha~:
is assumed to be: RealRange(Open(0),infinity)
Originally beta, renamed beta~:
is assumed to be: RealRange(Open(0),infinity)
 2005 Christopher R. Bilder
6.51
> f(x):=1/(beta^alpha*GAMMA(alpha))*
x^(alpha-1)*exp(-x/beta);
  x~ 


   
(   1 )
x~
e
f( x~ ) :=
 (  )
> simplify(int(f(x),x=0..infinity));
1
There are easier ways to use the gamma PDF in Maple
that will be discussed later.
 Below are a few comparative plots (gamma.xls).
Notice the x- and y-axis scales are fixed for
comparative purposes. Values of X could be greater
than 24!
=1, =1, =1, 2=1
=1, =2, =2, 2=4
Gamma PDF
Gamma PDF
1
1
0.9
0.8
0.7
0.6
0.9
0.8
f(x)
f(x)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10
12
14
16
18
20
22
24
0.5
0.4
0.3
0.2
0.1
0
0
2
x
4
6
8
10
12
x
 2005 Christopher R. Bilder
14
16
18
20
22
24
6.52
=1, =3, =3, 2=9
=2, =1, =2, 2=2
Gamma PDF
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
f(x)
f(x)
Gamma PDF
0
2
4
6
8
10
12
14
16
18
20
22
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
24
0
2
4
6
8
10
x
f(x)
f(x)
6
8
10
12
18
20
22
24
20
22
24
Gamma PDF
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
4
16
=4, =2, =8, 2=16
Gamma PDF
2
14
x
=4, =1, =4, 2=4
0
12
14
16
18
20
22
24
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
x
4
6
8
10
12
x
 2005 Christopher R. Bilder
14
16
18
6.53
=2.5, =2.5, =6.25, 2=15.625
Gamma PDF
1
0.9
f(x)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10
12
14
16
18
20
22
24
x
Questions:
 What happens if  and/or  are increased?
 What happens if  and/or  are decreased?
 Why would someone want to use different values of 
and/or ?
Theorem 6.3: The mean and variance of the gamma PDF
are: E(X) =  =  and Var(X) = 2 = 2.
pf:
1
1  x / 
x
e dx

 ()
1

 0 
x e x /  dx
 ()
(  1)1  1
 x / 

x
e dx

0
1

(  1)
 ()

E(X)  0 x
 2005 Christopher R. Bilder
6.54
1(  1) 
1
 x / 

x
e dx

0

1
 ()
 (  1)
1
Notice that 1
xe x /  is a gamma PDF with
 (  1)
+1 and  as its parameters! Thus,
1

xe x / dx = 1 and
0 1
 (  1)
1(  1)
()
E(X) 

1

  .

 ()
()
A similar proof can be done for the variance.
Maple code,
> E(X):=simplify(int(x*f(x),
x=0..infinity));
E( X ) :=  
> Var(X):=simplify(int((x-E(X))^2*f(x),
x=0..infinity));
Var ( X ) := 2 
Examine what happens to the PDF as values of  and 2
change the gamma PDF plots on the previous pages.
Example: Distribution of lifetimes (gamma_actuary.xls)
 2005 Christopher R. Bilder
6.55
Let X be a random variable denoting the lifetime of a
person in a particular population. An actuary uses the
PDF for X below to model the lifetimes of all people in
this population:
f(x) 
1
 x / 15
x
e
for x>0.
2
15
For this example, =15 and =2.
Gamma PDF for actuary example
0.03
0.025
f(x)
0.02
0.015
0.01
0.005
0
0
25
50
75
100
125
150
x
In Maple, the plot is
> plot(eval(f(x),[alpha=2,beta=15]),
x=0..150, title="Gamma PDF, alpha=2,
beta=15", labels=["x", "f(x)"]);
 2005 Christopher R. Bilder
6.56
This particular PDF may not be realistic for what we
would commonly perceive to be the distribution of
lifetimes in the United States.
Questions:
 What are the mean and variance?
The mean and variance are  =  = 215 = 30 and 2
= 2152 = 450. Thus, one would expect to live 30 years
on average for this population.
 What is the probability a person in the population lives
longer than 80 years?
The probability can be found from
 2005 Christopher R. Bilder
6.57

1
 x / 15
xe
dx . Notice that integration by
2
80 15
parts would be needed here. If the integration was
done in Maple,
P(X>80) = 
> P(X>80):=int(eval(f(x),
[alpha=2,beta=15]), x=80..infinity);
P( 80  X ) :=
19 ( -16/3 )
e
3
> evalf(P(X>80),4);
.03059
Also, note that P(X>80) = 1 - P(X<80) = 1 - F(80).
Thus, the CDF can be used to find the probability. The
GAMMADIST(x, , , TRUE) function in Excel can
simply be used here. Thus,
=1-GAMMADIST(80,2,15,TRUE)
results in a value of 0.0306.
Using the stats package in Maple,
> 1-stats[statevalf,cdf,gamma[2,15]](80);
.0305770166
 What is the median lifetime?
The value c needs to found such that the probability of
living less than c years is 0.5. Then we could use
 2005 Christopher R. Bilder
6.58
1
 x / 15
xe
dx
2
0 15
and solve for c. If the integration and solving was
done in Maple,
c
P(X  c)  0.5  
> solve(int(eval(f(x),[alpha=2,
beta=15]), x=0..c) = 0.5, c);
-11.52058571 , 25.17520485
Of course, the positive value for c would be the
answer. The GAMMAINV(prob., , ) function can
be used in Excel to find c. Thus,
=GAMMAINV(0.5,2,15)
results in c = 25.18.
Using the stats package in Maple,
> stats[statevalf,icdf, gamma[2,15]
](0.5);
25.17520485
There are a few important special cases of the gamma
PDF. One of them is the exponential PDF.
Exponential PDF: The continuous random variable X has an
exponential PDF, with parameter , if its PDF is given by
 2005 Christopher R. Bilder
6.59
 1 x /
 e
f(x)   
 0

where >0.
for x  0
otherwise
Notes:
 This is the gamma PDF with =1.
 In most realistic applications,  will not be known and it
will need to be estimated. How to do this will be
discussed in future chapters.
  controls the scale of the PDF since most of its
influence is for the spread of the PDF. In general, this
is what a plot of the PDF looks like.
1

f(x)
f(x) 
X
0
The height of the curve at a point xo is f(xo ) 
Notice that when xo=0, f(0) 
1  xo 
.
e

1 0 /  1
1
e
  1  since



e0=1.
 2005 Christopher R. Bilder
6.60
Theorem 6.3: The mean and variance for the exponential
PDF are: E(X) =  =  and Var(X) = 2 = 2.
pf: See the Chapter 4 examples with tire tread wear.
Substitute  in for 30. Also, see the proof used with the
gamma PDF earlier.
Example: Tire life (tire_wear.xls from Chapter 3)
The number of miles an automobile tire lasts before it
reaches a critical point in tread wear can be represented
by a PDF. Let X = the number of miles (in thousands)
an automobile is driven before it reaches the critical
tread wear point for one tire. Suppose the PDF for X is
 1  x
 e
f(x)   
 0

for x  0
for x  0
In Chapter 3 and 4, we used =30. Remember that we
found in Chapter 4 that E(X) =  = 30 and Var(X) = 2 =
302!
 2005 Christopher R. Bilder
6.61
In the spreadsheet, different values of  can be entered
into the  cell to see how it affects the PDF. Below is a
screen capture of the spreadsheet.
Note that the line on the plot should extend past
x=225.
Questions:
 What happens if  is increased? Explain why  has
this effect relative to it being called a “scale” parameter,
E(X)=, and Var(X) = 2.
 What happens if  is decreased? Explain why  has
this effect relative to it being called a “scale” parameter,
E(X)=, and Var(X) = 2.
 Why would someone want to use different values of ?
 2005 Christopher R. Bilder
6.62
Find the probability that a random selected tire will last
(will not get to the critical tread wear point) longer than
30,000 miles. In Chapter 3, we found the probability
through integration:
x
 1

P(X  30)  
e 30 dx  e1  0.3679
30 30
Using the relationship between the gamma and
exponential function, we can use the GAMMADIST()
function:
=1-GAMMADIST(30,1,30,TRUE)
Notes:
 Remember that if FALSE is used instead of TRUE in
the function, then f(x) is given as a result (the height of
the curve).
 Excel also has a function specifically for the
Exponential PDF: EXPONDIST(x,1/beta,TRUE) which
finds F(x). Please note that 1/beta corresponds to what
Excel defines as . Thus, Excel uses a PDF of
  e  x
f(x)  
 0
for x  0
otherwise
To find P(X>30), note that P(X>30) = 1 – P(X<30) = 1 –
F(30). In Excel,
 2005 Christopher R. Bilder
6.63
=1-EXPONDIST(30,1/30,TRUE)
To avoid confusion with  = 1/, I recommend using the
GAMMADIST() function instead.
Find the tire wear number of miles such that less than
0.95 of the total number of tires will reach the critical
point. In Chapter 3, we found the value of c as a
solution to
x
c 1

0.95  P(X  c)   e 30 dx
0 30
The value of c was 30ln(20)  89.87. We can use the
relationship between the gamma and exponential PDFs
to find the same answer with
=GAMMAINV(0.95,1,30)
Example: Exponential distribution with =10/3 (exp.xls)
 2005 Christopher R. Bilder
6.64
Exponential PDF
0.35
0.3
f(x)
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
8
10
12
14
16
18
x
To find the probability P(2<X<4), find the area
underneath that part of the plot. Note that P(2<X<4) =
P(X<4) – P(X<2) = F(4) = F(2) since
 2005 Christopher R. Bilder
6.65
=
The Excel functions to find the probability are
GAMMADIST(4,1,10/3,TRUE) –
GAMMADIST(2,1,10/3,TRUE)
and the answer is 0.2476.
Final notes:
o Go back to Chapter 3 and examine
example_sample_tire.xls. Notice how choosing  = 30
results in a very good fit of the PDF to the sampled values
displayed in the histogram!
o If you are an engineering major, I recommend examining
the Weibull PDF in Section 6.10.
o The chi-square PDF in Section 6.8 is an often used PDF
which we will discuss later in the course.
 2005 Christopher R. Bilder