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Chapter 6: Some Continuous Probability Distributions
Inference
Sample
Population
Take Sample
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.
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
for A x B
f(x;A,B) B A
0 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.
A
f(x;A,B)
1
B A
B x
Theorem 6.1 – The mean and variance of a random variable
X with a uniform PDF are
A B
E(X)
and
2
2
(B A)
Var(X)
Proving these are homework!
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
2
2
( x )
2
f(x; , ) e 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.
Normal PDF Example
0
20 21 22 23 24 25 26 27 28 29 30
x (MPG)
f(x)
2 4 .3 & 0 .6 2 4 .3 & 1 .3 2 3 .1 & 0.
A VERY IMPORTANT specific case of a normal PDF is
the standard normal PDF. This PDF has =0 and =1.
Therefore,
2
x
2
f(x) e for - 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.
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?
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.
and a standard deviation of =0.6. Let X denote the
MPG for one tank of gas. Answer the following
questions.
- 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
20 21 22 23 24 25 26 27 28 29 30
x (MPG)
f(x)
2 4 .3 & 0.
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 ]
> statevalfcdf,normald[24.3,0.6];
.
- Suppose is increased to =1.3. What do you
expect to happen to P(X<23)?
The Excel function is NORMDIST(23,24.3,1.3,TRUE)
Grand Am Normal PDF Example
0
20 21 22 23 24 25 26 27 28 29 30
x (MPG)
f(x)
2 4 .3 & 1.
- 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)
- Suppose =0.6 and =24.3 again. What is
P(2323)?
- Suppose =0.6 and =24.3 again. What is P(X<23 or
X>25)?
- 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:
2
2
(y 24.3)
2(0.6)
x
0.05 e dy
Equivalently,
2
2
(y 24.3)
x
2(0.6)
0.95 e dy
Notice the limits of integration used are in terms of y.
This is done to avoid confusion of integrating from
“x=x to ”.
Here are other ways to find the value of x in Maple:
> with(stats);
[ anova , describe , fit , importdata , random , statevalf , statplots , transform ]
> statevalficdf,normald[24.3,0.6];
> f:=1/(sqrt(2Pi)0.6)exp(-*
(y-mu)^2/(2sigma^2));*
f :=.
2 e
1 / 2
( y )
2
2
> solve(0.95 = eval(int(f, y =
-infinity..x), [mu=24.3, sigma=0.6], x);
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.
- What is the probability that a randomly chosen student
in the class received a grade of 90% or better?
Grading Normal PDF Example
0
50 55 60 65 70 75 80 85 90 95 100
x (Grade)
f(x)
Let X be a normal random variable with =73% and
=8%. Find P(X>90). Thus, we need to find
2
2
( x 73)
2(8)
90
P(X 90) e dx
The Excel function is 1-NORMDIST(90,73,8,TRUE)
and the answer is 0.0168.
- What percentage of students scored between a 70%
and 90%?
