Download Chapter Learning Objectives Continuous Random Variables

Chapter 4: Continuous Random
Variables and Probability Distributions
Continuous Random Variables
Probability Distributions and Probability Density Functions
Cumulative Distribution Functions
Mean and Variance of a Continuous Random Variable
Continuous Uniform Distribution
Normal Distribution
Normal Approximation to the Binomial and Poisson
Distributions
4-8 Exponential Distribution
4-9 Erlang and Gamma Distributions
4-10 Weibull Distribution
4-11 Lognormal Distribution
4-12 Beta Distribution
Chapter Learning Objectives
4-1
4-2
4-3
4-4
4-5
4-6
4-7
After careful study of this chapter you should be able to:
1.Determine probabilities from probability density functions
2.Determine probabilities from cumulative distribution functions and cumulative
distribution functions from probability density functions, and the reverse
3.Calculate means and variances for continuous random variables
4.Understand the assumptions for some common continuous probability
distributions
5.Select an appropriate continuous probability distribution to calculate
probabilities in specific applications
6.Calculate probabilities, determine means and variances for some common
continuous probability distributions
7.Standardize normal random variables
8.Use the table for the cumulative distribution function of a standard normal
distribution to calculate probabilities
9.Approximate probabilities for some binomial and Poisson distributions
1
2
Probability Distributions and
Probability Density Functions
Continuous Random Variables
3
4
Probability Density Function
Defined
Probability Density Functions
and Histograms
5
6
Another Probability Density
Function Example
A Probability Density Function
Example
Example 4-1
Example 4-2
Let the continuous random variable X denote the current measured in a thin copper wire in
milliamperes. Assume that the range of X is [0, 20 mA], and assume that the probability
density function of X is f(x) = 0.05 for 0 x 20. What is the probability that a current
measurement is less than 10 milliamperes?
The probability density function is shown in the figure below (it is assumed that f(x) =
0 wherever it is not specifically defined). The probability requested is indicated by the
shaded area in the figure, which can be computed using the equation from slide #5..
7
8
A Cumulative Distribution
Function Example
The Cumulative Distribution
Function in the Continuous Case
Example 4-4
9
Cumulative Distribution Function
Examples
• Here is the cdf for
Example 4-2:
10
Mean and Variance of a
Continuous Random Variable
• Here is the cdf for
Example 4-1:
11
12
The Mean of a Function of a
Continuous Random Variable
Example of Mean and Variance of a
Continuous Random Variable
Example 4-6
13
The Continuous Uniform
Distribution
14
The Mean and Variance of the
Continuous Uniform Distribution
• Use these formula for the situation in
example 4-1 (and 4-6) to verify the results
from slide #13 that when a=0 and b=20:
= 10.0
15
and
2 = 33.33
16
The cdf of the General
Continuous Uniform Distribution
The Normal Distribution
17
18
Well-Known Normal
Distribution Probabilities
Example of Normal
Distribution pdfs
• For any normal random variable:
19
20
How to Use a Table of the cdf of
the Standard Normal Distribution
The Standard Normal
Distribution
Example 4-11
Figure 4-13 Standard normal probability density function.
21
Cumulative Standard Normal Distribution
1.
Example 4-12: Standard Normal
Exercises
P(Z > 1.26) = 0.1038
2.
P(Z < -0.86) = 0.195
3.
P(Z > -1.37) = 0.915
4.
P(-1.25 < 0.37) =
22
0.5387
z
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
0.841345
0.864334
0.884930
0.903199
0.919243
0.933193
0.945201
0.955435
0.964070
0.971283
0.843752
0.866500
0.886860
0.904902
0.920730
0.934478
0.946301
0.956367
0.964852
0.971933
0.846136
0.868643
0.888767
0.906582
0.922196
0.935744
0.947384
0.957284
0.965621
0.972571
0.848495
0.870762
0.890651
0.908241
0.923641
0.936992
0.948449
0.958185
0.966375
0.973197
0.850830
0.872857
0.892512
0.909877
0.925066
0.938220
0.949497
0.959071
0.967116
0.973810
0.853141
0.874928
0.894350
0.911492
0.926471
0.939429
0.950529
0.959941
0.967843
0.974412
0.855428
0.876976
0.896165
0.913085
0.927855
0.940620
0.951543
0.960796
0.968557
0.975002
0.857690
0.878999
0.897958
0.914657
0.929219
0.941792
0.952540
0.961636
0.969258
0.975581
0.859929
0.881000
0.899727
0.916207
0.930563
0.942947
0.953521
0.962462
0.969946
0.976148
0.862143
0.882977
0.901475
0.917736
0.931888
0.944083
0.954486
0.963273
0.970621
0.976705
5.
P(Z -4.6) 0
6.
Find z for P(Z z) =
0.05, z = -1.65
7.
Find z for (-z < Z < z)
= 0.99, z = 2.58
23
Figure 4-14 Graphical displays for
standard normal distributions.
24
An Example on Standardizing a
Normal Random Variable
Using The Standard Normal cdf to Determine
Probabilities for Other Normal Distributions
Example 4-13
25
Another Example (4-14) on Standardizing A
Normal Random Variable
26
The Normal Approximation to
the Binomial Distribution
27
28
An Example (4-17 & 4-18) of a Normal
Approximation to the Binomial
Guidelines for Using the Normal
to Approximate the Binomial or
Hypergeometric
150+.05-160
-0.75)
0.75)
0.773
29
The Normal Approximation to
the Poisson Distribution
30
An Example of a Normal
Approximation to the Poisson
Example 4-20
31
32
An Exponential Distribution
Example (Example 4-21)
The Exponential Distribution
The exponential distribution cdf is:
F(x) = 1-e-Ox
33
34
An Exponential Distribution
Example (Example 4-21 continued)
An Exponential Distribution
Example (Example 4-21 continued)
35
36
An Unusual Property of the
Exponential Distribution
Exponential Application in
Reliability
Lack of Memory Property
• The reliability of electronic components is often modeled
by the exponential distribution. A chip might have mean
time to failure of 40,000 operating hours
• The memoryless property implies that the component does
not wear out – the probability of failure in the next hour is
constant, regardless of the component age
e.g.
In Example 4-21, suppose that there are no log-ons from
12:00 to 12:15; the probability that there are no log-ons from
12:15 to 12:21 is still 0.082. Because we have already been
waiting for 15 minutes, we feel that we are “due.” That is, the
probability of a log-on in the next 6 minutes should be
greater than 0.082. However, for an exponential
distribution this is not true.
• The reliability of mechanical components do have a
memory – the probability of failure in the next hour
increases as the component ages. The Weibull distribution
is used to model this situation
37
38
Example pdfs for the Erlang
Distribution
Correction!
The Erlang Distribution
2
Misc. Erlang pdfs for different values of r (the "order") {O=0.5}
The random variable X that equals the interval length until r
counts occur in a Poisson process with mean > 0 has and
Erlang random variable with parameters and r. The
probability density function of X is
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
for x > 0 and r =1, 2, 3, ….
0.1
0.0
0.0
39
1.0
2.0
r=1
3.0
r=2
4.0
r=4
5.0
r=8
6.0
7.0
8.0
40
Example pdfs for the Gamma
Distribution
The Gamma Distribution
Figure 4-25 Gamma
probability density
functions for selected
values of r and O.
41
42
Example pdfs for the Weibull
Distribution
The Weibull Distribution
Figure 4-26 Weibull
probability density
functions for selected
values of D and E.
43
44
Example pdfs for the Lognormal
Distribution
The Lognormal Distribution
45
Beta Distribution
Beta Shapes are Flexible
• A continuous distribution that is flexible, but bounded over
the [0, 1] interval is useful for probability models.
Examples are:
• Distribution shape
guidelines:
– Proportion of solar radiation absorbed by a material
– Proportion of the max time to complete a task
* D E D 1
E 1
for 0 d x d 1
x 1 x * D ˜ * E is a beta random variable with parameters D ! 0 and E ! 0.
If X has a beta distribution with
parameters and , the mean
and variance of X are:
P
EX V2 V X Figure 4-28 Beta probability
density functions for selected
values of the parameters and – If = , symmetrical
about x = 0.5
– If = = 1, uniform
– If = < 1, symmetric
& U- shaped
– If = > 1, symmetric
& mound-shaped
– If , skewed
The random variable X with probability density function
f x =
46
D
D E
DE
D E D E 1
2
47
48
Extended Range for the Beta
Distribution
The beta random variable X is defined for the [0, 1]
interval. That interval can be changed to [a, b].
Then the random variable W is defined as a linear
function of X:
W = a + (b –a)X
With mean and variance:
E(W) = a + (b –a) E(X)
V(W) = (b - a)2V(X)
49