Transcript EGR252S10Lecture 4 Chapter3 JMB v2 publish

Chapter 3: Random Variables and
Probability Distributions
 Definition and nomenclature
 A random variable is a function that associates a real
number with each element in the sample space.
 We use a capital letter such as X to denote the
random variable.
 We use the small letter such as x for one of its
values.
 Example: Consider a random variable Y which takes
on all values y for which y > 5.
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 1
Defining Probabilities: Random Variables
 Examples:
 Out of 100 heart catheterization procedures
performed at a local hospital each year, the
probability that more than five of them will result in
complications is
P(X > 5)
 Drywall anchors are sold in packs of 50 at the local
hardware store. The probability that no more than 3
will be defective is
P(Y < 3)
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 2
Discrete Random Variables
 Problem 2.53 Page 55 Modified
 Assume someone spends $75 to buy 3 envelopes.
The sample space describing the presence of $10
bills (H) vs bills that are not $10 (N) is:
 S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN,
HHH}
 The random variable associated with this
situation, X, reflects the outcome of the
experiment
 X is the number of envelopes that contain $10
 X = {0, 1, 2, 3}
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 3
Discrete Probability Distributions 1
 The probability that the envelope contains a
$10 bill is 275/500 or .55
 What is the probability that there are no $10
bills in the group?
P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125
P(X = 1) = 3 * (0.55)*(1-0.55)* (1-0.55) = 0.334125
 Why 3 for the X = 1 case?
 Three items in the sample space for X = 1
 NNH NHN HNN
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 4
Discrete Probability Distributions 2
P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125
P(X = 1) = 3*(0.55)*(1-0.55)* (1-0.55) = 0.334125
P(X = 2) = 3*(0.55^2*(1-0.55)) = 0.408375
P(X = 3) = 0.55^3 = 0.166375
 The probability distribution associated with the
number of $10 bills is given by:
x
0
1
2
3
P(X = x)
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 5
Example 3.8, pg 80
 Shipment: 8 computers of which 3 are defective
 Random purchase of 2 computers
 What is the probability distribution for the random variable X =
defective computers purchased?
Possibilities: X = 0 X =1 X = 2
Let’s start with P(X=0)
P = specified target / all possible
(0 defectives
and
2 nondefectives are selected)
(all ways to get 0 out of 3 defectives)
∩ (all ways to get 2 out of 5 nondefectives)
(all ways to choose 2 out of 8 computers)
(all ways to choose 2 out of 8 computers)
 3  5 
   
1 1 15
P( X  1 )      
28
8
 
 2
 3  5
   
0 2 10
P( X  0)      
28
8
 
 2
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 6
Discrete Probability Distributions
 The discrete probability distribution function
(pdf)
 f(x) = P(X = x) ≥ 0
 Σx f(x) = 1
 The cumulative distribution, F(x)
 F(x) = P(X ≤ x) = Σt ≤ x f(t)
 Note the importance of case: F not same as f
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 7
Probability Distributions
 From our example, the probability that no more
than 2 of the envelopes contain $10 bills is
 P(X ≤ 2) = F (2) = _________________
 F(2) = f(0) + f(1) + f(2) = .833625
  (OR 1 - f(3))
 The probability that no fewer than 2 envelopes
contain $10 bills is
 P(X ≥ 2) = 1 - P(X ≤ 1) = 1 – F (1) = ________
 1 – F(1) = 1 – (f(0) + f(1)) = 1 - .425 = .575
  (OR f(2) + f(3))
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 8
Another View
 The probability histogram
0.45
0.40
0.35
f(x)
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
2
1
3
x
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 9
Your Turn …

The output of the same type of circuit board from two assembly lines is
mixed into one storage tray. In a tray of 10 circuit boards, 6 are from line
A and 4 from line B. If the inspector chooses 2 boards from the tray, show
the probability distribution function associated with the selected boards
being from line A.
 6  4 
  
0 2
1* 6
 0.133
f (0)  P ( X  0)     
45
10 
x
P(x)
 
2
 6  4 
  
1 1
6*4
 0.533
f (1)  P ( X  1)     
10
45
 
 
2
 6  4 
  
2 0
15 *1
 0.333
f ( 2)  P ( X  2)     
45
10 
 
2
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
0
1
2
Slide 10
Continuous Probability Distributions
b
In general,
P (a  X  b)   f ( x )dx
a
 The probability that the average daily temperature
in Georgia during the month of August falls
between 90 and 95 degrees is
 The probability that a given part will fail before 1000
hours of use is
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 11
Visualizing Continuous Distributions
 The probability that the
average daily
temperature in Georgia
during the month of
August falls between 90
and 95 degrees is
-5
-3
-1
1
3
5
 The probability that a
given part will fail before
1000 hours of use is
0
JMB Chapter 3 Lecture 1 v2
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EGR 252.001 Spring 2010
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Slide 12
Continuous Probability Calculations
 The continuous probability density function (pdf)
f(x) ≥ 0, for all x ∈ R

 f ( x )dx  1

b
P (a  X  b)   f ( x )dx
a
 The cumulative distribution, F(x)
x
F ( x )  P( X  x ) 
 f (t )dt

JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 13
Example: Problem 3.7, pg. 88
The total number of hours, measured in units of 100 hours
x,
0<x<1
f(x) =
2-x,
1≤x<2
0,
elsewhere
{
a) P(X < 120 hours) = P(X < 1.2)
= P(X < 1) + P (1 < X < 1.2)
NOTE: You will need to integrate two different functions
over two different ranges.
b) P(50 hours < X < 100 hours) =
Which function(s) will be used?
JMB Chapter 3 Lecture 1 v2
EGR 252.001 Spring 2010
Slide 14