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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 9th ed
EGR 252.001 2011
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 9th ed
EGR 252.001 2011
Slide 2
Discrete Random Variables
Problem 2.51 Page 59 (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 9th ed
EGR 252.001 2011
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 9th ed
EGR 252.001 2011
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 9th ed
EGR 252.001 2011
Slide 5
Another View
The probability histogram
0.45
0.40
0.35
0.30
f(x)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
x
JMB Chapter 3 Lecture 1 9th ed
EGR 252.001 2011
Slide 6
Another Discrete Probability Example
Given:
A shipment consists of 8 computers
3 of the 8 are defective
Experiment: Randomly select 2 computers
Definition: random variable X = # of defective computers selected
What is the probability distribution for X?
Possible values for X: X = 0 X =1 X = 2
Let’s start with P(X=0) [0 defectives and 2 nondefectives are selected]
Recall that P = specified target / all possible
(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
0 2 10
P( X 0)
28
8
2
JMB Chapter 3 Lecture 1 9th ed
EGR 252.001 2011
Slide 7
Discrete Probability Example
What is the probability distribution for X?
Possible values for X: X = 0 X =1 X = 2
Let’s calculate P(X=1) [1 defective and 1 nondefective are selected]
(all ways to get 1 out of 3 defectives)
∩ (all ways to get 1 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
2 0
3
P ( X 2)
28
8
2
x
P(X = x)
JMB Chapter 3 Lecture 1 9th ed
0
1
EGR 252.001 2011
2
Slide 8
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 9th ed
EGR 252.001 2011
Slide 9
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
Another way to calculate F(2) (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
Another way to calculate P(X ≥ 2) is f(2) + f(3)
JMB Chapter 3 Lecture 1 9th ed
EGR 252.001 2011
Slide 10
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 for the number of selected boards
coming 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)
10
45
2
JMB Chapter 3 Lecture 1 9th ed
EGR 252.001 2011
0
1
2
Slide 11
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 9th ed
EGR 252.001 2011
Slide 12
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 9th ed
5
EGR 252.001 2011
10
15
20
25
30
Slide 13
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 9th ed
EGR 252.001 2011
Slide 14
Example: Problem 3.7, pg. 92
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 9th ed
EGR 252.001 2011
Slide 15