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4.1 Mathematical Expectation
Example: Repair costs for a particular machine
are represented by the following probability
distribution:
x
$50
$200
$350
P(X = x)
0.3
0.2
0.5
What is the expected value of the repairs?
 That is, over time what do we expect repairs to cost
on average?
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 1
Expected Value – Repair Costs
μ = E(X)
 μ = mean of the probability distribution
For discrete variables,
μ = E(X) = ∑ x f(x)
So, for our example,
E(X) = 50(0.3) + 200(0.2) + 350(0.5) =
$230
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 2
Another Example – Investment
By investing in a particular stock, a person can
take a profit in a given year of $4000 with a
probability of 0.3 or take a loss of $1000 with a
probability of 0.7. What is the investor’s
expected gain on the stock?
X
$4000
P(X) 0.3
-$1000
0.7
E(X) = $4000 (0.3) -$1000(0.7) = $500
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 3
Expected Value - Continuous Variables
For continuous variables,
μ = E(X) = E(X) = ∫ x f(x) dx
Vacuum cleaner example: problem 7 pg. 88
f(x) =
{
x,
2-x,
0,
0<x<1
1≤x<2
elsewhere
(in hundreds of hours.)
1
E(X)   x dx  
2
0
2
1
x3 1  2 x3 
x 2  x  dx 
  x  
|
3 0 
3
|
2
1
= 1 * 100 = 100.0 hours of operation annually, on average
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 4
Functions of Random Variables
 Ex 4.4. pg. 111: Probability of X, the number of cars
passing through a car wash in one hour on a sunny
Friday afternoon, is given by
x
P(X = x)
4
5
1/12 1/12
6
7
8
9
1/4
1/4
1/6
1/6
 Let g(X) = 2X -1 represent the amount of money paid
to the attendant by the manager. What can the
attendant expect to earn during this hour on any given
sunny Friday afternoon?
E[g(X)] = Σ g(x) f(x) = Σ (2X-1) f(x)
= (2*4-1)(1/12) +(2*5-1)(1/12) …+(2*9-1)(1/6) = $12.67
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 5
4.2 Variance of a Random Variable
Recall our example: Repair costs for a particular
machine are represented by the following
probability distribution:
x
$50
200
350
P(X = x)
0.3
0.2
0.5
What is the variance of the repair cost?
– That is, how might we quantify the spread of costs?
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 6
Variance – Discrete Variables
 For discrete variables,
 σ2 = E [(X - μ)2] = ∑ (x - μ)2 f(x)
= E (X2) - μ2
 Recall, for our example, μ = E(X) = $230
 Preferred method of calculation:
 σ2
= [E(X2)] – μ2
= 502 (0.3) + 2002 (0.2) + 3502 (0.5) – 2302 = $17,100
 Alternate method of calculation:
 σ2 = E(X- μ)2 f(x)
= (50-230)2 (0.3) + (200-230)2 (0.2) + (350-230)2 (0.5)
= $17,100
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 7
Variance - Investment Example
 By investing in a particular stock, a person can take a
profit in a given year of $4000 with a probability of 0.3
or take a loss of $1000 with a probability of 0.7. What
are the variance and standard deviation of the
investor’s gain on the stock?
 E(X) = $4000 (0.3) -$1000 (0.7) = $500
 σ2 = [∑(x2 f(x))] – μ2
= (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000
 σ = $2291.29
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 8
Variance of Continuous Variables
 For continuous variables,
σ2 = E [(X - μ)2] =[∫ x2 f(x) dx] – μ2
 Recall our vacuum cleaner example pr. 7 pg. 88
{
f(x) =
x,
2-x,
0,
0<x<1
1≤x<2
elsewhere
(in hundreds of hours of operation.)
 What is the variance of X? The variable is
continuous, therefore we will need to evaluate the
integral.
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 9
Variance Calculations for Continuous Variables
b
2

  2
  (X) 
x
f
(
x
)
dx
a

(Preferred calculation)
2
1
2
3
2
2


 (X)   x dx   x 2  x  dx  
 0

1
2
4
x
 2 (X) 
4
4 
 2 x3
x
|0   3  4 


1
2
|
1
 12  0.1667
What is the standard deviation?
 σ = 0.4082 hours
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 10
Covariance/ Correlation
A measure of the nature of the association
between two variables
Describes a potential linear relationship
Positive relationship
 Large values of X result in large values of Y
Negative relationship
 Large values of X result in small values of Y
“Manual” calculations are based on the joint
probability distributions
Statistical software is often used to calculate
the sample correlation coefficient (r)
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 11
What if the distribution is unknown?
Chebyshev’s theorem:
The probability that any random variable X will
assume a value within k standard deviations of
the mean is at least 1 – 1/k2. That is,
P(μ – kσ < X < μ + kσ) ≥ 1 – 1/k2
“Distribution-free” theorem – results are weak
If we believe we “know” the distribution, we do
not use Chebyshev’s theorem to characterize
variability
JMB Chapter 4 Lecture 1
EGR 252 2012
Slide 12