Agenda  Review Homework • •   Chapter 6: 1, 2, 3, 4, 13 Chapter 7 - 2, 5, 11 Probability Control charts for attributes Week 12  Week 13 Assignment • • Read Chapter 10: “Reliability” Homework • • Chapter.

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Transcript Agenda  Review Homework • •   Chapter 6: 1, 2, 3, 4, 13 Chapter 7 - 2, 5, 11 Probability Control charts for attributes Week 12  Week 13 Assignment • • Read Chapter 10: “Reliability” Homework • • Chapter.

Agenda

Review Homework
•
•


Chapter 6: 1, 2, 3, 4, 13
Chapter 7 - 2, 5, 11
Probability
Control charts for
attributes
Week 12

Week 13 Assignment
•
•
Read Chapter 10:
“Reliability”
Homework
•
•
Chapter 8: 5, 9,10,
20, 26, 33, 34
Chapter 9: 9, 23
Probability
Chapter Eight
Probability
Probability theorems



Probability is expressed as a number
between 0 and 1
Sum of the probabilities of the events of a
situation equals 1
If P(A) is the probability that an event will
occur, then the probability the event will
not occur is
•
1.0 - P(A)
Probability
Probability theorems


For mutually exclusive events, the
probability that either event A or event B
will occur is the the sum of their
respective probabilities.
When events A and B are not mutually
exclusive events, the probability that
either event A or event B will occur is
•
P(A or B or both) = P(A) + P(B) - P(both)
Probability
Probability theorems

If A and B are dependent events, the
probability that both A and B will occur is
•

P(A and B) = P(A) x P(B|A)
If A and B are independent events, then
the probability that both A and B will
occur is
•
P(A and B) = P(A) x P(B)
Probability
Permutations and
combinations


A permutation is the number of
arrangements that n objects can have
when r of them are used.
When the order in which the items are
used is not important, the number of
possibilities can be calculated by using
the formula for a combination.
Probability
Discrete probability
distributions

Hypergeometric - random samples from
small lot sizes.
•
•


Population must be finite
samples must be taken randomly without
replacement
Binomial - categorizes “success” and
“failure” trials
Poisson - quantifies the count of discrete
events.
Probability
Continuous probability
distributions






Normal
Uniform
Exponential
Chi Square
F
student t
Probability
Fundamental concepts




Probability = occurrences/trials
0<P<1
The sum of the simple probabilities for all
possible outcomes must equal 1
Complementary rule - P(A) + P(A’) = 1
Probability
Addition rule

P(A + B) = P(A) + P(B) - P(A and B)
•
If mutually exclusive; just P(A) + P(B)
P(AandB)
P(A)
P(B)
Probability
Addition rule example


P(A + B) = P(A) + P(B) - P(A and B)
Roll one die
•
•
Probability of even and divisible by 1.5?
Sample space {1,2,3,4,5,6}
•
•
•

Event A - Even {2,4,6}
Event B - Divisible by 1.5 {3,6}
Event A and B ?
Solution?
Probability
Conditional probability rule


P(A|B) = P(A and B) / P(B)
A die is thrown and the result is known to
be an even number. What is the
probability that this number is divisible by
1.5?
•
•
P(/1.5|Even)=P(/1.5 and even)/P(even)
1/6 / 3/6 = 1/3
Probability
Compound or joint
probability


The probability of the simultaneous
occurrence of two or more events is
called the compound probability or,
synonymously, the joint probability.
Mutually exclusive events cannot be
independent unless one of them is zero.
Probability
Multiplication for
independent events

P(A and B) = P(A) x P(B)
•
P(ace and heart) = P(ace) x P(heart)
•
1/13 x 1/4 = 1/52
Probability
Computing conditional
probabilities


P(A|B) = P(A and B)/P(B)
P(Own and Less than 2 years)?
Number of credit applicants by category
On present job 2
years or less
On present job
more than 2 years
Own Home
20
40
Rent Home
80
60
Total
100
100
Probability
Computing conditional
probabilities

P(A|B) = P(A and B)/P(B)
P(AandB)
P(A)
P(B)
Probability
Joint probability
table
On present job
2 years or less
On present job
more than 2
years
Marginal
probability
Own Home
.10
.20
.30
Rent Home
.40
.30
.70
Marginal probability
.50
.50
1.00
Probability
Conditional probability
Satisfied Not Satisfied
Totals
New
300
100
400
Used
450
150
600
Total
750
250
1000
S=satisfied N= bought new car
P(N|S) = ?
Probability
Just for fun

60 business students
from a large university
are surveyed with the
following results:
•
•
•
•
•
•
•



19 read Business Week
18 read WSJ
50 read Fortune
13 read BW and WSJ
11 read WSJ and Fortune
13 read BW and Fortune
9 read all three

How many read none?
How many read only
Fortune?
How many read BW, the
WSJ, but not Fortune?
Hint: Try a Venn
diagram.
Probability
Probability
Distributions
Probability
Learning objectives




Know the difference between discrete
and continuous random variables.
Provide examples of discrete and
continuous probability distributions.
Calculate expected values and
variances.
Use the normal distribution table.
Probability
Random variables

A random variable is a numerical quantity
whose value is determined by chance.
•
•
“A random variable assigns a number to
every possible outcome or event in an
experiment”.
For non-numerical outcomes such as a coin
flip you must assign a random variable that
associates each outcome with a unique real
number.
Probability
Random variable types

Discrete random variable - assumes a
limited set of values; non-continuous,
generally countable
•
•
•
number of Mark McGwire homeruns in a
season
number of auto parts passing assembly-line
inspection
GRE exam scores
Probability
Random variable types

Continuous random variable - random
variable with an infinite set of values.
0.000
Baseball player’s batting average
1.000
Can occur anywhere on a continuous number scale
Probability
Random variables and
probability distributions

The relationship between a random
variable’s values and their probabilities is
summarized by its probability distribution.
Probability
Probability distribution

Whether continuous or discrete, the
probability distribution provides a
probability for each possible value of a
random variable, and follows these rules:
•
•
•
The events are mutually exclusive
The individual probability values are
between 0 and 1.
The total value of the probability values sum
to 1
Probability
Probability distribution for
rates of return

Possible rate of
return
•
•
•
•
•
•
•
•
10%
11%
12%
13%
14%
15%
16%
17%

Probability
•
•
•
•
•
•
•
•

.05
.15
.20
.35
.10
.10
.03
.02
Total = 1.0
Probability
Describing distributions

Measures of
central tendency
•

Measures of
variability
expected value
•
(weighted
average)
•
•
Probability
variance
standard deviation
Expected value of a discrete
random variable

For discrete random variables, the
expected value is the sum of all the
possible outcomes times the probability
that they occur.
E(X) =  {xi * P(xi)}
Probability
Example: A fair die

Roll 1 die:
Can you sketch
the distribution?
x
1
2
3
4
5
6
P(x) x*P(x) E(x)=?
1/6
1/6
1/6
2/6
1/6
3/6
1/6
4/6
1/6
5/6
1/6
6/6
Probability
Fair die illustrates a discrete
“uniform distribution”

The random variable, x, has n possible
outcomes and each outcome is equally
likely. Thus, x is distributed uniform.
Probability
Probability distribution
P(x)
1/6
1
2
3
Probability
4
5
6
x
Example: An unfair die

Roll 1 die:
Can you sketch
the distribution?
x
1
2
3
4
5
6
P(x) x*P(x) E(x)=?
1/12
1/12
2/12
4/12
2/12
6/12
2/12
8/12
2/12
10/12
3/12
18/12
Probability
Expected value of a bet


Suppose I offer you the following wager:
You roll 1 die. If the result is even, I pay
you $2.00. Otherwise you pay me
$1.00.
E(your winnings)=.5 ($2.00) + .5 (-1.00)
= 1.00 - .50 = $0.50
Probability
Expected Value of a Bet


Suppose I offer you the following wager:
You roll 1 die. If the result is 5 or 6 I pay
you $3.00. Otherwise you pay me $2.00.
What is your expected value?
Probability
Variance of a discrete
random variable
The variance of a random variable is a
measure of dispersion calculated by
squaring the differences between the
expected value and each random
variable and multiplying by its associated
probability.

2
{(xi-E(x))
Probability
* P(xi)}
Example: A fair die

Roll 1 die: [x- E(X)] 2
1 - 21/6 6.25
2 - 21/6 2.25
3 - 21/6 .25
4 - 21/6 .25
5 - 21/6 2.25
6 - 21/6 6.25
2.91
Probability
P(x) *P(x)
1/6 1.04
1/6
.375
1/6
.04
1/6
.04
1/6
.375
1/6 1.04
Probability distributions for
continuous random variables



A continuous mathematical function
describes the probability distribution.
It’s called the probability density function
and designated ƒ(x)
Some well know continuous probability
density functions:
•
•
Normal
Exponential
Beta
Student t
Probability
Continuous probability
density function - Uniform
If a random variable, x, is distributed
uniform over the interval [a,b], then its
pdf is given by
1
f (x) 
b a
1
b-a
a
b
Probability
Uniform
What is the probability of x?
1
b-a
a
x
Probability
b
Uniform
1
b-a
Area under the rectangle = base*height
= (b-a)* 1 = 1
b-a
a
b
Probability
Uniform
P(c<x<b) = Area of brown rectangle
1 * (b-c) Ht x Width)
= b-a
1
b-a
a
c
b
Probability
Uniform
P(2<x<5) = Brown rectangle
1 =1/4
5-1
1 * (5-2) =(1/4) *3 =
= 5-1
1
2
3/4
5
Probability
Uniform distribution
If a random variable, x, is distributed
uniform over the interval [a,b], then its pdf
is given by
1
f (x) 
And, the mean and variance are
(a+b)
( b-a )2
E(x) = ------- Var(x)=--------2
12
Probability
b a
Uniform
Mean?
Variance?
3
8
Probability
Calculate uniform mean,
variance
So, if a = 3 and b = 8
1
f (x ) 
5
And, the mean and variance are
(a+b)
( b-a )2
25
E(x) = ------ = 5.5
V(x)=--------- = ----- = 2.08
2
12
12
Probability
Continuous pdf - Normal
If x is a normally distributed variable, then
f (x) 
1
2
2
e
 ( x   )2
2 2
is the pdf for x. The expected value is  and
the variance is 2.
Probability
One standard deviation


68.3%
Probability
Two standard deviations
2 2
95.5%
Probability
Three standard deviations
3
3
99.73%
Probability
Continuous PDF - Standard
Normal
If z is distributed standard normal,
then and 
f (z ) 
f (x) 
1
e
2
1
2
2
e
 z2
2
 ( x   )2
2 2
Probability