engineering probability and statistics dr. leonore findsen department of statistics

Engineering Probability and Statistics

Dr. Leonore FindsenDepartment of Statistics

Outline

• Sets and Operations• Counting Sets• Probability• Random Variables• Standard Distribution Functions• Statistical Treatment of Data• Statistical Inference

Sets and Operations

– A set is a collection of objects.– An element of the set is one of the objects.– The empty set, , contains no objects.

• Venn Diagrams

Set Operations

Union, U, A or B or both Intersection, ∩, A and B, AB

Complement, Ac, everything but A.

Set Operations/Product Sets

• Set Operations (de Morgan’s Laws)– (A U B)c = Ac ∩ Bc (A ∩ B)C = Ac U Bc

• Product Sets – Cartesian Product– The set of all ordered pairings of the elements of

two sets.– Example: A = {1,2}, B = {3,4}

A X B ={(1,3), (2,3), (1,4), (2,4)}

Copyright Kaplan AEC Education, 2008

Basic Set Theory

Example

PF

G

F ∩ P ∩ GC = 6

Copyright Kaplan AEC Education, 2008

Solution

Counting Sets

• Finding the number of possible outcomes.• Counting the number of possibilities• Ways of counting

– Sampling with or without replacement– Ordered or unordered– Product Rule– Permutations– Combinations– Complicated

Product Rule

• Ordered Pairs with replacement• Formula: n1 ∙∙∙ nm

• Examples: – Number of ways that you can combine

alphanumerics into a password.– Number of ways that you can combine different

components into a circuit.

Permutations

• An ordered subset without replacement• Formula

• Examples:– Number of ways that you can combine

alphanumerics into a password if you can not repeat any symbols.

– Testing of fuses to see which one is good or bad.– Choosing officers in a club.

k,n

n!P n(n 1) (n k 1)

(n k)!

Copyright Kaplan AEC Education, 2008

Combinations – ordered with and without replacement

With replacement: each letter can be repeated. # of airports =(26)(26)(26) = 17,576 airports

Without replacement: each letter can not be repeated# of airports =(26)(25)(24) = 15,600 airports

Combinations

• An unordered subset without replacement• Formula

• Examples:– Choosing members of a club to see who will be

going to a national conference.– Selecting 3 red cards from a deck of 52 cards.

k,nk,n

n P n!C

k k! k!(n k)!

Copyright Kaplan AEC Education, 2008

Combinations - Example

a) # of teams = (15)(12)(8)(5) = 7,200

b) # of teams =15 12 5

900,9004 2 2

Complicated Counting

How many different ways can you get a full house?

4 4(13)(12) 3,744

3 2

Probability

• Definitions– The probability of an event is the ratio of the

number of times that it occurs to the number of times that everything occurs

– N(E)P(E)

N(everything)

Probability - Properties

• 0 P(E) 1– P() = 0, P(everything) = 1

• P(E) = 1 – P(Ec)– Example: Consider the following system of

components connected in a series. Let E = the event that the system fails. What is P(E)?

P(E) = 1 – P(SSSSS)

54321

Joint Probability

• P(A U B) = P(A) + P(B) – P(A ∩ B)

• P(A ∩ B) = P(A)P(B) if A and B are independent

Joint Probability - Example2-54: Given the following odds:

In favor of event A 2:1In favor of event B 1:5In favor of event A or event B or both 5:1

Find the probability of event AB occurring?

P(A U B) = P(A) + P(B) – P(A ∩ B)

P(A ∩ B) = 0

5 2 1P(A B)

6 3 6

Joint Probability - ExampleThe probability that a defective part is generated from

Machine A is 0.01; the probability that a defective part is generated from Machine B is 0.02, What is the probability that both machines have defective parts?

P(A ∩ B) = P(A)P(B) = (0.01)(0.02) = 0.0002

Conditional Probability

• Conditional Probability Definition

• General Multiplication– P(A ∩ B) = P(A|B)P(B) = P(A)P(B|A)– P(A ∩ B ∩ C) = P(A)P(B|A)P(C|A ∩ B)

• Bayes’ Theorem

P(A B)P(A|B)

P(B)

i i ii

j j

P(A B) P(B|A )P(A )P(A |B)

P(B) P(B|A )P(A )

Copyright Kaplan AEC Education, 2008

Bayes’ Theorem 1

P(D) = 0.6 P(E) = 0.2 P(F) = 0.2P(B|D) = 0.12 P(B|E) = 0.04 P(B|F) = 0.10P(B) = P(D)P(B|D) + P(E)P(B|E) + P(F)P(B|F) = 0.10

Copyright Kaplan AEC Education, 2008

Bayes’ Theorem 2

Given that the car has bad tires, what is the probability that it was rented from Agency E?

P(D) = 0.6 P(E) = 0.2 P(F) = 0.2P(B|D) = 0.12 P(B|E) = 0.04 P(B|F) = 0.10P(B) = P(D)P(B|D) + P(E)P(B|E) + P(F)P(B|F) = 0.10

P(B|E)P(E) (0.04)(0.2)P(E|B) 0.08

P(B) 0.10

Random Variables

• Definition– A random variable is any rule that associates a

number with each outcome in your total sample space.

– A random variable is a function.

Probability Density Functions• The area under a pdf curve for an interval is the

probability that an event mapped into that interval will occur.

P(a X b)

b

aP(a X b) f(x)dx

Cumulative Distribution Functions

• P(X a) = F(a)

Properties of pdfs

• Percentilesp = F(a)

• Mean

• E(h(x))

• Varianceσ2 = Var(X) = E[(X – μ)2] = E(X2) – [E(X)]2

E(X) xf(x)dx

E(h(X)) h(x)f(x)dx

Copyright Kaplan AEC Education, 2008

Cumulative Distribution Function - Example

a)

b)

x 2 3

0F(X) 3y dy x

Copyright Kaplan AEC Education, 2008

Cumulative Distribution Function – Example (cont)

c) 1

1 12 3 4

0 00

3 3x3x dx 3x dx x

4 4

11 12 2 2 5 5

0 00

3 3E(X ) x 3x dx 3x dx x

5 5

22 3 3 3

0.03755 4 80

Standard Distribution Functions

There are some standard distributions that are commonly used. These are determined from either from the experiment or from analysis of the data.

Binomial Distribution

Experimental Conditions – BInS1. B: Each trial can have only two outcomes

(binary).2. I: The trials are independent.3. n: Know the number of trials4. S: The probability of success is constant.Want to find the number of successes.Formula:

r n rnP(X r) p (1 p)

r

Copyright Kaplan AEC Education, 2008

Binomial Discrete Distributions - Example

Let X = number of cars out of five that get a green light.

X ~ B(n,p) = B(5,0.7)P(X 3) = 1 – P(X < 3) = 1 – P(X 2) = 1 – F(2) = 1 – 0.1631

= 0.8369

Other Discrete Distributions

• Hypergeometric– Like binomial but without replacement

• Poisson– Like a binomial but with very low probability of

success• Negative Binomial

– Like binomial but want to know how many trials until a certain number of successes.

Normal Distribution Function

• Continuous• This is the most commonly occurring

distribution.– Systematic errors– A large number of small values equally likely to be

positive or negative

Normal Distribution Function (cont)

• The parameters of the normal distribution are μ and σ

• The normal distribution can not be integrated so we use z-tables which are for the standard normal with μ = 0, σ = 1. The z-tables contain the cdf, (z).

• To convert our distribution to the standard normal,

XZ

Copyright Kaplan AEC Education, 2008

Normal Distribution Function - Example

F(c*) = 0.01 ==> c* = -2.327 σ = 0.86 kN/sq. mc

c*

30 32

2.327

Statistical Treatment of Data

• Most people need to visualize the data to get a feel for what it looks like.

• In addition, summarizing the data using numerical methods is also helpful in analyzing the results.

Frequency Distribution

• Frequency table• Histogram• Example

– 100 married couples between 30 and 40 years of age are studied to see how many children each couple have. The table below is the frequency table of this data set.

Frequency Distribution –

Example (cont)

Kids # of Couples Rel. Freq1 11 0.112 22 0.223 30 0.304 11 0.115 1 0.016 0 0.007 1 0.01

100 1.00

1 2 3 4 5 6 70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Number of Kids

Numerical Statistical Measures

• Measures of the central value– Mean– Median– Mode

• Measures of variability– Range– Variance (standard deviation)– Interquartile range

ixx

n

2i2

2

i2i

(x x)s

n 1

xx

nn 1

Copyright Kaplan AEC Education, 2008

Measures of Dispersion

Copyright Kaplan AEC Education, 2008

Solution

Statistical InferenceConfidence Intervals

• t- Distribution– Used when the population distribution is normal

but σ is unknown– Tables will have to be provided if necessary

• Confidence Intervals for μ

• General form:point estimator critical value SEestimator

/2

sX z

n /2,n 1

sX t

n

Copyright Kaplan AEC Education, 2008

Interval Estimates

Copyright Kaplan AEC Education, 2008

Solution

Copyright Kaplan AEC Education, 2008

Solution (continued)

Copyright Kaplan AEC Education, 2008

Solution (continued)

Statistical InferenceHypothesis Testing - Procedure

• Hypotheses– Ho: null hypothesis, = 0

– HA: alternative hypothesis, 0, > 0, < 0

• Test statistic

–

• Decision 0:P(|T|>ts), > 0:P(T>ts), < 0:P(T<ts)

0s df

x

xt t ( ')

SE

Statistical InferenceHypothesis Testing - Errors

calculated/true Ho true Ho falsefail to reject Ho correct Type II, βreject Ho Type I, α correct

Conclusion

• Sets and Operations• Counting Sets• Probability• Random Variables• Standard Distribution Functions• Statistical Treatment of Data• Statistical Inference