known probability distributions

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JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 1 Known Probability Distributions Engineers frequently work with data that can be modeled as one of several known probability distributions . Being able to model the data allows us to: model real systems design predict results Key discrete probability distributions include: binomial negative binomial hypergeometric Poisson

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Known Probability Distributions. Engineers frequently work with data that can be modeled as one of several known probability distributions . Being able to model the data allows us to: model real systems design predict results Key discrete probability distributions include: binomial - PowerPoint PPT Presentation

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Page 1: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 1

Known Probability Distributions

Engineers frequently work with data that can be modeled as one of several known probability distributions.

Being able to model the data allows us to:model real systemsdesign predict results

Key discrete probability distributions include:binomialnegative binomialhypergeometricPoisson

Page 2: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 2

Discrete Uniform DistributionSimplest of all discrete distributions

All possible values of the random variable have the same probability, i.e.,

f(x; k) = 1/ k, x = x1 , x2 , x3 , … , xk

Expectations of the discrete uniform distribution

k

xand

k

xk

ii

k

ii

1

2

21)(

Discrete Uniform Distribution

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 2 4 6 8

Page 3: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 3

Binomial & Multinomial Distributions Bernoulli Trials

Inspect tires coming off the production line. Classify each as defective or not defective. Define “success” as defective. If historical data shows that 95% of all tires are defect-free, then P(“success”) = 0.05.

Signals picked up at a communications site are either incoming speech signals or “noise.” Define “success” as the presence of speech. P(“success”) = P(“speech”)

Bernoulli Processn repeated trials the outcome may be classified as “success” or “failure” the probability of success (p) is constant from trial to trial repeated trials are independent

Page 4: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 4

Binomial Distribution

Example:Historical data indicates that 10% of all bits transmitted through a digital transmission channel are received in error. Let X = the number of bits in error in the next 4 bits transmitted. Assume that the transmission trials are independent. What is the probability that Exactly 2 of the bits are in error? At most 2 of the 4 bits are in error? More than 2 of the 4 bits are in error?

The number of successes, X, in n Bernoulli trials is called a binomial random variable.

Page 5: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 5

Binomial DistributionThe probability distribution is called the binomial

distribution.b(x; n, p) = , x = 0, 1, 2, …, n

where p = probability of success

q = probability of failure = 1-p

For our example,

b(x; n, p) =

xnxqpx

n

4,3,2,1,0,9.01.04 4

xx

xx

Page 6: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 6

For Our Example … What is the probability that exactly 2 of the bits are in

error?

At most 2 of the 4 bits are in error?

More than 2 of the 4 bits are in error?

9963.0)2( XP

0486.09.01.02

4)2( 242

XP

223140 9.01.02

49.01.0

1

49.01.0

0

4

Page 7: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 7

Expectations of the Binomial DistributionThe mean and variance of the binomial

distribution are given by

μ = np

σ2 = npqSuppose, in our example, we check the next 20

bits. What are the expected number of bits in error? What is the standard deviation?

μ = 20 (0.1) = 2

σ 2 = 20 (0.1) (0.9) = 1.8 σ = 1.34

Page 8: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 8

Another example A worn machine tool produces 1% defective parts. If we

assume that parts produced are independent, what is the mean number of defective parts that would be expected if we inspect 25 parts?

μ = 25 (0.01) = 0.25

What is the expected variance of the 25 parts?

σ 2 = 25 (0.01) (0.99) = 0.2475

Note that 0.2475 does not equal 0.25.

Page 9: Known Probability Distributions

JMB Chapter 5 Part 1 EGR 252.001 Spring 2010 Slide 9

Helpful Hints … Suppose we inspect the next 5 parts …b(x ; 5, 0.01)

Sometimes it helps to draw a picture.P(at least 3) ________________

0 1 2 3 4 5

P(2 ≤ X ≤ 4) ________________0 1 2 3 4 5

P(less than 4) ________________0 1 2 3 4 5

Appendix Table A.1 (pp. 742-747) lists Binomial Probability Sums, ∑r

x=0 b(x; n, p)