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 JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

Discrete Uniform Distribution Simplest 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 draw the uniform distribution JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

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 Process n 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 JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

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. remember, YOU define what a “success” is … e.g., votes, defects, errors JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

Binomial Distribution The 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) = p = q = b(x; n, p) = (4 choose x)(0.1)x(0.9)4-x , x = 0,1,2,3,4 JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

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? b(x; n, p) = (4 choose x)(0.1)x(0.9)4-x , x = 0,1,2,3,4 P(X = 2) = (4 choose 2) (0.1)2(0.9)2 = 0.0486 P(X < 2) = P(0) + P(1) + P(2) = (4 choose 0) (0.1)0(0.9)4 + (4 choose 1) (0.1)1(0.9)3 + (4 choose 2) (0.1)2(0.9)2 = 0.9963 JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

Expectations of the Binomial Distribution The mean and variance of the binomial distribution are given by μ = np σ2 = npq Suppose, 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 μ = np = 20(0.1) = 2 σ2 = npq = 20(0.1)(0.9)= 1.8, σ = 1.34 JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

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. μ = np = 25(0.01) = 0.25 σ2 = npq = 25(0.01)(0.99)= 0.2475** NOTE: 0.2475 ≠ 0.25 JMB Chapter 5 Part 1 EGR 252.001 Spring 2008

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)  ________________ P(less than 4)  ________________ Appendix Table A.1 (pp. 742-747) lists Binomial Probability Sums, ∑rx=0 b(x; n, p) JMB Chapter 5 Part 1 EGR 252.001 Spring 2008