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 See examples in Chapter 4 (pp. 123-126) MDH Chapter 4 - 5 EGR 252 2015

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 Note that the 9th edition eliminated section 5.2 Discrete Uniform Distribution Section 5.2 is now binomial and multinomial MDH Chapter 4 - 5 EGR 252 2015

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 MDH Chapter 4 - 5 EGR 252 2015

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 MDH Chapter 4 - 5 EGR 252 2015

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 MDH Chapter 4 - 5 EGR 252 2015

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 MDH Chapter 4 - 5 EGR 252 2015

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 MDH Chapter 4 - 5 EGR 252 2015

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 MDH Chapter 4 - 5 EGR 252 2015

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. 726-731) lists Binomial Probability Sums, ∑ rx=0 b(x; n, p) MDH Chapter 4 - 5 EGR 252 2015