Known probability distributions

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1 Known probability distributions Engineers frequently wor with data that can be modeled as one of several nown probability distributions. Being able to model the data allows us to: model real systems design predict results Key discrete probability distributions include: binomial / multinomial negative binomial hypergeometric Poisson Discrete uniform distribution Simplest of all discrete distributions All possible values of the random variable have the same probability, i.e., f(x; ) = /, x = x, x, x 3,, x Expectations of the discrete uniform distribution i x i and i ( x i )

2 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 ) = Signals piced up at a communications site are either incoming speech signals or noise. Define success as the presence of speech. P( success ) = P( speech ) Administer a test drug to a group of patients with a specific condition. P( success ) = 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. 3 Binomial distribution Example: Historical data indicates that 0% 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 of the bits are in error? At most of the 4 bits are in error? more than of the 4 bits are in error? The number of successes, X, in n Bernoulli trials is called a binomial random variable. 4

3 Binomial distribution The probability distribution is called the binomial distribution. b(x; n, p) = n x p x q n x, x = 0,,,, n where p = For our example, q = b(x; n, p) = 5 For our example What is the probability that exactly of the bits are in error? At most of the 4 bits are in error? 6 3

4 Your turn What is the probability that more than of the 4 bits are in error? 7 Expectations of the binomial distribution The mean and variance of the binomial distribution are given by μ = np σ = npq Suppose, in our example, we chec the next 0 bits. What are the expected number of bits in error? What is the standard deviation? μ = σ =, σ = 8 4

5 Another example A worn machine tool produces % 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 5 parts? What is the expected variance of the 5 parts? 9 Helpful hints Sometimes it helps to draw a picture. Suppose we inspect the next 5 parts P(at least 3) P( X 4) P(less than 4) Appendix Table A. (pp ) lists Binomial Probability Sums, r x=0b(x; n, p) 0 5

6 Your turn Use Table A. to determine. b(x; 5, 0.4), P(X 8) =. b(x; 5, 0.4), P(X < 8) = 3. b(x;, 0.), P( X 5) = 4. b(x; 4, 0.), P(X > ) = Multinomial experiments What if there are more than possible outcomes? (e.g., acceptable, scrap, rewor) That is, suppose we have: n independent trials outcomes that are mutually exclusive (e.g.,,,, ) Then exhaustive (i.e., all p i = ) f(x, x,, x ; p, p,, p, n) = x, x,..., x n p x p x... p x 6

7 Example Loo at problem 5., pg. 5 x = p = x = p = n = x 3 = p 3 = f(,, ;,,, ) = = 3 Hypergeometric distribution Example*: Automobiles arrive in a dealership in lots of 0. Five out of each 0 are inspected. For one lot, it is now that out of 0 do not meet prescribed safety standards. What is probability that at least out of the 5 tested from that lot will be found not meeting safety standards? *from Complete Business Statistics, 4th ed (McGraw-Hill) 4 7

8 This example follows a hypergeometric distribution: A random sample of size n is selected without replacement from N items. of the N items may be classified as successes and N- are failures. The probability associated with getting x successes in the sample (given successes in the lot.) Where, = number of successes = N = the lot size = 0 P ( X x) h( x; N, n, ) = 5 x N n N n x n = number in sample x = number found = or 5 Hypergeometric distribution In our example, P( X x) P( X ) P( X ) 0 0 h(; 0,5,) h(; 0,5,) = 6 8

9 Expectations of the hypergeometric distribution The mean and variance of the hypergeometric distribution are given by n N N N n * n * N ( N ) What are the expected number of cars that fail inspection in our example? What is the standard deviation? μ = σ =, σ = 7 Your turn A worn machine tool produced defective parts for a period of time before the problem was discovered. Normal sampling of each lot of 0 parts involves testing 6 parts and rejecting the lot if or more are defective. If a lot from the worn tool contains 3 defective parts:. What is the expected number of defective parts in a sample of six from the lot?. What is the expected variance? 3. What is the probability that the lot will be rejected? 8 9

10 Binomial approximation Note, if N >> n, then we can approximate this with the binomial distribution. For example: Automobiles arrive in a dealership in lots of out of each 00 are inspected. /0 (p=0.) are indeed below safety standards. What is probability that at least out of 5 will be found not meeting safety standards? Recall: P(X ) = P(X < ) = P(X = 0) Hypergeometric distribution Binomial distribution 9 (Compare to example 5.5, pg. 55) Negative binomial distribution Example: Historical data indicates that 30% of all bits transmitted through a digital transmission channel are received in error. An engineer is running an experiment to try to classify these errors, and will start by gathering data on the first 0 errors encountered. What is the probability that the 0 th error will occur on the 5 th trial? 0 0

11 This example follows a negative binomial distribution: Repeated independent trials. Probability of success = p and probability of failure = q = -p. Random variable, X, is the number of the trial on which the th success occurs. The probability associated with the th success occurring on trial x is given by, x x b *( x;, p) p q, x,, Where, = success number = 0 x = trial number on which occurs = 5 p = probability of success (error) = 0.3 q = p = 0.7,... Negative binomial distribution In our example, b *(5; 0,0.3) 5 0 (0.3) 0 (0.7) 5 0 =

12 Geometric distribution Example: In our example, what is the probability that the st bit received in error will occur on the 5 th trial? This is an example of the geometric distribution, which is a special case of the negative binomial in which =. The probability associated with the st success occurring on trial x is given by g( x; p) x pq = 3 Your turn A worn machine tool produces % defective parts. If we assume that parts produced are independent:. What is the probability that the nd defective part will be the 6th one produced?. What is the probability that the st defective part will be seen before 3 are produced? 3. How many parts can we expect to produce before we see the st defective part? (Hint: see Theorem 5.4, pg. 6) 4

13 Poisson process The number of occurrences in a given interval or region with the following properties: memoryless P(occurrence) during a very short interval or small region is proportional to the size of the interval and doesn t depend on number occurring outside the region or interval. P(X>) in a very short interval is negligible 5 Poisson process Examples: Number of bits transmitted per minute. Number of calls to customer service in an hour. Number of bacteria in a given sample. Number of hurricanes per year in a given region. 6 3

14 Poisson process Example An average of.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. To determine personnel and equipment needs to maintain a desired level of service, the plant manager needs to be able to determine the probabilities associated with numbers of service calls. What is the probability that fewer than calls will be received in any given minute? 7 Poisson distribution The probability associated with the number of occurrences in a given period of time is given by, t x e ( t) p( x; t), x x! 0,,,... Where, λ = average number of outcomes per unit time or region =.7 t = time interval or region = minute 8 4

15 Our example The probability that fewer than calls will be received in any given minute is P(X < ) = P(X = 0) + P(X = ) = The mean and variance are both λt, so μ = Note: Table A., pp , gives Σ t p(x;μ) 9 Poisson distribution If more than 6 calls are received in a 3-minute period, an extra service technician will be needed to maintain the desired level of service. What is the probability of that happening? μ = λt = P(X > 6) = P(X < 6) = 30 5

16 Frequency Poisson distribution Calls per minute 3 Poisson distribution The effect of λ on the Poisson distribution

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