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Systems Engineering Program Department of Engineering Management, Information and Systems EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Discrete Probability Distributions Binomial, Negative Binomial, Geometric Distributions Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering 1 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Binomial Distribution 2 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Binomial Model: Let X be the number of successes in n trials. If 1. The trials are identical and independent 2. Each trial results in one of two possible outcomes success or failure 3. The probability of success on a single trial is p, and is constant from trial to trial, 3 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Binomial Model: then X has the Binomial Distribution with Probability Mass Function given by: n x n -x b( x) b( x; n, p) p q x x 0,1,..., n where n n! x x!n x ! and q 1 p 4 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Binomial Distribution Rule: n x p b( x 1; n, p) b( x; n, p) x 1 q for x = 0, …, n - 1 Rule: n x n x 1 p q p q x 0 x n n 5 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Binomial Distribution • Mean or Expected Value = np • Standard Deviation σ npq 1 2 6 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Binomial Model in Excel • Excel provides an easier way of finding probabilities: – Click the Insert button on the menu bar (at the top of the Excel page) • • • • Go to the function option Choose Statistical from the Function Category window (a list of all available statistical functions will appear in the Function Name window) Choose the BINOMDIST function Type in parameters: – – – – number_s => X Trials => N Probability_s => p Cumulative (logical) => TRUE for cumulative function, FALSE for mass function. Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 7 Example - Binomial Distribution When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defects is 5%. Let X=the number of defective boards in a random sample size n=25, so X~B(25,0.05). a) Determine P(X 2). b) Determine P(X 5). c) Determine P(1 X 4). d) What is the probability that none of the 25 boards are defective? e) Calculate the expected value and standard deviation of X. 8 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Solution – Binomial Distribution 9 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Negative Binomial Distribution 10 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Negative Binomial Model: The negative binomial distribution is based on an experiment satisfying the following conditions: 1. The experiment consists of a sequence of independent trials 2. Each trial can result in either a success S, or a failure F 3. The probability of success is constant from trial to trial, so P(S on trial i) = p for i = 1, 2, 3, … 11 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Negative Binomial Model: 4. The experiment continues (trials are performed) until a total of r successes have been observed, where r is a specified positive integer. The associated random variable is: X = number of failures that precede the rth success X is called the negative binomial random variable because, in contrast to the binomial random variable, the number of successes is fixed and the number of trials is random. Possible values of X are x = 0, 1, 2, ... Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 12 The Negative Binomial Model: The probability mass function of the negative binomial random variable X with parameters r = number of successes and p = probability of success on a single trial x r 1 r x p 1 p , x 0,1,2,... NBx; r , p r 1 13 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Negative Binomial Model: The negative binomial model can also be expressed as: X = total number of trials to get k successes k = number of successes and p = probability of success on a single trial x 1 k x k p 1 p , x 0,1,2,... NBx; k , p k 1 14 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Negative Binomial Probability Mass Function Derivation th • Consider the probability of a success on the x trial preceded by k 1 successes and x k failures in some specified order. • Since the trials are independent, we can multiply all the probabilities corresponding to each desired outcome. • Each success occurs with probability failure with probability q 1 p pand each • Therefore, the probability for the specified order, ending in a success, is p k 1q xk p p k q xk 15 Stracener_EMIS 7370/STAT 5340_Sum 07_06.05.07 • The total number of sample points in the experiment ending in a success, after the occurrence of k 1successes and x k failures in any order, is equal to the number of partitions of x 1 trials into two groups with k 1 successes corresponding to one group and x k failures corresponding to the other group. • The sample space consists of x 1 points, each k 1 mutually exclusive and occurring with equal k xk p q probability k xk p • We obtain the general formula by q multiplying by x 1 k 1 16 Stracener_EMIS 7370/STAT 5340_Sum 07_06.05.07 The Negative Binomial Model: Example Find the probability that a person tossing three coins will get either all heads or all tails for the second time on the fifth toss. Solution p Psuccess P[( HHH )or (TTT )] 1 1 8 8 1 4 17 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Since a success must occur on the fifth toss of the 3 coins, the following outcomes are possible: probability Toss 1 S F F F 2 F S F F 3 F F S F 4 F F F S 5 S S S S 27 1024 27 1024 27 1024 27 1024 18 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 Therefore, the probability of obtaining all heads or all tails for the second time on the fifth toss is 27 27 4. 1024 256 19 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Negative Binomial Model: Example Solution continued Or, using the Negative Binomial Distribution with r = 2, p = 0.25, and x = 3 gives 4 27 3 2 NB3;2,0.25 0.25 0.75 0.105 256 1 20 Stracener_EMIS 7370/STAT 5340_Sum 07_06.05.07 The Negative Binomial Model If X is a negative binomial random variable with probability mass function nb(x;r,p) then r 1 p EX p and r 1 p Var X 2 p 21 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Negative Binomial Model Note: By expanding the binomial coefficient in front of pr(1 - p)x and doing some cancellation, it can be seen that NB(x;r,p) is well defined even when r is not an integer. This generalized negative binomial distribution has been found to fit the observed data quite well in a wide variety of applications. 22 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Negative Binomial Model in Excel • Similar to the Binomial Dist in Excel: – Click the Insert button on the menu bar (at the top of the Excel page) • • • • Go to the function option Choose Statistical from the Function Category window (a list of all available statistical functions will appear in the Function Name window) Choose the NEGBINOMDIST function Type in parameters: – number_f => X – Number_s => r – Probability_s => p 23 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Geometric Distribution If repeated independent trials can result in a success with probability p and a failure with probability q = 1 - p, then the probability mass function of the random variable x, the number of the trial on which the first success occurs, is: g(x;p) = pqx-1, x = 1,2,3… 24 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Geometric Distribution The mean and variance of a random variable following the geometric distribution are 1 μ , p 1 p σ 2 p 2 25 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Geometric Distribution - Example At “busy time” a telephone exchange is very near capacity, so callers have difficulty placing their calls. It may be on interest to know the number of attempts necessary in order to gain a connection. Suppose that we let p = 0.05 be the probability of a connection during a busy time. We are interested in knowing the probability that 5 attempts are necessary for a successful call. 26 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08 The Geometric Distribution - Example Solution The random variable X is the number of attempts for a successful call. Then X~G(0.05), So that for with x = 5 and p = 0.05 yields: P(X=x) = g(5;0.05) = (0.05)(0.95)4 = 0.041 1 And the expected number of attempts is μ 20 0.05 27 Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08