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6-1. Discrete Distributions Probability Models Probability Models Discrete Distributions Discrete Distributions Uniform Distribution Uniform Distribution.

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Presentation on theme: "6-1. Discrete Distributions Probability Models Probability Models Discrete Distributions Discrete Distributions Uniform Distribution Uniform Distribution."— Presentation transcript:

1 6-1

2 Discrete Distributions Probability Models Probability Models Discrete Distributions Discrete Distributions Uniform Distribution Uniform Distribution Bernoulli Distribution Bernoulli Distribution Binomial Distribution Binomial Distribution Poisson Distribution Poisson Distribution Chapter 66 McGraw-Hill/Irwin© 2008 The McGraw-Hill Companies, Inc. All rights reserved.

3 6-3 Probability Models A random (or stochastic) process is a repeatable random experiment.A random (or stochastic) process is a repeatable random experiment. Probability can be used to analyze random (or stochastic) processes and to understand business processes.Probability can be used to analyze random (or stochastic) processes and to understand business processes. For example, each call arriving at the L.L. Bean order center is a random experiment in which the variable of interest is the amount of the order.For example, each call arriving at the L.L. Bean order center is a random experiment in which the variable of interest is the amount of the order.  Probability Models

4 6-4 Discrete Distributions A random variable is a function or rule that assigns a numerical value to each outcome in the sample space of a random experiment.A random variable is a function or rule that assigns a numerical value to each outcome in the sample space of a random experiment.  Random Variables Nomenclature: - Capital letters are used to represent random variables (e.g., X, Y). - Lower case letters are used to represent values of the random variable (e.g., x, y).Nomenclature: - Capital letters are used to represent random variables (e.g., X, Y). - Lower case letters are used to represent values of the random variable (e.g., x, y). A discrete random variable has a countable number of distinct values.A discrete random variable has a countable number of distinct values.

5 6-5  Probability Distributions A discrete probability distribution assigns a probability to each value of a discrete random variable X.A discrete probability distribution assigns a probability to each value of a discrete random variable X. To be a valid probability, each probability must be betweenTo be a valid probability, each probability must be between 0  P( x i )  1 and the sum of all the probabilities for the values of X must be equal to unity.and the sum of all the probabilities for the values of X must be equal to unity. Discrete Distributions

6 6-6 When you flip a coin three times, the sample space has eight equally likely simple events. They are:  Example: Coin Flips 1 st Toss2 nd Toss 3 rd Toss HHHHHHHHHHHH HHTHHTHHTHHT HTHHTHHTHHTH HTTHTTHTTHTT THHTHHTHHTHH THTTHTTHTTHT TTHTTHTTHTTH TTTTTTTTTTTT Discrete Distributions

7 6-7 If X is the number of heads, then X is a random variable whose probability distribution is as follows: Possible Events x P(x)P(x)P(x)P(x) TTT01/8 HTT, THT, TTH 13/8 HHT, HTH, THH 23/8 HHH31/8 Total1 Discrete Distributions  Example: Coin Flips

8 6-8 Note that the values of X need not be equally likely. However, they must sum to unity. Note also that a discrete probability distribution is defined only at specific points on the X-axis. Discrete Distributions  Example: Coin Flips

9 6-9 The expected value E(X) of a discrete random variable is the sum of all X-values weighted by their respective probabilities.The expected value E(X) of a discrete random variable is the sum of all X-values weighted by their respective probabilities. The E(X) is a measure of central tendency.The E(X) is a measure of central tendency. If there are n distinct values of X,If there are n distinct values of X, Discrete Distributions  Expected Value

10 6-10 The probability distribution of emergency service calls on Sunday by Ace Appliance Repair is: What is the average or expected number of service calls? x P(x)P(x)P(x)P(x) Total1.00 Discrete Distributions  Example: Service Calls

11 6-11 The sum of the xP(x) column is the expected value or mean of the discrete distribution. xP(x)P(x)xP(x) Total First calculate x i P(x i ): Discrete Distributions  Example: Service Calls

12 6-12 This particular probability distribution is not symmetric around the mean  = However, the mean is still the balancing point, or fulcrum.  = 2.75 Because E(X) is an average, it does not have to be an observable point. Discrete Distributions  Example: Service Calls

13 6-13 Expected value is the basis of life insurance.Expected value is the basis of life insurance. For example, what is the probability that a 30-year- old white female will die within the next year?For example, what is the probability that a 30-year- old white female will die within the next year? Based on mortality statistics, the probability is and the probability of living another year is = Based on mortality statistics, the probability is and the probability of living another year is = What premium should a life insurance company charge to break even on a $500,000 1-year term policy?What premium should a life insurance company charge to break even on a $500,000 1-year term policy? Discrete Distributions  Application: Life Insurance

14 6-14 Let X be the amount paid by the company to settle the policy. EventxP(x)P(x)xP(x) Live Die500, Total Source: Centers for Disease Control and Prevention, National Vital Statistics Reports, 47, no. 28 (1999). The total expected payout is So, the premium should be $295 plus whatever return the company needs to cover administrative overhead and profit. Discrete Distributions  Application: Life Insurance

15 6-15 Expected value can be applied to raffles and lotteries.Expected value can be applied to raffles and lotteries. If it costs $2 to buy a ticket in a raffle to win a new car worth $55,000 and 29,346 raffle tickets are sold, what is the expected value of a raffle ticket?If it costs $2 to buy a ticket in a raffle to win a new car worth $55,000 and 29,346 raffle tickets are sold, what is the expected value of a raffle ticket? If you buy 1 ticket, what is the chance you willIf you buy 1 ticket, what is the chance you will win = 1 29,346 29,346 lose = 29,345 29,345 29,346 29,346 Discrete Distributions  Application: Raffle Tickets

16 6-16 Now, calculate the E(X):Now, calculate the E(X): E(X)= (value if you win)P(win) + (value if you lose)P(lose) = (55,000) 129,346 + (0) 29,34529,346 = (55,000)( ) + (0)( ) = $1.87 The raffle ticket is actually worth $1.87. Is it worth spending $2.00 for it?The raffle ticket is actually worth $1.87. Is it worth spending $2.00 for it? Discrete Distributions  Application: Raffle Tickets

17 6-17 If there are n distinct values of X, then the variance of a discrete random variable is:If there are n distinct values of X, then the variance of a discrete random variable is: The variance is a weighted average of the dispersion about the mean and is denoted either as  2 or V(X).The variance is a weighted average of the dispersion about the mean and is denoted either as  2 or V(X). The standard deviation is the square root of the variance and is denoted .The standard deviation is the square root of the variance and is denoted . Discrete Distributions  Variance and Standard Deviation

18 6-18 The probability distribution of room rentals during February is: The Bay Street Inn is a 7-room bed-and-breakfast in Santa Theresa, Ca. xP(x)P(x) Total1.00 Discrete Distributions  Example: Bed and Breakfast

19 6-19 First find the expected value xP(x)P(x) Total1.00 = 4.71 rooms x P(x)  = 4.71 Discrete Distributions  Example: Bed and Breakfast

20 6-20 The E(X) is then used to find the variance: x P(x)P(x)P(x)P(x) x P(x) Total1.00  = 4.71 = rooms 2 The standard deviation is:  = = rooms = rooms [x]2 P(x)[x]2 P(x)[x]2 P(x)[x]2 P(x)  2 = [x]2[x]2[x]2[x] Discrete Distributions  Example: Bed and Breakfast

21 6-21 The histogram shows that the distribution is skewed to the left and bimodal.  = 2.06 indicates considerable variation around . The mode is 7 rooms rented but the average is only 4.71 room rentals. Discrete Distributions  Example: Bed and Breakfast

22 6-22  Characteristics of the Uniform Distribution The uniform distribution describes a random variable with a finite number of integer values from a to b (the only two parameters).The uniform distribution describes a random variable with a finite number of integer values from a to b (the only two parameters). Each value of the random variable is equally likely to occur.Each value of the random variable is equally likely to occur. Consider the following summary of the uniform distribution:Consider the following summary of the uniform distribution: Uniform Distribution

23 6-23 Parameters a = lower limit b = upper limit PDF Range a  x  b a  x  b Mean Std. Dev. Random data generation in Excel =a+INT((b-a+1)*RAND()) Comments Used as a benchmark, to generate random integers, or to create other distributions. Uniform Distribution

24 6-24 The number of dots on the roll of a die form a uniform random variable with six equally likely integer values: 1, 2, 3, 4, 5, 6The number of dots on the roll of a die form a uniform random variable with six equally likely integer values: 1, 2, 3, 4, 5, 6 PDF for one die CDF for one die What is the probability of rolling any of these?What is the probability of rolling any of these? Uniform Distribution  Example: Rolling a Die

25 6-25 The PDF for all x is:The PDF for all x is: Calculate the standard deviation as:Calculate the standard deviation as: Calculate the mean as:Calculate the mean as: Uniform Distribution  Example: Rolling a Die

26 6-26  On a gas pump, the last two digits (pennies) displayed will be a uniform random integer (assuming the pump stops automatically). PDF CDF The parameters are: a = 00 and b = 99 Uniform Distribution  Application: Pumping Gas

27 6-27 The PDF for all x is:The PDF for all x is: Calculate the standard deviation as:Calculate the standard deviation as: Calculate the mean as:Calculate the mean as: Uniform Distribution  Application: Pumping Gas

28 6-28 A random experiment with only 2 outcomes is a Bernoulli experiment.A random experiment with only 2 outcomes is a Bernoulli experiment. One outcome is arbitrarily labeled a “success” (denoted X = 1) and the other a “failure” (denoted X = 0).One outcome is arbitrarily labeled a “success” (denoted X = 1) and the other a “failure” (denoted X = 0).  is the P(success), 1 –  is the P(failure). Note that P(0) + P(1) = (1 –  ) +  = 1 and 0 <  < 1.Note that P(0) + P(1) = (1 –  ) +  = 1 and 0 <  < 1. “Success” is usually defined as the less likely outcome so that  <.5 for convenience.“Success” is usually defined as the less likely outcome so that  <.5 for convenience.  Bernoulli Experiments Bernoulli Distribution

29 6-29 Bernoulli ExperimentPossible OutcomesProbability of “Success” Flip a coin1 = heads 0 = tails  =.50 Consider the following Bernoulli experiments: Inspect a jet turbine blade1 = crack found 0 = no crack found  =.001 Purchase a tank of gas1 = pay by credit card 0 = do not pay by credit card  =.78 Do a mammogram test1 = positive test 0 = negative test  =.0004  Bernoulli Experiments Bernoulli Distribution

30 6-30 The expected value (mean) of a Bernoulli experiment is calculated as:The expected value (mean) of a Bernoulli experiment is calculated as: The variance of a Bernoulli experiment is calculated as:The variance of a Bernoulli experiment is calculated as: The mean and variance are useful in developing the next model.The mean and variance are useful in developing the next model. Bernoulli Distribution  Bernoulli Experiments

31 6-31 The binomial distribution arises when a Bernoulli experiment is repeated n times.The binomial distribution arises when a Bernoulli experiment is repeated n times. Each Bernoulli trial is independent so the probability of success  remains constant on each trial.Each Bernoulli trial is independent so the probability of success  remains constant on each trial. In a binomial experiment, we are interested in X = number of successes in n trials. So,In a binomial experiment, we are interested in X = number of successes in n trials. So, X = x 1 + x x n The probability of a particular number of successes P(X) is determined by parameters n and .The probability of a particular number of successes P(X) is determined by parameters n and . Binomial Distribution  Characteristics of the Binomial Distribution

32 6-32 The mean of a binomial distribution is found by adding the means for each of the n Bernoulli independent events:The mean of a binomial distribution is found by adding the means for each of the n Bernoulli independent events:  +  + … +  = n  The variance of a binomial distribution is found by adding the variances for each of the n Bernoulli independent events:The variance of a binomial distribution is found by adding the variances for each of the n Bernoulli independent events:  (1-  )+  (1-  ) + … +  (1-  ) = n  (1-  ) The standard deviation isThe standard deviation is n  (1-  ) Binomial Distribution  Characteristics of the Binomial Distribution

33 6-33 Parametersn = number of trials  = probability of success PDF Excel function =BINOMDIST(k,n, ,0) RangeX = 0, 1, 2,..., n Mean n  Std. Dev. Random data generation in Excel Sum n values of =1+INT(2*RAND()) or use Excel’s Tools | Data Analysis Comments Skewed right if .50, and symmetric if  =.50. Binomial Distribution

34 6-34 On average, 20% of the emergency room patients at Greenwood General Hospital lack health insurance. In a random sample of 4 patients, what is the probability that at least 2 will be uninsured? X = number of uninsured patients (“success”) P(uninsured) =  = 20% or.20 P(insured) = 1 –  = 1 –.20 =.80 n = The range is X = 0, 1, 2, 3, 4 patients. 4 patients Binomial Distribution  Application: Uninsured Patients

35 6-35 What is the mean and standard deviation of this binomial distribution?What is the mean and standard deviation of this binomial distribution? Mean =  = n  = (4)(.20) = 0.8 patients Standard deviation =  = = 4(.20(1-.20) = 0.8 patients Binomial Distribution  Application: Uninsured Patients


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