Chapter 5 Discrete Probability Distributions Yandell – Econ 216

Chapter Goals After completing this chapter, you should be able to: Distinguish between discrete and continuous probability distributions Apply the Binomial probability distribution to find probabilities Use PHStat to find Binomial probabilities Yandell – Econ 216

Introduction to Probability Distributions Random Variable Represents a possible numerical value from a random event Random Variables Discrete Random Variable Continuous Random Variable Yandell – Econ 216

Discrete Random Variables Can only assume a countable number of values Examples: Roll a die twice Let X be the number of times 4 comes up (then X could be 0, 1, or 2 times) Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5) Yandell – Econ 216

Discrete Probability Distribution Experiment: Toss 2 Coins. Let X = # heads. 4 possible outcomes Probability Distribution T T X Value Probability 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25 T H listen H T .50 .25 Probability H H 0 1 2 X Yandell – Econ 216

Discrete Probability Distribution A list of all possible [ Xi , P(Xi) ] pairs Xi = Value of Random Variable (Outcome) P(Xi) = Probability Associated with Value Xi’s are mutually exclusive (no overlap) Xi’s are collectively exhaustive (nothing left out) 0 £ P(Xi) £ 1 for each Xi S P(Xi) = 1 Yandell – Econ 216

Discrete Random Variable Summary Measures Expected Value of a discrete distribution (Weighted Average) E(X) = Xi P(Xi) Example: Toss 2 coins, X = # of heads, compute expected value of X: E(X) = (0 x .25) + (1 x .50) + (2 x .25) = 1.0 X P(X) 0 .25 1 .50 2 .25 Yandell – Econ 216

Discrete Random Variable Summary Measures (continued) Standard Deviation of a discrete distribution where: E(X) = Expected value of the random variable X = Values of the random variable P(X) = Probability of the random variable having the value of X Yandell – Econ 216

Discrete Random Variable Summary Measures (continued) Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(X) = 1) Possible number of heads = 0, 1, or 2 Yandell – Econ 216

Two Discrete Random Variables Expected value of the sum of two discrete random variables: E(X + Y) = E(X) + E(Y) =  X P(X) +  Y P(Y) (The expected value of the sum of two random variables is the sum of the two expected values) Yandell – Econ 216

Covariance Covariance between two discrete random variables: σXY =  [Xi – E(X)][Yj – E(Y)]P(XiYj) where: Xi = possible values of the X discrete random variable Yj = possible values of the Y discrete random variable P(Xi ,Yj) = joint probability of the values of Xi and Yj occurring Yandell – Econ 216

Interpreting Covariance Covariance between two discrete random variables: XY > 0 X and Y tend to move in the same direction XY < 0 X and Y tend to move in opposite directions XY = 0 X and Y do not move closely together Yandell – Econ 216

Probability Distributions Ch. 5 Discrete Probability Distributions Continuous Probability Distributions Ch. 6 Binomial Normal Poisson Uniform Hypergeometric Exponential Yandell – Econ 216

Continuous Probability Distributions A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches These can potentially take on any value, depending only on the ability to measure accurately Yandell – Econ 216

Discrete Probability Distributions A discrete random variable is a variable that can assume only a countable number of values Many possible outcomes: number of complaints per day number of TV’s in a household number of rings before the phone is answered Only two possible outcomes: gender: male or female defective: yes or no spreads peanut butter first vs. spreads jelly first Yandell – Econ 216

Application: Return on Investment Let X measure the return on an investment measured in percent and let P(X) be the probability associated with each return. The list of outcomes and their associated probabilities is given in the table on the next slide. Yandell – Econ 216

A Discrete Probability Distribution X (%) P(X) 9 0.05 10 0.15 11 0.30 12 0.20 13 14 0.10 15 This row means that the probability of a 10% return is 15%. Note that these probabilities sum to one, so this list is exhaustive. Yandell – Econ 216

Graphically Yandell – Econ 216

Problems The probability of at least a 14% return is: P(X  14) = P(X=14) + P(X=15) = 0.10 + 0.05 = 0.15 X (%) P(X) 9 0.05 10 0.15 11 0.30 12 0.20 13 14 0.10 15 Yandell – Econ 216

A Useful Trick What is the probability of earning less than 14%? The hard way to find this is: P(X < 14) = P(X=9) + P(X=10) + P(X=11) + P(X=12) + P(X=13) The easy way is: P(X < 14) = 1 - P(X  14) = 1.0 - 0.15 = 0.85 X (%) P(X) 9 0.05 10 0.15 11 0.30 12 0.20 13 14 0.10 15 Yandell – Econ 216

The Mean The Mean of a discrete distribution is the weighted average of the outcomes: Which is: Yandell – Econ 216

Calculating the Mean listen X (%) P(X) X  P(X) 9 0.05 0.45 10 0.15 1.50 11 0.30 3.30 12 0.20 2.40 13 1.95 14 0.10 1.40 15 0.75 Mean = 11.75 listen Yandell – Econ 216

The Variance The Variance of a discrete distribution is the weighted average of the squared deviations from the mean: Which becomes: Yandell – Econ 216

Calculating the Variance X - m (X - m) 2 (X - m) 2  P(X) 9 - 11.75 7.5625 0.378 10 - 11.75 3.0625 0.459 11 - 11.75 0.5625 0.169 12 - 11.75 0.0625 0.013 13 - 11.75 1.5625 0.234 14 - 11.75 5.0625 0.506 15 - 11.75 10.5625 0.528 Variance = 2.288 (PHStat will do the calculations for the mean and variance for us when we use the Binomial distribution) Yandell – Econ 216

The Standard Deviation The standard deviation is the square root of the variance: Empirical Rule: the mean  2 standard deviations includes about 95% of the observations Here, So we know that about 95% of the values from this distribution fall between 8.75 and 14.75. Yandell – Econ 216

Mean  2 Standard Deviations +/- 2σ m Yandell – Econ 216

Practice The probability distribution for damage claims paid per year for collision insurance is: X ($) P(X) 0.90 400 0.04 1000 0.03 2000 0.01 4000 6000 Yandell – Econ 216

Question 1 Find the expected collision payment to determine the annual collision insurance premium that will enable the company to break even. Yandell – Econ 216

Question 2 The insurance company charges an annual rate of $260 for collision coverage. What is the expected value of the collision coverage to a policyholder? (expected payments from the company minus cost of the coverage) Yandell – Econ 216

Another type of customer Consider a different class of customer with the following claim distribution: X ($) P(X) 0.80 400 0.06 1000 0.05 2000 0.03 4000 6000 Yandell – Econ 216

Questions 3&4 How do your answers change for questions 1&2 using this different type of customer? How do these customers compare? Yandell – Econ 216

Probability Distributions The Binomial Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson Hypergeometric Yandell – Econ 216

The Binomial Distribution Characteristics of the Binomial Distribution: A trial has only two possible outcomes – “success” or “failure” There is a fixed number, n, of identical trials The trials of the experiment are independent of each other The probability of a success, π, remains constant from trial to trial If π represents the probability of a success, then (1- π) is the probability of a failure Yandell – Econ 216

Binomial Distribution Settings A manufacturing plant labels items as either defective or acceptable A firm bidding for a contract will either get the contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer or reject it Yandell – Econ 216

Counting Rule for Combinations A combination is an outcome of an experiment where x objects are selected from a group of n objects where: n! =n(n - 1)(n - 2) . . . (2)(1) x! = x(x - 1)(x - 2) . . . (2)(1) 0! = 1 (by definition) Yandell – Econ 216

Binomial Distribution Formula ! x n - x P(x) = π (1- π) x ! ( n - x ) ! listen P(X = x) = probability of x successes in n trials, with probability of success π on each trial x = number of ‘successes’ in sample, (x = 0, 1, 2, ..., n) π = probability of “success” per trial 1 - π = probability of “failure” n = number of trials (sample size) Example: Flip a coin four times, let X = # heads: n = 4 π = 0.5 1 - π = (1 - .5) = .5 x = 0, 1, 2, 3, 4 Yandell – Econ 216

Don’t Worry We will use PHStat to get Binomial Probabilities More examples and sample PHStat output appear soon Yandell – Econ 216

Binomial Distribution The shape of the binomial distribution depends on the values of π and n Mean P(x) n = 5, π = 0.1 .6 .4 .2 Here, n = 5 and π = .1 X 1 2 3 4 5 n = 5, π = 0.5 P(x) listen .6 .4 Here, n = 5 and π = .5 .2 X 1 2 3 4 5 Yandell – Econ 216

Binomial Distribution Characteristics Mean Variance and Standard Deviation Where n = sample size π = probability of success 1 – π = probability of failure Yandell – Econ 216

Binomial Characteristics Examples n = 5, π = 0.1 Mean P(x) .6 .4 .2 X 1 2 3 4 5 n = 5, π = 0.5 P(x) .6 .4 .2 X 1 2 3 4 5 Yandell – Econ 216

Using Binomial Tables Examples: n x π=.15 π=.20 π=.25 π=.30 π=.35 π=.40 π=.45 π=.50 10 0 1 2 3 4 5 6 7 8 9 10 0.1969 0.3474 0.2759 0.1298 0.0401 0.0085 0.0012 0.0001 0.0000 0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0055 0.0008 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0368 0.0090 0.0014 0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0689 0.0212 0.0043 0.0005 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.1115 0.0425 0.0106 0.0016 0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.1596 0.0746 0.0229 0.0042 0.0003 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 0 10 π=.85 π=.80 π=.75 π=.70 π=.65 π=.60 π=.55 x n Examples: n = 10, π = .35, x = 3: P(X = 3|n =10, π = .35) = .2522 n = 10, π = .75, x = 2: P(X = 2|n =10, π = .75) = .0004 Yandell – Econ 216

Using PHStat In Excel 2007 Select PHStat / Probability & Prob. Distributions / Binomial… Yandell – Econ 216

In Excel 2010 or 2013 Yandell – Econ 216

Using PHStat Enter desired values in dialog box Here: n = 10 π = .35 (continued) Enter desired values in dialog box Here: n = 10 π = .35 Output for X = 0 to X = 10 will be generated by PHStat Optional check boxes for additional output Yandell – Econ 216

PHStat Output P(X = 3 | n = 10, π = .35) = .2522 Yandell – Econ 216

Sample Problem: A Light Bulb Shipment Consider a large shipment of light bulbs. We will select a sample of five bulbs and determine whether to accept or reject the whole shipment. Binomial Assumptions are satisfied: ‘n’ Identical Trials 5 light bulbs randomly taken from a shipment 2 Mutually Exclusive Outcomes in Each Trial defective or not defective light bulb Constant Probability For Each Trial each light bulb has the same probability, π, of being not defective Yandell – Econ 216

The Light Bulb Problem (continued) Suppose that we randomly select five (n = 5) light bulbs from a shipment. We will accept the shipment if at least four bulbs work. If each bulb has a π = .90 chance of working, what is the probability that we will accept the shipment? That is, what is P(X  4) ? Yandell – Econ 216

PhStat Demo Click here to view the PHStat demo for the light bulb problem: Click here to watch the demonstration Yandell – Econ 216

Probability Distributions The Poisson Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson Hypergeometric Yandell – Econ 216

The Poisson Distribution Characteristics of the Poisson Distribution: The outcomes of interest are rare relative to the possible outcomes The average number of outcomes of interest per unit is  (lambda) The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest The probability of that an outcome of interest occurs in a given segment is the same for all segments Yandell – Econ 216

Poisson Distribution Formula where: x = number of successes per unit  = expected number of successes e = base of the natural logarithm system (2.71828...) Yandell – Econ 216

Poisson Distribution Characteristics Mean Variance and Standard Deviation where  = expected number of successes per unit Yandell – Econ 216

Using Poisson Tables Example: Find P(X = 2) if  = .50 X  0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1 2 3 4 5 6 7 0.9048 0.0905 0.0045 0.0002 0.0000 0.8187 0.1637 0.0164 0.0011 0.0001 0.7408 0.2222 0.0333 0.0033 0.0003 0.6703 0.2681 0.0536 0.0072 0.0007 0.6065 0.3033 0.0758 0.0126 0.0016 0.5488 0.3293 0.0988 0.0198 0.0030 0.0004 0.4966 0.3476 0.1217 0.0284 0.0050 0.4493 0.3595 0.1438 0.0383 0.0077 0.0012 0.4066 0.3659 0.1647 0.0494 0.0111 0.0020 Example: Find P(X = 2) if  = .50 Yandell – Econ 216

Graph of Poisson Probabilities Graphically:  = .50 X  = 0.50 1 2 3 4 5 6 7 0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 P(X = 2) = .0758 Yandell – Econ 216

Poisson Distribution Shape The shape of the Poisson Distribution depends on the parameter  :  = 0.50  = 3.00 Yandell – Econ 216

Probability Distributions The Hypergeometric Distribution Probability Distributions Discrete Probability Distributions Binomial Poisson Hypergeometric Yandell – Econ 216

The Hypergeometric Distribution “n” trials in a sample taken from a finite population of size N Sample taken without replacement Trials are dependent Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population Yandell – Econ 216

Hypergeometric Distribution Formula (Two possible outcomes per trial) Where N = Population size A = number of successes in the population N – A = number of failures in the population n = sample size X = number of successes in the sample n – X = number of failures in the sample Yandell – Econ 216

Hypergeometric Distribution Formula Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective? N = 10 n = 3 A = 4 X = 2 Yandell – Econ 216

Hypergeometric Distribution in PHStat Select: Add Ins / PHStat / Probability & Prob. Distributions / Hypergeometric … Yandell – Econ 216

Hypergeometric Distribution in PHStat (continued) Complete dialog box entries and get output … N = 10 n = 3 A = 4 X = 2 P(X = 2) = 0.3 Yandell – Econ 216

Chapter Summary Distinguished between discrete and continuous probability distributions Examined discrete probability distributions and their summary measures (Binomial, Poisson, and Hypergeometric) Yandell – Econ 216