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## Lesson 4:

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**Lesson 4:**Discrete Probability Distributions**Random Variables and probability distribution**• A random variable is a numerical value determined by the outcome of an experiment. A random variable is often denoted by a capital letter, e.g., X or Y. • A probability distribution is the listing of all possible outcomes of an experiment and the corresponding probability.**Types of Probability Distributions**• A discrete probability distribution can assume only certain outcomes (need not be finite) – for random variables that take discrete values. • The number of students in a class. • The number of children in a family. • A continuous probability distribution can assume an infinite number of values within a given range – for random variables that take continuous values. • The time it takes an executive to drive to work. • The amount of money spent on your last haircut. • The distance students travel to class.**Types of Probability Distributions**Probability distribution may be classified according to the number of random variables it describes.**Discrete Random Variables**• Can only take on a countable number of values Examples: • Roll a die once Let X be the number of times “4” comes up (then X could be 0, or 1 time) • Roll a die twice Let X be the number of times “4” comes up (then X could be 0, 1, or 2 times)**Discrete Random Variables**• Can only take on a countable number of values Examples: • Toss a coin once. Let X be the number of heads (then X = 0 or 1) • Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5)**Example 1: Tossing coin(s) Discrete Probability Distribution**Experiment: Toss 1 Coin. Let X = # heads. Show P(x) , i.e., P(X = x) , for all values of x: 2 possible outcomes Probability Distribution T H**Example 1: Tossing coin(s) Discrete Probability Distribution**Experiment: Toss 2 Coins. Let X = # heads. Show P(x) , i.e., P(X = x) , for all values of x: 4 possible outcomes Probability Distribution T T T H H T H H**Example 1: Tossing coin(s) Discrete Probability Distribution**• Consider a random experiment in which a coin is tossed three times. Let x be the number of heads. Let H represent the outcome of a head and T the outcome of a tail. • The possible outcomes for such an experiment will be: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH. • Thus the possible values of x (number of heads) are P(x=0) =1/8 P(x=1) =3/8 P(x=2) =3/8 P(x=3) =1/8 x=0: TTTx=1: TTH, THT, HTT x=2: THH, HTH, HHTx=3: HHH**Features of a Univariate Discrete Distribution**• Let x1,…,xN be the list of all possible outcomes (N of them). • The main features of a discrete probability distribution are: • The probability of a particular outcome, P(xi), is between 0 and 1.00. • The sum of the probabilities of the various outcomes is 1.00. That is, P(x1) + … + P(xN) = 1 • The outcomes are mutually exclusive. That is, P(x1 and x2) = 0 and P(x1 or x2) = P(x1)+ P(x2) • Generally, for all i not equal to k. P(xi and xk) = 0. P(xi or xk) = P(xi)+ P(xk)**Features of a Univariate Discrete Distribution**Can the following be a probability distribution of a random variable?**Cumulative Probability Function**• The cumulative probability function, denoted F(c), shows the probability that X is less than or equal to c F(c) = P(X≤c) • In other words, F(c) = ∑x≤c P(X=x)**Moments of a random variable**The n-th moment is defined as the expectation of the n-th power of a random variable: E(Xn) The n-th centralized moment is defined as: E[X-E(X)]n**The Expectation (Mean) of Discrete Probability Distribution**• The expectation or mean of a discrete random variable X, is computed by the formula: • Example: Toss 2 coins, x = # of heads, what is the expected value of x? • P(0) = 0.25, P(1)=0.5, P(2)=0.25. • E(X) = P(0)*0 + P(1)*1 + P(2)*2 = 1**Transformation of Random variables**• A transformation of random variable(s) results in a new random variable. • For example, if X is a random variable, the following are also random variables: • Z=2X • Z=3+2X • Z=X2 • Z=log(X) Example: X = amount of insurance policy underwritten by an agent for the month Z = income earned for the month**Transformation of Random variables**• Example: Z=3+2X • X = amount of insurance policy underwritten by an agent for the month • Z = income earned for the month P(x) = P(3+2x)**Expectation of a linear transformed random variable**• If a and b are constants and X is a random variable, then E(a) = a E(bX) = bE(X) E(a+bX) = a+bE(X)**Expectation of a general functions of random variables**• If P(x) is the probability function of a discrete random variable X, and g(X) is some function of X , then the expected value of function g is E[g(x)] = g( E(x) ) = g( ∑x xP(x) ) ✓ E[g(x)] = E[g(x)] = ∑x g(x)P(x)**Expectation of a general functions of random variables**Experiment: Toss 2 Coins. Let X = # heads. g(x) = x2 E[g(x)] = g( E(x) ) = g( ∑x xP(x) ) = (0*0.25 + 1*0.5 + 2*0.25)2=12=1 ✓ E[g(x)] = E[g(x)] = ∑x g(x)P(x) = 02*0.25 + 12*0.5 + 22*0.25=1.5**Variance of a discrete random variable**• For univariate discrete probability distribution • Example: Toss 2 coins, x = # of heads, what is the variance of x? • P(0) = 0.25, P(1)=0.5, P(2)=0.25. • E(X) = P(0)*0 + P(1)*1 + P(2)*2 = 1 • V(X) = P(0)*(0-1)2 + P(1)*(1-1)2 + P(2)*(2-1)2 = 0.5**Variance of a linear transformed random variable**• If a and b are constants and X is a random variable, then Var(a) = 0 Var(bX) = b2Var(X) Var(a+bX) = b2Var(X)**The Binomial Distribution**Probability Distributions Discrete Probability Distributions Binomial Hypergeometric Poisson**Bernoulli Distribution**• Consider only two outcomes: “success” or “failure” • Let P denote the probability of success • Let 1 – P be the probability of failure • Define random variable X: x = 1 if success, x = 0 if failure • Then the Bernoulli probability function is**Bernoulli DistributionMean and Variance**• The mean is µ = P • The variance is σ2 = P(1 – P)**Sequences of x Successes inn Trials**• The number of sequences with x successes in n independent trials is: C(n,x) = n!/([x!(n-x)!] where n! = n*(n – 1)*(n – 2)* . . . *1 and 0! = 1 • These sequences are mutually exclusive, since no two can occur at the same time**Binomial Probability Distribution**• The binomial distribution has the following characteristics: • An outcome of an experiment is classified into one of two mutually exclusive categories, such as a success or failure. • The data collected are the results of counts in a series of trials. • The probability of success stays the same for each trial. • The trials are independent. • For example, tossing an unfair coin three times. • H is labeled success and T is labeled failure. • The data collected are number of H in the three tosses. • The probability of H stays the same for each toss. • The results of the tosses are independent.**Binomial Probability Distribution**• To construct a binomial distribution, let • n be the number of trials • x be the number of observed successes • pbe the probability of success on each trial • The formula for the binomial probability distribution is: P(x) = C(n,x) px(1- p)n-x**Binomial Probability Distribution**• The formula for the binomial probability distribution is: P(x) = C(n,x) px(1- p)n-x C(n,x) = n!/([x!(n-x)!] TTT, TTH, THT, THH, HTT, HTH, HHT, HHH. • X=number of heads • The coin is fair, i.e., P(head) = 1/2. • P(x=0) = C(3,0) 0.50(1- 0.5)3-0 =3!/(0!3!) (1) (1/8)=1/8 • P(x=1) = C(3,1) 0.51(1- 0.5)3-1 =3!/(1!2!) (1) (1/8)= 3/8 • P(x=2) = C(3,2) 0.52(1- 0.5)3-2 =3!/(2!1!) (1) (1/8)= 3/8 • P(x=3) = C(3,3) 0.53(1- 0.5)3-3 =3!/(3!0!) (1) (1/8)= 1/8 When the coin is not fair, simple counting rule will not work.**The density functions of binomial distributions with n=20**and different success rates p P(x) = C(n,x) px(1- p)n-x**Example: Binomial**P(x) = C(n,x) px(1- p)n-x x = number of students who will receive scholarships among those who have a semester GPA>3.5 , q = 1 – p = 1 - 0.1 = 0.9 , p = 0.1 n = 4 Find the probability that 2 of the 4 students among those who have a semester GPA>3.5 will receive scholarships.**Example: Binomial**P(x) = C(n,x) px(1- p)n-x After Asian Financial Crisis, Thailand’s Department of Labor reports that 20% of the workforce is unemployed. From a sample of 14 workers, calculate the following probabilities: • Exactly three are unemployed. • At least three are unemployed. • At least one are unemployed.**Example: Binomial**P(x) = C(n,x) px(1- p)n-x • The probability of exactly 3: • The probability of at least 3 is: • The probability of at least one being unemployed:**Mean & Variance of the Binomial Distribution**P(x) = C(n,x) px(1- p)n-x • The mean is found by: • The variance is found by:**Example: Binomial**P(x) = C(n,x) px(1- p)n-x • Since p =.2 and n=14. • Hence, the meanis: =np = 14(.2) = 2.8. • The varianceis: 2= np(1- p ) = (14)(.2)(.8) =2.24.**Using Binomial Tables**P(x) = C(n,x) px(1- p)n-x Examples: n = 10, x = 3, P = 0.35: P(x = 3|n =10, p = 0.35) = .2522 n = 10, x = 8, P = 0.45: P(x = 8|n =10, p = 0.45) = .0229**Finite Population**• A finite population is a population consisting of a fixed number of known individuals, objects, or measurements. Examples include: • The number of students in this class. • The number of cars in the parking lot.**Hypergeometric Distribution**• The hypergeometric distribution has the following characteristics: • There are only 2 possible outcomes. • The probability of a success is not the same on each trial. • It results from a count of the number of successes in a fixed number of trials.**Example: Hypergeometric Distribution**In a bag containing 7 red chips and 5 blue chips you select 2 chips one after the other without replacement. 6/11 R2 R1 7/12 B2 5/11 R2 7/11 B1 5/12 4/11 B2 The probability of a success (red chip) is not the same on each trial.**Hypergeometric Distribution**• The formula for finding a probability using the hypergeometric distribution is: where Nis the size of the population, S is the number of successes in the population,xis the number of successes in a sample of n observations.**Hypergeometric Distribution**• Use the hypergeometric distribution to find the probability of a specified number of successes or failures if: • the sample is selected from a finite population without replacement (recall that a criteria for the binomial distribution is that the probability of success remains the same from trial to trial) • the size of the sample n is greater than 5% of the size of the population N .**The density functions of hypergeometric distributions with**N=100, n=20 and different success rates p (=S/N).**Example: Hypergeometric Distribution**• The University Examination Board has a list of 10 reported cheating cases. Suppose only 4 of the reported cheating cases are real cheating and the University Examination Board will only be able to investigate five of the cheating cases. What is the probability that three of five reported cheating cases randomly selected to be investigated are real cheating cases? • Number of reported cases, N=10 • Number of real cheating cases out of 10, S=4 • Number of cases investigated, n=5**Example: Hypergeometric Distribution**• 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded? • N = 10, S=4, n=3, x=2 The probability that 2 of the 3 selected computers have illegal software loaded is 0.30, or 30%.**Poisson Probability Distribution**The formula for the binomial probability distribution is: P(x) = C(n,x) px(1- p)n-x • The binomial distribution becomes more skewed to the right (positive) as the probability of success become smaller. • The limiting form of the binomial distribution where the probability of success p is small and n is large is called the Poisson probability distribution. Let l=np, i.e., p=l/n. limn C(n,x) (l/n)x(1- l/n)n-x = [lxe-l]/x!**The Poisson Distribution**• Apply the Poisson Distribution when: • You wish to count the number of times an event occurs in a given continuous interval • The probability that an event occurs in one subinterval is very small and is the same for all subintervals • The number of events that occur in one subinterval is independent of the number of events that occur in the other subintervals • There can be no more than one occurrence in each subinterval • The average number of events per unit is (lambda)**Poisson Probability Distribution**ThePoissondistribution can be described mathematically using the formula: where • is the mean number of successes in a particular interval of time, • e is the constant 2.71828, and • x is the number of successes.**Poisson Distribution Characteristics**• Mean • Variance and Standard Deviation where = expected number of successes per unit**Poisson Probability Distribution**• The mean number of successes can be determined in binomial situations by np, where n is the number of trials and p the probability of a success. • The variance of the Poisson distribution is also equal to np. • X, the number of success generally has no specific upper limit. • Probability distribution always skewed to the right. • Becomes symmetrical when gets large.**Example: Poisson Probability Distribution**• The Sylvania Urgent Care facility specializes in caring for minor injuries, colds, and flu. For the evening hours of 6-10 PM the mean number of arrivals is 4.0 per hour. What is the probability of 2 arrivals in an hour?