Discrete Distributions

Applications of the Binomial Distribution A manufacturing plant labels items as either defective or acceptable A firm bidding for contracts will either get a 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 Your team either wins or loses the football game at the company picnic

Bernoulli Random Variable If an experiment consists of a single trial and the outcome of the trial can only be either a success * or a failure, then the trial is called a Bernoulli trial. The number of success X in one Bernoulli trial, which can be 1 or 0, is a Bernoulli random variable. Note: If p is the probability of success in a Bernoulli experiment, the E(X) = p and V(X) = p(1 p). * The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a success, although it is not a positive result.

The Binomial Random Variable Consider a Bernoulli Process in which we have a sequence of n identical trials satisfying the following conditions: 1. Each trial has two possible outcomes, called success *and failure. The two outcomes are mutually exclusive and exhaustive.. The probability of success, denoted by p, remains constant from trial to trial. The probability of failure is denoted by q, where q = 1-p. 3. The n trials are independent. That is, the outcome of any trial does not affect the outcomes of the other trials. A random variable, X, that counts the number of successes in n Bernoulli trials, where p is the probability of success* in any given trial, is said to follow the binomial probability distribution with parameters n (number of trials) and p (probability of success). We call X the binomial random variable. * The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a success, although it is not a positive result.

The Binomial Distribution: Properties A fixed number of observations, n ex. 15 tosses of a coin; ten light bulbs taken from a warehouse Two mutually exclusive and collectively exhaustive categories ex. head or tail in each toss of a coin; defective or not defective light bulb; having a boy or girl Generally called success and failure Probability of success is p, probability of failure is 1 p Constant probability for each observation ex. Probability of getting a tail is the same each time we toss the coin

The Binomial Distribution: Properties Observations are independent The outcome of one observation does not affect the outcome of the other Two sampling methods Infinite population without replacement Finite population with replacement

Binomial Probabilities (Introduction) Suppose we toss a single fair and balanced coin five times in succession, and let X represent the number of heads. There are 5 = 3 possible sequences of H and T (S and F) in the sample space for this experiment. Of these, there are 10 in which there are exactly heads (X=): HHTTT HTHTT HTTHT HTTTH THHTT THTHT THTTH TTHHT TTHTH TTTHH The probability of each of these 10 outcomes is p q 3 = (1/) (1/) 3 =(1/3), so the probability of heads in 5 tosses of a fair and balanced coin is: P(X = ) = 10 * (1/3) = (10/3) = 0.315 10 (1/3) Number of outcomes with heads Probability of each outcome with heads

Binomial Probabilities (continued) P(X=) = 10 * (1/3) = (10/3) =.315 Notice that this probability has two parts: 10 (1/3) Number of outcomes with heads Probability of each outcome with heads In general: 1. The probability of a given sequence of x successes out of n trials with probability of success p and probability of failure q is equal to: p x q (n-x). The number of different sequences of n trials that result in exactly x successes is equal to the number of choices of x elements out of a total of n elements. This number is denoted: n ncx x n! x!( n x)!

The Binomial Probability Distribution The binomial probability distribution: Px () n x pq n! x!( n x)! pq x ( nx) x ( nx) where : p is the probability of success in a single trial, q = 1-p, n is the number of trials, and x is the number of successes. N umber of successes, x Probability P(x) n! n 0 p q 0!( n 0)! n! 1 ( n 1) 1 p q 1!( n 1)! n! n p q!( n )! n! n 3 p q 3!( n 3)! n! n p q n!( n n)! 0 ( 0) ( ) 3 ( 3) n ( n n) 1.00

The Cumulative Binomial Probability Table n=5 p x 0.01 0.05 0.10 0.0 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99 0.951.774.590.38.168.078.031.010.00.000.000.000.000 1.999.977.919.737.58.337.187.087.031.007.000.000.000 1.000.999.991.94.837.683.500.317.163.058.009.001.000 3 1.000 1.000 1.000.993.969.913.813.663.47.63.081.03.001 4 1.000 1.000 1.000 1.000.998.990.969.9.83.67.410.6.049 Cumulative Binomial Probability Distribution and Binomial Probability Distribution of H,the Number of Heads Appearing in Five Tosses of a Fair Coin h F(h) P(h) 0 0.031 0.031 1 0.187 0.156 0.500 0.313 3 0.813 0.313 4 0.969 0.156 5 1.000 0.031 1.000 Deriving Individual Probabilities from Cumulative Probabilities Fx ( ) PX ( x) Pi ( ) P() 3 F() 3 F(). 813. 500. 313 all ix P(X) = F(x) - F(x - 1) For example:

Calculating Binomial Probabilities - Example 60% of of Brooke shares are are owned by by LeBow. A random sample of of 15 15 shares is is chosen. What is is the the probability that at at most three of of them will be be found to to be be owned by by LeBow? Check by by excel if if table values are are not available. n=15 p.50.60.70 0.000.000.000 1.000.000.000.004.000.000 3.018.00.000 4.059.009.001............ F ( x) F (3) P( X P( X x) alli x 3) 0.00 P( i)

Mean, Variance, and Standard Deviation of the Binomial Distribution Mean of a binomial distribution: E( X) np Variance of a binomial distribution: V ( X ) npq Standard deviation of a binomial distribution: For example, if H counts the number of heads in five tosses of a fair coin : E( H ) (5)(.5).5 H V ( H ) (5)(.5)(.5) 1.5 H = SD(X) = npq H SD( H ) 1.5 1.118

Shape of the Binomial Distribution p = 0.1 p = 0.3 p = 0.5 Binomial Probability: n=4 p=0.1 Binomial Probability: n=4 p=0.3 Binomial Probability: n=4 p=0.5 0.7 0.7 0.7 0.6 0.6 0.6 n = 4 P(x) 0.5 0.4 0.3 P(x) 0.5 0.4 0.3 P(x) 0.5 0.4 0.3 0. 0. 0. 0.1 0.1 0.1 0.0 0 1 x 3 4 0.0 0 1 x 3 4 0.0 0 1 x 3 4 Binomial Probability: n=10 p=0.1 Binomial Probability: n=10 p=0.3 Binomial Probabil i ty: n=10 p=0.5 05. 05. 05. n = 10 P(x) 04. 03. 0. P(x) 04. 03. 0. P(x) 04. 03. 0. 01. 01. 01. 00. 0 1 3 4 5 x 6 7 8 9 10 00. 0 1 3 4 5 x 6 7 8 9 10 00. 0 1 3 4 5 x 6 7 8 9 10 Binomial Probability: n=0 p=0.1 Binomial Probability: n=0 p=0.3 Binomial Probability: n=0 p=0.5 n = 0 P(x) 0. 0.1 P(x) 0. 0.1 P(x) 0. 0.1 0.0 0 1 3 4 5 6 7 8 9 10111131415161718190 x 0.0 0 1 3 4 5 6 7 8 9 10111131415161718190 x 0.0 0 1 3 4 5 6 7 8 9 10111131415161718190 x Binomial distributions become more symmetric as n increases and as p 0.5.

Example What is the probability of one success in five observations if the probability of success is.1? X = 1, n = 5, and p =.1 P(X 1) n! p X!(n X)! 5! (.1) 1!(5 1)! X 1 (1 p) (1.1) nx 51 (5)(.1)(.9) 4.3805

Example Suppose the probability of purchasing a defective computer is 0.0. What is the probability of purchasing defective computers is a lot of 10?

X =, n = 10, and p =.0 P(X ) n! X!(n X)! p X (1 p) nx 10!!(10 (.0) )! (1.0) 10 (45)(.0004)(.8508).01531

Problem A manufacturing company of south Maharashtra found that after launching a golden handshake scheme for voluntary retirement, 10% of workers are unemployed. What is the probability of obtaining three or fewer unemployed workers in a random sample of 30 in a survey conducted by the company?

We have to find out the probability of getting (a) zero unemployed, x = 0; (b) one unemployed, x = 1; (c) two unemployed x = ; and (d) three unemployed, x = 3 workers.

Problem According to the U.S. Census Bureau, approximately 6% of all workers in jackson, Mississippi, are unemployed. In conducting a random telephone survey in Jackson, what is the probability of getting two or fewer unemployed workers in a sample of 0?

Answer n 0 p. 06 q. 94 PX ( ) PX ( 0) PX ( 1) PX ( ). 901. 3703. 46. 8850 P( X 0) 0! 0!(0 0)!. 06. 94 0 00 ()()(. 1 1 901). 901 P( X 1) 0! 1!( 0 1)!. 06. 94 1 01 ( 0)(. 06)(. 3086). 3703 P( X ) 0!!( 0 )!. 06. 94 0 ( 190)(. 0036)(. 383). 46

The Binomial Distribution Using Binomial Tables n = 10 x p=.0 p=.5 p=.30 p=.35 p=.40 p=.45 p=.50 0 1 3 4 5 6 7 8 9 10 0.1074 0.684 0.300 0.013 0.0881 0.064 0.0055 0.0008 0.0001 0.0563 0.1877 0.816 0.503 0.1460 0.0584 0.016 0.0031 0.0004 0.08 0.111 0.335 0.668 0.001 0.109 0.0368 0.0090 0.0014 0.0001 0.0135 0.075 0.1757 0.5 0.377 0.1536 0.0689 0.01 0.0043 0.0005 0.0060 0.0403 0.109 0.150 0.508 0.007 0.1115 0.045 0.0106 0.0016 0.0001 0.005 0.007 0.0763 0.1665 0.384 0.340 0.1596 0.0746 0.09 0.004 0.0003 0.0010 0.0098 0.0439 0.117 0.051 0.461 0.051 0.117 0.0439 0.0098 0.0010 10 9 8 7 6 5 4 3 1 0 Examples: p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x n = 10, p =.35, x = 3: P(x = 3 n =10, p =.35) =.5 n = 10, p =.75, x = : P(x = n =10, p =.75) =.0004

Poisson Distribution: Applications Arrivals at queuing systems airports -- people, airplanes, automobiles, baggage banks -- people, automobiles, loan applications computer file servers -- read and write operations The number of scratches in a car s paint The number of mosquito bites on a person The number of computer crashes in a day Defects in manufactured goods number of defects per 1,000 feet of extruded copper wire number of blemishes per square foot of painted surface number of errors per typed page

The Poisson Distribution The Poisson probability distribution is useful for determining the probability of a number of occurrences over a given period of time or within a given area or volume. That is, the Poisson random variable counts occurrences over a continuous interval of time or space. It can also be used to calculate approximate binomial probabilities when the probability of success is small (p 0.05) and the number of trials is large (n 0). Poisson Distribution: x e P( x) for x = 1,,3,... x! where is the mean of the distribution (which also happens to be the variance) and e is the base of natural logarithms (e=.7188...).

The Poisson Distribution Properties Apply the Poisson Distribution when: You wish to count the number of times an event occurs in a given area of opportunity The probability that an event occurs in one area of opportunity is the same for all areas of opportunity The number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunity The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smaller The average number of events per unit is (lambda)

Example Suppose that, on average, 5 cars enter a parking lot per minute. What is the probability that in a given minute, 7 cars will enter? So, X = 7 and λ = 5 P(7) λ e λ X! x e 5 5 7! 7 0.104 So, there is a 10.4% chance 7 cars will enter the parking in a given minute.

Bank customer arrive randomly on weekday afternoons at an average of 3. customer every 4 minutes. A. What is the probability of having 10 customers every 8 minutes. B. What is the probability of having 6 customers every 8 minutes. C. What is the probability of having more than 7 customers in a 4-minute interval on a weekday afternoon. 3. customers / 4 minutes X = 10 customers / 8 minutes Adjusted = 64. customers / 8 minutes P(X) = X e X! PX ( = 10)= 6.4 10 e 10! 64. 0. 058 3. customers / 4 minutes X = 6 customers / 8 minutes Adjusted = 64. customers / 8 minutes P(X) = X e PX ( = 6)= X! 6.4 6 e 6! 64. 0. 1586

Poisson Distribution: Probability Table X 0.5 1.5 1.6 3.0 3. 6.4 6.5 7.0 8.0 0 0.6065 0.31 0.019 0.0498 0.0408 0.0017 0.0015 0.0009 0.0003 1 0.3033 0.3347 0.330 0.1494 0.1304 0.0106 0.0098 0.0064 0.007 0.0758 0.510 0.584 0.40 0.087 0.0340 0.0318 0.03 0.0107 3 0.016 0.155 0.1378 0.40 0.6 0.076 0.0688 0.051 0.086 4 0.0016 0.0471 0.0551 0.1680 0.1781 0.116 0.1118 0.091 0.0573 5 0.000 0.0141 0.0176 0.1008 0.1140 0.1487 0.1454 0.177 0.0916 6 0.0035 0.0047 0.0504 0.0608 0.1586 0.1575 0.1490 0.11 7 0.0008 0.0011 0.016 0.078 0.1450 0.146 0.1490 0.1396 8 0.0001 0.000 0.0081 0.0111 0.1160 0.1188 0.1304 0.1396 9 0.007 0.0040 0.085 0.0858 0.1014 0.141 10 0.0008 0.0013 0.058 0.0558 0.0710 0.0993 11 0.000 0.0004 0.0307 0.0330 0.045 0.07 1 0.0001 0.0001 0.0164 0.0179 0.063 0.0481 13 0.0081 0.0089 0.014 0.096 14 0.0037 0.0041 0.0071 0.0169 15 0.0016 0.0018 0.0033 0.0090 16 0.0006 0.0007 0.0014 0.0045 17 0.000 0.0003 0.0006 0.001 18 0.0001 0.0001 0.000 0.0009

Poisson Distribution: Using the Poisson Tables X 0.5 1.5 1.6 3.0 0 0.6065 0.31 0.019 0.0498 1 0.3033 0.3347 0.330 0.1494 0.0758 0.510 0.584 0.40 3 0.016 0.155 0.1378 0.40 4 0.0016 0.0471 0.0551 0.1680 5 0.000 0.0141 0.0176 0.1008 6 0.0035 0.0047 0.0504 7 0.0008 0.0011 0.016 8 0.0001 0.000 0.0081 9 0.007 10 0.0008 11 0.000 1 0.0001 16. PX ( 4) 0. 0551

Poisson Approximation of the Binomial Distribution Binomial probabilities are difficult to calculate when n is large. Under certain conditions binomial probabilities may be approximated by Poisson probabilities. If n 0 and n p 7, the approximation is acceptable. Use n p. Poisson approximation

The Poisson Distribution - Example Example Telephone manufacturers now offer 1000 different choices for a telephone (as combinations of color, type, options, portability, etc.). A company is opening a large regional office, and each of its 00 managers is allowed to order his or her own choice of a telephone. Assuming independence of choices and that each of the 1000 choices is equally likely, what is the probability that a particular choice will be made by none, one, two, or three of the managers? n = 00 = np = (00)(0.001) = 0. p = 1/1000 = 0.001. e P( 0) 0!. e P() 1 1!. e P( )!. e P() 3 3! 0. 1.. 3. = 0.8187 = 0.1637 = 0.0164 = 0.0011

The Poisson Distribution (continued) =1.0 =1.5 0.4 0.4 0.3 0.3 P(x) 0. P(x) 0. 0.1 0.1 0.0 0 1 X 3 4 0.0 0 1 3 X 4 5 6 7 =4 =10 0. 0.15 0.10 P(x) 0.1 P(x) 0.05 0.0 0.00 0 1 3 4 5 6 7 8 9 10 0 1 3 4 5 6 7 8 9 10111131415161718190 X X

The Poisson Distribution Using Poisson Tables X 0.10 0.0 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0 1 3 4 5 6 7 0.9048 0.0905 0.0045 0.000 0.8187 0.1637 0.0164 0.0011 0.0001 0.7408 0. 0.0333 0.0033 0.0003 0.6703 0.681 0.0536 0.007 0.0007 0.0001 0.6065 0.3033 0.0758 0.016 0.0016 0.000 0.5488 0.393 0.0988 0.0198 0.0030 0.0004 0.4966 0.3476 0.117 0.084 0.0050 0.0007 0.0001 0.4493 0.3595 0.1438 0.0383 0.0077 0.001 0.000 0.4066 0.3659 0.1647 0.0494 0.0111 0.000 0.0003 Example: Find P(X = ) if =.50 P(X ) λ e λ X! X e 0.50 (0.50)!.0758

The Poisson Distribution Shape 0.70 =.50 0.60 X 0 P(X) 0.6065 P(x) 0.50 0.40 0.30 1 3 0.3033 0.0758 0.016 0.0 0.10 4 5 6 7 0.0016 0.000 0.00 0 1 3 4 5 6 7 P(X = ) =.0758 x

The Poisson Distribution Shape The shape of the Poisson Distribution depends on the parameter : 0.70 = 0.50 = 3.00 0.5 0.60 0.50 0.0 P(x) 0.40 0.30 P(x) 0.15 0.10 0.0 0.10 0.05 0.00 0 1 3 4 5 6 7 x 0.00 1 3 4 5 6 7 8 9 10 11 1 x

Poisson Approximation of the Binomial Distribution Binomial X Poisson 15. n 50 p. 03 Error 0 0.31 0.181-0.0051 1 0.3347 0.337 0.005 0.510 0.555 0.0045 3 0.155 0.164 0.0009 4 0.0471 0.0459-0.0011 5 0.0141 0.0131-0.0010 6 0.0035 0.0030-0.0005 7 0.0008 0.0006-0.000 8 0.0001 0.0001 9 X Poisson 30. Binomial n 10, 000 p. 0003 Error 0 0.0498 0.0498 1 0.1494 0.1493 0.40 0.41 3 0.40 0.41 4 0.1680 0.1681 5 0.1008 0.1008 6 0.0504 0.0504 7 0.016 0.016 8 0.0081 0.0081 9 0.007 0.007 10 0.0008 0.0008 11 0.000 0.000 1 0.0001 0.0001 13

Discrete and Continuous Random Variables - Revisited A discrete random variable: counts occurrences has a countable number of possible values has discrete jumps between successive values has measurable probability associated with individual values probability is height A continuous random variable: measures (e.g.: height, weight, speed, value, duration, length) has an uncountably infinite number of possible values moves continuously from value to value has no measurable probability associated with individual values probability is area For example: Binomial n=3 p=.5 x P(x) 0 0.15 1 0.375 0.375 3 0.15 1.000 P(x) 0.4 0.3 0. 0.1 0.0 0 Binomial: n=3 p=.5 1 C1 3 For example: In this case, the shaded area epresents the probability that the task takes between and 3 minutes. P(x) 0.3 0. 0.1 0.0 MinutestoCompleteTask 1 3 4 5 6 Minutes

From a Discrete to a Continuous Distribution The time it takes to complete a task can be subdivided into: Half-Minute Intervals Quarter-Minute Intervals Eighth-Minute Intervals 0.15 Minutes to Complete Task: By Half-Minutes MinutestoCompleteTask:FourthsofaMinute Minutes to Complete Task: Eighths of aminute 0.10 P(x) P(x) P(x) 0.05 0.00 0.0. 1.0 1.5.0.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 0 1 3 4 5 6 7 Minutes Minutes 0 1 3 4 5 6 7 Minutes Or even infinitesimally small intervals: Minutes to Complete Task: Probability Density Function f(z) When When a a continuous continuous random random variable variable has has been been subdivided subdivided into into infinitesimally infinitesimally small small intervals, intervals, a a measurable measurable probability probability can can only only be be associated associated with with an an interval interval of of values, values, and and the the probability probability is is given given by by the the area area beneath beneath the the probability probability density density function function corresponding corresponding to to that that interval. interval. In In this this example, example, the the shaded shaded area area represents represents P( P( X ). ). 0 1 3 Minutes 4 5 6 7

Continuous Random Variables A continuous random variable is is a a random random variable variable that that can can take take on on any any value value in in an an interval interval of of numbers. The The probabilities associated with with a a continuous random random variable variable X are are determined by by the the probability density density function of of the the random random variable. The The function, denoted denoted f(x), f(x), has has the the following properties. 1. 1. f(x) f(x) 0 for for all all x. x... The The probability that that X will will be be between two two numbers a and and b is is equal equal to to the the area area under under f(x) f(x) between a and and b. b. 3. 3. The The total total area area under under the the curve curve of of f(x) f(x) is is equal equal to to 1.00. 1.00. The The cumulative distribution function of of a a continuous random random variable: F(x) F(x) = P(X P(X x) x) =Area =Area under under f(x) f(x) between between the the smallest smallest possible possible value value of of X (often (often-) -) and and the the point point x. x.

Probability Density Function and Cumulative Distribution Function F(x) 1 F(b) F(a) } P(a X b)=f(b) - F(a) f(x) 0 a b x P(a X b) = Area under f(x) between a and b = F(b) - F(a) 0 a b x