© The McGraw-Hill Companies, Inc., 2000 6-1 Chapter 6 Probability Distributions.

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© The McGraw-Hill Companies, Inc., Chapter 6 Probability Distributions

© The McGraw-Hill Companies, Inc., Outline 6-1 Introduction 6-2 Probability Distributions 6-3 Mean, Variance, and Expectation 6-4 The Binomial Distribution

© The McGraw-Hill Companies, Inc., Objectives Construct a probability distribution for a random variable. Find the mean, variance, and expected value for a discrete random variable. Find the exact probability for X successes in n trials of a binomial experiment.

© The McGraw-Hill Companies, Inc., Objectives Find the mean, variance, and standard deviation for the variable of a binomial distribution.

© The McGraw-Hill Companies, Inc., Probability Distributions variable A variable is defined as a characteristic or attribute that can assume different values. random variable A variable whose values are determined by chance is called a random variable.

© The McGraw-Hill Companies, Inc., Probability Distributions variable discrete variable If a variable can assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called a discrete variable. Discrete variables have values that can be counted.

© The McGraw-Hill Companies, Inc., Probability Distributions variable continuous variable.Example - If a variable can assume all values in the interval between two given values then the variable is called a continuous variable. Example - temperature between 68 0 to Continuous random variables are obtained from data that can be measured rather than counted.

© The McGraw-Hill Companies, Inc., Probability Distributions - Tossing Two Coins First Toss T H H T H T Second Toss

© The McGraw-Hill Companies, Inc., HH, HT, TH, TT. From the tree diagram, the sample space will be represented by HH, HT, TH, TT. X X 0, 1, or 2 If X is the random variable for the number of heads, then X assumes the value 0, 1, or Probability Distributions - Tossing Two Coins

© The McGraw-Hill Companies, Inc., Probability Distributions - Tossing Two Coins TTTHHTHH 012 Sample SpaceNumber of Heads

© The McGraw-Hill Companies, Inc., Probability Distributions - Tossing Two Coins

© The McGraw-Hill Companies, Inc., probability distribution A probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation. 6-2 Probability Distributions

© The McGraw-Hill Companies, Inc., Probability Distributions -- Graphical Representation NUMBER OF HEADS P R O B A B I L I T Y Experiment: Toss Two Coins

© The McGraw-Hill Companies, Inc., Mean, Variance, and Expectation for Discrete Variable

© The McGraw-Hill Companies, Inc., Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below. 6-3 Mean for Discrete Variable Mean for Discrete Variable - Example

© The McGraw-Hill Companies, Inc., Mean for Discrete Variable Mean for Discrete Variable - Example  ·   ·  ·  ·  ·  ·  ·  XPX() (/)(/)(/)(/) (/)(/) / That is, when a die is tossed many times, the theoretical mean will be 3.5. That is, when a die is tossed many times, the theoretical mean will be 3.5.

© The McGraw-Hill Companies, Inc., In a family with two children, find the mean number of children who will be girls. The probability distribution is given below. 6-3 Mean for Discrete Variable Mean for Discrete Variable - Example

© The McGraw-Hill Companies, Inc., =.    XPX() (/)(/)(/) That is, the average number of girls in a two-child family is 1. That is, the average number of girls in a two-child family is Mean for Discrete Variable Mean for Discrete Variable - Example

© The McGraw-Hill Companies, Inc., Formula for the Variance of a Probability Distribution The variance of a probability distribution is found by multiplying the square of each outcome by its corresponding probability, summing these products, and subtracting the square of the mean.

© The McGraw-Hill Companies, Inc., Formula for the Variance of a Probability Distribution  = 2 Theformulafortheofa probabilitydistributionis XPX Thestandarddeviationofa probabilitydistributionis variance   222  ·   ()..

© The McGraw-Hill Companies, Inc., Variance of a Probability Distribution Variance of a Probability Distribution - Example The probability that 0, 1, 2, 3, or 4 people will be placed on hold when they call a radio talk show with four phone lines is shown in the distribution below. Find the variance and standard deviation for the data.

© The McGraw-Hill Companies, Inc., Variance of a Probability Distribution Variance of a Probability Distribution - Example

© The McGraw-Hill Companies, Inc., Variance of a Probability Distribution Variance of a Probability Distribution - Example  = 1.59  X 2  P(X)P(X) =3.79  2 = 3.79 – = 1.26  2 = 3.79 – = 1.26

© The McGraw-Hill Companies, Inc., 2000 Now,  = (0)(0.18) + (1)(0.34) + (2)(0.23) + (3)(0.21) + (4)(0.04) =  X 2 P(X) = (0 2 )(0.18) + (1 2 )(0.34) + (2 2 )(0.23) + (3 2 )(0.21) + (4 2 )(0.04) = = 2.53 (rounded to two decimal places).  2 = 3.79 – 2.53 = 1.26  = = Variance of a Probability Distribution Variance of a Probability Distribution - Example

© The McGraw-Hill Companies, Inc., Expectation le expected. ()  · 

© The McGraw-Hill Companies, Inc., Expectation Expectation - Example A ski resort loses $70,000 per season when it does not snow very much and makes $250,000 when it snows a lot. The probability of it snowing at least 75 inches (i.e., a good season) is 40%. Find the expected profit.

© The McGraw-Hill Companies, Inc., The expected profit = ($250,000)(0.40) + (–$70,000)(0.60) = $58, Expectation Expectation - Example Profit, X250,000–70,000 P(X)

© The McGraw-Hill Companies, Inc., The Binomial Distribution binomial experiment A binomial experiment is a probability experiment that satisfies the following four requirements: Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. Each outcome can be considered as either a success or a failure.

© The McGraw-Hill Companies, Inc., The Binomial Distribution There must be a fixed number of trials. The outcomes of each trial must be independent of each other. The probability of success must remain the same for each trial.

© The McGraw-Hill Companies, Inc., The Binomial Distribution binomial distribution. The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.

© The McGraw-Hill Companies, Inc., The Binomial Distribution Notation for the Binomial Distribution: P(S) = p P(S) = p, probability of a success P(F) = 1 – p = q P(F) = 1 – p = q, probability of a failure n n = number of trials X X = number of successes.

© The McGraw-Hill Companies, Inc., Binomial Probability Formula Inabinomialtheprobabilityof exactlyXsuccessesinntrialsis PX n nXX pq Xn X experiment, () ! ()!!   

© The McGraw-Hill Companies, Inc., Binomial Probability Binomial Probability - Example If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices. Solution: Solution: n = 5, X = 3, and p = 1/5. Then, P(3) = [5!/((5 – 3)!3! )](1/5) 3 (4/5)

© The McGraw-Hill Companies, Inc., Binomial Probability Binomial Probability - Example A survey from Teenage Research Unlimited (Northbrook, Illinois.) found that 30% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs.

© The McGraw-Hill Companies, Inc., Binomial Probability Binomial Probability - Example Solution: Solution: n = 5, X = 3, 4, and 5, and p = 0.3. Then, P(X  3) = P(3) + P(4) + P(5) = = NOTE: NOTE: You can use Table B in the textbook to find the Binomial probabilities as well.

© The McGraw-Hill Companies, Inc., Binomial Probability Binomial Probability - Example A report from the Secretary of Health and Human Services stated that 70% of single- vehicle traffic fatalities that occur on weekend nights involve an intoxicated driver. If a sample of 15 single-vehicle traffic fatalities that occurred on a weekend night is selected, find the probability that exactly 12 involve a driver who is intoxicated.

© The McGraw-Hill Companies, Inc., Binomial Probability Binomial Probability - Example Solution: Solution: n = 15, X = 12, and p = 0.7. From Table B, P(X =12) = 0.170

© The McGraw-Hill Companies, Inc., Mean, Variance, Standard Deviation for the Binomial Distribution Mean, Variance, Standard Deviation for the Binomial Distribution - Example A coin is tossed four times. Find the mean, variance, and standard deviation of the number of heads that will be obtained. Solution: Solution: n = 4, p = 1/2, and q = 1/2.  = n  p = (4)(1/2) = 2.  2 = n  p  q = (4)(1/2)(1/2) = 1.  = = 1.