MTH120_Chapter6
Upcoming SlideShare
Loading in...5
×
 

MTH120_Chapter6

on

  • 308 views

 

Statistics

Views

Total Views
308
Views on SlideShare
308
Embed Views
0

Actions

Likes
0
Downloads
4
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

MTH120_Chapter6 MTH120_Chapter6 Presentation Transcript

  • 6- 1 Chapter Six Discrete Probability DistributionsGOALSWhen you have completed this chapter, you will be able to: ONE Define the terms random variable and probability distribution. TWO Distinguish between a discrete and continuous probability distributions. THREE Calculate the mean, variance, and standard deviation of a discrete probability distribution. FOUR Describe the characteristics and compute probabilities using the binomial probability distribution.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 2 Chapter Six continued A Survey of Probability ConceptsGOALSWhen you have completed this chapter, you will be able to:FIVEDescribe the characteristics and compute probabilities using thehypergeometric distribution.SIXDescribe the characteristics and compute the probabilities usingthe Poisson distribution.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 3 Random Variables A random variable is a numerical value determined by the outcome of an experiment. A probability distribution is the listing of all possible outcomes of an experiment and the corresponding probability.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 4 Types of Probability Distributions A discrete probability distribution can assume only certain outcomes. A continuous probability distribution can assume an infinite number of values within a given range.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 5 Types of Probability Distributions Examples of a discrete distribution are: The number of students in a class. The number of children in a family. The number of cars entering a carwash in a hour. Number of home mortgages approved by Coastal Federal Bank last week.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 6 Types of Probability Distributions Examples of a continuous distribution include: The distance students travel to class. The time it takes an executive to drive to work. The length of an afternoon nap. The length of time of a particular phone call.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 7 Features of a Discrete Distribution The main features of a discrete probability distribution are: The sum of the probabilities of the various outcomes is 1.00. The probability of a particular outcome is between 0 and 1.00. The outcomes are mutually exclusive.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 8 Example 1 EXAMPLE 1: 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.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 9 EXAMPLE 1 continued 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 0,1,2,3.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 10 EXAMPLE 1 continued  The outcome of zero heads occurred once.  The outcome of one head occurred three times.  The outcome of two heads occurred three times.  The outcome of three heads occurred once.  From the definition of a random variable, x as defined in this experiment, is a random variable.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 11 The Mean of a Discrete Probability Distribution  The mean: reports the central location of the data. is the long-run average value of the random variable. is also referred to as its expected value, E(X), in a probability distribution. is a weighted average.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 12 The Mean of a Discrete Probability Distribution The mean is computed by the formula: µ = Σ[ xP( x)] where w represents the mean and P(x) is the probability of the various outcomes x.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 13 The Variance of a Discrete Probability Distribution The variance measures the amount of spread (variation) of a distribution. The variance of a discrete distribution is denoted by the Greek letter d2 (sigma squared). The standard deviation is the square root of 2.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 14 The Variance of a Discrete Probability Distribution The variance of a discrete probability distribution is computed from the formula: 2 2 σ = Σ[( x − µ ) P ( x)]McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 15 EXAMPLE 2 Dan Desch, owner of College Painters, # of Houses Weeks studied his records for Painted 10 5 the past 20 weeks and reports the following 11 6 number of houses painted per week: 12 7 13 2McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 16 EXAMPLE 2 continued Probability Distribution: Number of houses Probability, P(x) painted, x 10 .25 11 .30 12 .35 13 .10 Total 1.00McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 17 EXAMPLE 2 continued Compute the mean number of houses painted per week: µ = E ( x) = Σ[ xP( x)] = (10)(.25) + (11)(.30) + (12)(.35) + (13)(.10) = 11.3McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 18 EXAMPLE 2 continued Compute the variance of the number of houses painted per week: 2 2 σ = Σ[( x − µ ) P( x)] = (10 − 11.3) 2 (.25) + ... + (13 − 11.3) 2 (.10) = 0.4225 + 0.0270 + 0.1715 + 0.2890 = 0.91McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 19 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. The probability of success stays the same for each trial. The trials are independent.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 20 Binomial Probability Distribution To construct a binomial distribution, let  n be the number of trials  x be the number of observed successes  be the probability of success on each trialMcGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 21 Binomial Probability Distribution The formula for the binomial probability distribution is: x n− x P( x)= n C xπ (1 − π )McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 22 EXAMPLE 3 The Alabama Department of Labor reports that 20% of the workforce in Mobile is unemployed. From a sample of 14 workers, calculate the following probabilities: Exactly three are unemployed. At least three are unemployed. At least none are unemployed.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 23 EXAMPLE 3 continued  The probability of exactly 3: P (3)=14 C 3 (.20) 3 (1 − .20)11 = (364)(.0080)(.0859) = .2501  The probability of at least 3 is: 3 11 14 0 P( x ≥ 3)= 14 C3 (.20) (.80) + ...+ 14 C14 (.20) (.80) = .250 + .172 + ... + .000 = .551McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 24 Example 3 continued The probability of at least one being unemployed. P( x ≥ 1) = 1 − P(0) 0 14 = 1− 14 C0 (.20) (1 − .20) = 1 − .044 = .956McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 25 Mean & Variance of the Binomial Distribution The mean is found by: µ = nπ The variance is found by: σ = nπ (1 − π ) 2McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 26 EXAMPLE 4 From EXAMPLE 3, recall that , =.2 and n=14. π Hence, the mean is: i = n = 14(.2) = 2.8. The variance is: i 2 = n (1- ( ) = (14)(.2)(.8) =2.24.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 27 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. The number of homes built in Blackmoor.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 28 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.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 29 Hypergeometric Distribution  The formula for finding a probability using the hypergeometric distribution is: ( S Cx )( N −S Cn −x ) P( x ) = N Cn where N is the size of the population, S is the number of successes in the population, x is the number of successes in a sample of n observations.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 30 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 .McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 31 EXAMPLE 5  The National Air Safety Board has a list of 10 reported safety violations. Suppose only 4 of the reported violations are actual violations and the Safety Board will only be able to investigate five of the violations. What is the probability that three of five violations randomly selected to be investigated are actually violations?McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 32 EXAMPLE 5 continued ( 4 C 3 )(10− 4 C 5− 2 P(3) = 10 C 5 ( 4 C 3 )( 6 C 2 ) 4(15) = = = .238 10 C 5 252McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 33 Poisson Probability Distribution 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 w is small and n is large is called the Poisson probability distribution.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 34 Poisson Probability Distribution The Poisson distribution can be described mathematically using the formula: µe x −u P( x) = x! where wis the mean number of successes in a particular interval of time, e is the constant 2.71828, and x is the number of successes.McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 35 Poisson Probability Distribution The mean number of successes T can be determined in binomial situations by n , where n is the number of trials and the probability of a success. The variance of the Poisson distribution is also equal to n .McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
  • 6- 36 EXAMPLE 6  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 4 arrivals in an hour? x −u −4 µ e 2 4 e P( x) = = = .1465 x! 2!McGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.