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Chap06 01

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Chap06 01

1. 1. Chapter SixDiscrete Probability Distributions 6.1 Probability Distributions
2. 2. A random variable is a numericalmeasure of the outcome from aprobability experiment, so its value isdetermined by chance. Randomvariables are denoted using letters suchas X.
3. 3. A discrete random variable is a randomvariable that has values that has either afinite number of possible values or acountable number of possible values.A continuous random variable is arandom variable that has an infinitenumber of possible values that is notcountable.
4. 4. EXAMPLE Distinguishing Between Discrete and Continuous Random VariablesDetermine whether the following randomvariables are discrete or continuous. Statepossible values for the random variable.(a) The number of light bulbs that burn out in aroom of 10 light bulbs in the next year.(b) The number of leaves on a randomly selectedOak tree.(c) The length of time between calls to 911.(d) A single die is cast. The number of pipsshowing on the die.
5. 5. • We use capital letter , like X, to denote the random variable and use small letter to list the possible values of the random variable.• Example. A single die is cast, X represent the number of pips showing on the die and the possible values of X are x=1,2,3,4,5,6.
6. 6. A probability distribution provides thepossible values of the random variable andtheir corresponding probabilities. Aprobability distribution can be in the form ofa table, graph or mathematical formula.
7. 7. The table below shows the probability distributionfor the random variable X, where X representsthe number of DVDs a person rents from a videostore during a single visit.
8. 8. EXAMPLE Identifying Probability DistributionsIs the following a probability distribution?
9. 9. EXAMPLE Identifying Probability DistributionsIs the following a probability distribution?
10. 10. Answer:0.16 + 0.18 + 0.22 + 0.10 + 0.3 + 0.01 = 0.97 <1 , Not a probability distribution.
11. 11. EXAMPLE Identifying Probability DistributionsIs the following a probability distribution?
12. 12. Answer:• 0.16 + 0.18 + 0.22 + 0.10 + 0.3 + 0.04 = 1• It is a probability distribution
13. 13. A probability histogram is a histogram inwhich the horizontal axis corresponds tothe value of the random variable and thevertical axis represents the probability ofthat value of the random variable.
14. 14. EXAMPLE Drawing a Probability HistogramDraw a probability histogram of the followingprobability distribution which represents thenumber of DVDs a person rents from a video storeduring a single visit.
15. 15. x prob Probability Distribution 0 0.06 0.7 0.6 1 0.58 0.5probabilities 0.4 2 0.22 0.3 0.2 0.1 3 0.1 0 1 2 3 4 5 6 4 0.03 random variable values 5 0.01
16. 16. EXAMPLE The Mean of a Discrete Random VariableCompute the mean of the following probabilitydistribution which represents the number of DVDsa person rents from a video store during a singlevisit.
17. 17. Mean=0*0.06+1*0.58+2 *0.22+3* 0.1+4 *0.03+5*0.01= 1.49
18. 18. The following data represent the number of DVDsrented by 100 randomly selected customers in a singlevisit. Compute the mean number of DVDs rented.
19. 19. x1 + x2 + ... + x100X = = 1.49 100
20. 20. EXAMPLE Variance and StdCompute the variance and standard deviation ofthe following probability distribution whichrepresents the number of DVDs a person rentsfrom a video store during a single visit.
21. 21. • The variance=(0-1.49)^2*0.06+(1-1.49)^2*0.58 +(2-1.49)^2*0.22+(3-1.49)^2*0.1 +(4-1.49)^2*0.03+(5-1.49)^2*0.01 =0.8699• Standard Deviation= 0.932684
22. 22. EXAMPLE Expected ValueA term life insurance policy will pay a beneficiary acertain sum of money upon the death of the policyholder. These policies have premiums that must bepaid annually. Suppose a life insurance companysells a \$250,000 one year term life insurance policyto a 49-year-old female for \$520. According to theNational Vital Statistics Report, Vol. 47, No. 28, theprobability the female will survive the year is0.99791. Compute the expected value of this policyto the insurance company.
23. 23. • Correct: E(X) = (520-250,000)*(1- 0.99791)+520*0.99791 = -2.5• Wrong: E(X) = -250,000*(1-0.99791)+520*0.99791 = -3.5868
24. 24. Chapter SixDiscrete Probability Distributions 6.2The Binomial Probability Distribution
25. 25. Criteria for a Binomial Probability ExperimentAn experiment is said to be a binomial experimentprovided1. The experiment is performed a fixed number oftimes. Each repetition of the experiment is called atrial.2. The trials are independent. This means theoutcome of one trial will not affect the outcome of theother trials.3. For each trial, there are two mutually exclusiveoutcomes, success or failure.4. The probability of success is fixed for each trial ofthe experiment.
26. 26. Notation Used in the Binomial Probability Distribution• There are n independent trials of the experiment• Let p denote the probability of success so that1 – p is the probability of failure.• Let x denote the number of successes in nindependent trials of the experiment. So, 0 < x < n.
27. 27. EXAMPLE Identifying Binomial ExperimentsWhich of the following are binomial experiments?(a) A player rolls a pair of fair die 10 times. The numberX of 7’s rolled is recorded.(b) The 11 largest airlines had an on-time percentage of84.7% in November, 2001 according to the Air TravelConsumer Report. In order to assess reasons fordelays, an official with the FAA randomly selects flightsuntil she finds 10 that were not on time. The number offlights X that need to be selected is recorded.(c ) In a class of 30 students, 55% are female. Theinstructor randomly selects 4 students. The number Xof females selected is recorded.
28. 28. EXAMPLE Constructing a Binomial Probability DistributionAccording to the Air Travel Consumer Report,the 11 largest air carriers had an on-timepercentage of 84.7% in November, 2001.Suppose that 4 flights are randomly selectedfrom November, 2001 and the number of on-timeflights X is recorded. Construct a probabilitydistribution for the random variable X using atree diagram.
29. 29. • X=(x1,x2,x3,x4)• X=the number of on-time
30. 30. EXAMPLE Using the Binomial Probability Distribution FunctionAccording to the United States Census Bureau,18.3% of all households have 3 or more cars.(a) In a random sample of 20 households, what isthe probability that exactly 5 have 3 or more cars?(b) In a random sample of 20 households, what isthe probability that less than 4 have 3 or morecars?(c) In a random sample of 20 households, what isthe probability that at least 4 have 3 or more cars?
31. 31. EXAMPLE Finding the Mean and Standard Deviation of a Binomial Random VariableAccording to the United States Census Bureau,18.3% of all households have 3 or more cars. In asimple random sample of 400 households,determine the mean and standard deviationnumber of households that will have 3 or morecars.
32. 32. EXAMPLE Constructing Binomial Probability Histograms(a) Construct a binomial probability histogram withn = 8 and p = 0.15.(b) Construct a binomial probability histogram withn = 8 and p = 0. 5.(c) Construct a binomial probability histogram withn = 8 and p = 0.85.For each histogram, comment on the shape of thedistribution.
33. 33. Construct a binomial probabilityhistogram with n = 15 and p = 0.8.Comment on the shape of thedistribution.
34. 34. Construct a binomial probabilityhistogram with n = 25 and p = 0.8.Comment on the shape of thedistribution.
35. 35. Construct a binomial probabilityhistogram with n = 50 and p = 0.8.Comment on the shape of thedistribution.
36. 36. Construct a binomial probabilityhistogram with n = 70 and p = 0.8.Comment on the shape of thedistribution.
37. 37. As the number of trials n in a binomialexperiment increase, the probabilitydistribution of the random variable Xbecomes bell-shaped. As a general rule ofthumb, if np(1 – p) > 10, then the probabilitydistribution will be approximately bell-shaped.
38. 38. EXAMPLE Using the Mean, Standard Deviationand Empirical Rule to Check for Unusual Resultsin a Binomial ExperimentAccording to the United States Census Bureau, in2000, 18.3% of all households have 3 or morecars. A researcher believes this percentage hasincreased since then. He conducts a simplerandom sample of 400 households and found that82 households had 3 or more cars. Is this resultunusual if the percentage of households with 3 ormore cars is still 18.3%?
39. 39. EXAMPLE Using the Binomial Probability Distribution Function to Perform InferenceAccording to the United States Census Bureau, in2000, 18.3% of all households have 3 or morecars. A researcher believes this percentage hasincreased since then. He conducts a simplerandom sample of 20 households and found that 5households had 3 or more cars.Is this result unusual if the percentage ofhouseholds with 3 or more cars is still 18.3%?
40. 40. EXAMPLE Using the Binomial Probability Distribution Function to Perform InferenceAccording to the United States Census Bureau, in2000, 18.3% of all households have 3 or morecars. One year later, the same researcherconducts a simple random sample of 20households and found that 8 households had 3 ormore cars.Is this result unusual if the percentage ofhouseholds with 3 or more cars is still 18.3%?
41. 41. Chapter SixDiscrete Probability Distributions Section 6.3 The Poisson Probability Distribution
42. 42. A random variable X, the number of successes ina fixed interval, follows a Poisson processprovided the following conditions are met1. The probability of two or more successes in anysufficiently small subinterval is 0.2. The probability of success is the same for anytwo intervals of equal length.3. The number of successes in any interval isindependent of the number of successes in anyother interval provided the intervals are notoverlapping.
43. 43. EXAMPLE A Poisson ProcessThe Food and Drug Administration sets a FoodDefect Action Level (FDAL) for various foreignsubstances that inevitably end up in the food weeat and liquids we drink. For example, the FDALlevel for insect filth in chocolate is 0.6 insectfragments (larvae, eggs, body parts, and so on)per 1 gram.
44. 44. • For a sufficiently small interval, theprobability of two successes is 0.• The probability of insect filth in oneregion of a candy bar is equal to theprobability of insect filth in some otherregion of the candy bar.• The number of successes in any randomsample is independent of the number ofsuccesses in any other random sample.
45. 45. EXAMPLE Computing Poisson ProbabilitiesThe Food and Drug Administration sets a Food Defect Action Level (FDAL) for various foreign substances that inevitably end up in the food we eat and liquids we drink. For example, the FDAL level for insect filth in chocolate is 0.6 insect fragments (larvae, eggs, body parts, and so on) per 1 gram.(a)Determine the mean number of insect fragments in a 5 gram sample of chocolate.(b) What is the standard deviation?
46. 46. Probability Histogram of a Poisson Distribution with µ = 1
47. 47. Probability Histogram of a Poisson Distribution with µ = 3
48. 48. Probability Histogram of a Poisson Distribution with µ = 7
49. 49. Probability Histogram of a Poisson Distribution with µ = 15
50. 50. EXAMPLE Poisson ParticlesIn 1910, Ernest Rutherford and Hans Geigerrecorded the number of α-particles emitted from apolonium source in eighth-minute (7.5 second)intervals. The results are reported in the table on thenext slide. Does a Poisson probability functionaccurately describe the number of α-particlesemitted?Source: Rutherford, Sir Ernest; Chadwick, James; and Ellis, C.D.. Radiations fromRadioactive Substances. London, Cambridge University Press, 1951, p. 172.
51. 51. The Poisson probability distribution functioncan be used to approximate binomialprobabilities provided the number of trialsn > 100 and np < 10. In other words, thenumber of independent trials of the binomialexperiment should be large and theprobability of success should be small.
52. 52. EXAMPLE Using the Poisson Distribution to Approximate Binomial ProbabilitiesAccording to the U.S. National Center for HealthStatistics, 7.6% of male children under the age of15 years have been diagnosed with AttentionDeficit Disorder (ADD). In a random sample of120 male children under the age of 15 years,what is the probability that at least 4 of thechildren have ADD?