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Special Probability DistributionsSpecial Probability DistributionsChapter 5Chapter 5

““A throw of the dice will never abolish chance.”A throw of the dice will never abolish chance.”Stéphane Mallarmé, French poet Stéphane Mallarmé, French poet

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Goals for Chapter 5Goals for Chapter 5• Probability Distributions to understand:Probability Distributions to understand:

– possible outcomes (“choosing” formulas)– Bernoulli Trials– Binomial– Poisson– Uniform– Exponential– Normal

• Standard Normal Random Variable; Z-scoresStandard Normal Random Variable; Z-scores

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Probability from Possible Outcomes Probability from Possible Outcomes

• If we know the number of ways that all possible If we know the number of ways that all possible events can occur,W, and if we know the number of events can occur,W, and if we know the number of ways a specific event, A, can occur, W(A), and if each ways a specific event, A, can occur, W(A), and if each way (or possible outcome) is equally likely, then the way (or possible outcome) is equally likely, then the probability of the event A is given byprobability of the event A is given by

P(A) = W(A) / WP(A) = W(A) / W• Example: what is the probability of throwing “7” in Example: what is the probability of throwing “7” in

craps (two fair dice): 36 = 6x6 possible outcomes;craps (two fair dice): 36 = 6x6 possible outcomes;

(i.e. Six ways each of the two dice can land); for “7” (i.e. Six ways each of the two dice can land); for “7” can get 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 can get 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1, so and 1, so P(throw=7) = 6/36 = 1/6P(throw=7) = 6/36 = 1/6..

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““Choosing” Formulas for Probability--1Choosing” Formulas for Probability--1

• Use combinatorial (“choosing”) formulas to Use combinatorial (“choosing”) formulas to calculate number of ways for all outcomes calculate number of ways for all outcomes and for event of interest to occur:and for event of interest to occur:

• P(A) = W(A) / W,P(A) = W(A) / W,

where W(A) is the number of ways that where W(A) is the number of ways that event A can occur, and W is the total number event A can occur, and W is the total number of ways that all events can occur.of ways that all events can occur.

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““Choosing” Formulas for Probability--2Choosing” Formulas for Probability--2• Example: Ex.5.3, taste test--8 glasses, 4 containing Example: Ex.5.3, taste test--8 glasses, 4 containing

house brand, 4 national brand. Tester is to identify four house brand, 4 national brand. Tester is to identify four glasses containing house brand; how many different glasses containing house brand; how many different choices can he make? choices can he make? W, number of ways to choose 4 things from eight objects is 8! / [4!4!] = 70 (denoted as (8

4)--”8, choose 4”)

• Ex. 5.4, How many of these choices include 3 correct Ex. 5.4, How many of these choices include 3 correct (house brand) and 1 incorrect? (house brand) and 1 incorrect?

W(A) = (43)(4

1) = 42 = 16.

• What is probability that taster gets 3 correct (out of 4) What is probability that taster gets 3 correct (out of 4) by chance? by chance?

P(A) = 16 / 70 = 0.228 or about 23%.

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Bernouilli Trials-1Bernouilli Trials-1• Count number of successes in a series of similar Count number of successes in a series of similar

events: events:

• number of heads in n coin tosses; number of heads in n coin tosses;

• number of defective parts in an assembly number of defective parts in an assembly line; line;

• number who vote party-line in the total of number who vote party-line in the total of voters at a polling place;voters at a polling place;

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Bernouilli Trials-2Bernouilli Trials-2If the If the following conditions are met, each event is a, each event is a

Bernoulli TrialBernoulli Trial::

1 1 There are only two possible outcomes for each There are only two possible outcomes for each eventevent----Yes or No; Success or Failure, Test + or -.Yes or No; Success or Failure, Test + or -.

2 2 statistical independence of successstatistical independence of success..

The probability of success in one event does not depend on The probability of success in one event does not depend on whether a success or failure occurred in previous events whether a success or failure occurred in previous events

3 3 probability of success (probability of success () is constant.) is constant. The probability of success in any event remains the same for any The probability of success in any event remains the same for any event event ..

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Examples of Examples of nonnon-Bernoulli Trials-Bernoulli Trials• Survey households through the year to determine Survey households through the year to determine

employment status of head of householdemployment status of head of household::

Not Bernoulli trials--probability of employment will Not Bernoulli trials--probability of employment will show a seasonal variation;show a seasonal variation;

• Analyze grade distribution for a graduate course (possible Analyze grade distribution for a graduate course (possible grades: distinction, pass, fail):grades: distinction, pass, fail):

Not Bernoulli trials--more than two outcomes Not Bernoulli trials--more than two outcomes possiblepossible;;

• Look at the price of a stock daily over a period of 1 month;Look at the price of a stock daily over a period of 1 month;

check whether stock price has increased (a success)check whether stock price has increased (a success)

Not Bernoulli trials--success of one event not Not Bernoulli trials--success of one event not independent of previous events.independent of previous events.

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Bernoulli Trials and the Binomial DistributionBernoulli Trials and the Binomial Distribution• For a series of n Bernoulli trials, can calculate the For a series of n Bernoulli trials, can calculate the

probability of “x” successes (e.g x = 5 heads in 10 probability of “x” successes (e.g x = 5 heads in 10 tosses) if the order of success events is not critical.tosses) if the order of success events is not critical.

• PPXX(x; n) = ((x; n) = (nnxx) ) xx(1- (1- ) ) n-xn-x

is the probability of x successes in n trials, if is the probability of x successes in n trials, if is the probability of success in an individual trial.is the probability of success in an individual trial.

• The coefficient (The coefficient (nnxx) can be derived as follows: if the order of ) can be derived as follows: if the order of

successful trials is not important, then all we have to do is successful trials is not important, then all we have to do is count the number of ways in which x successes can occur in n count the number of ways in which x successes can occur in n trials; these are the number of branches in a probability tree trials; these are the number of branches in a probability tree (see board demonstration) and give the total probability for x (see board demonstration) and give the total probability for x successes.successes.

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The Binomial Distribution--ExampleThe Binomial Distribution--Example• Ex. 5.9-11. 3% discount given to cash customers at motel; Ex. 5.9-11. 3% discount given to cash customers at motel;

experience indicates that 30% of all customers will take the experience indicates that 30% of all customers will take the discount (discount ( = 0.30) rather than use a credit card/ = 0.30) rather than use a credit card/– Probability that exactly 5 of next 20 customers pay cash?

PX(5; 20) = (205) 0.35(0.7)15 = 0.1789, (n=20; x =5; = 0.3);

– Probability that 5 or fewer of the next 20 customers pay cash? Need to calculate cumulative probability: FX(5; 20)= PX(x; 20), (where the

sum is taken from x=0 to x= 5). The cumulative probability value can be given by software or table values (see App. 1): FX(5; 20) = 0.0008+0.0068+0.0278+0.0716+0.1304 + 0.1789 = 0.4163

– What is the probability that at least 5 of the next 20 customers pay by credit card? P(X 5) = 1 - FX(4; 20) = 1 - (0.4163-0.1789) = 0.7626

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The Binomial Distribution--ExpectationThe Binomial Distribution--Expectation• For the binomial distribution, the For the binomial distribution, the expectation value of xexpectation value of x, ,

E(x), the mean or long-term expected value for the number E(x), the mean or long-term expected value for the number of successes, is given byof successes, is given by

E(x) = n E(x) = n • The variance of xThe variance of x is given by is given by

V(x) = n V(x) = n ( 1 - ( 1 - ))• Previous example (5.9-5.11): n = 20, Previous example (5.9-5.11): n = 20, = 0.3, so = 0.3, so

E(x) = 20 E(x) = 20 0.3 = 6, the average number of people who 0.3 = 6, the average number of people who would pay by cash (out of 20);would pay by cash (out of 20);

V(x) = 20 V(x) = 20 0.3 0.3 ( 1 -0.3) = 4.2 = ( 1 -0.3) = 4.2 = XX2 2 = (2.1) = (2.1) 22;;

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The Poisson DistributionThe Poisson Distribution• The Poisson distribution is useful in considering the The Poisson distribution is useful in considering the

frequency of events that occur randomly over time or frequency of events that occur randomly over time or over spatial dimensions. (e.g. the frequency of over spatial dimensions. (e.g. the frequency of telephone calls to 911, the number of chocolate bits in a telephone calls to 911, the number of chocolate bits in a Toll-house cookie; the frequency that ocean liners hit Toll-house cookie; the frequency that ocean liners hit icebergs) icebergs)

• Assumptions and conditionsAssumptions and conditions::– Events occur infrequently--two events do not occur

simultaneously– Events occur randomly during a time period or in a

spatial interval; the probability of occurrence is not affected by previous occurrences.

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The Poisson Distribution--ContinuedThe Poisson Distribution--Continued• The Poisson distribution is given by the formulaThe Poisson distribution is given by the formula

PPXX(x) = e(x) = e - - xx / (x!), where x, the number of events / (x!), where x, the number of events counted during the time interval, can have values counted during the time interval, can have values x = 0, 1, 2, 3, …., and the quantity x = 0, 1, 2, 3, …., and the quantity = E(x) = V(x). = E(x) = V(x).

(note that (note that is denoted by is denoted by in many other texts; note also that 0! in many other texts; note also that 0! = 1 and and that anything to the 0 power = 1, so that = 1 and and that anything to the 0 power = 1, so that PPXX(0) = e(0) = e - - .) .)

• Changing interval value changes µ by proportionate Changing interval value changes µ by proportionate amountamountExample; Ex. 5.31, mean rate of tire failures for logging Example; Ex. 5.31, mean rate of tire failures for logging trucks is 4.0 per 10,000 miles; therefore µ= 0.4 failures trucks is 4.0 per 10,000 miles; therefore µ= 0.4 failures per 1000 miles of driving.per 1000 miles of driving.

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The Poisson Distribution--ContinuedThe Poisson Distribution--ContinuedExampleExample: Ex. 5.32--: Ex. 5.32--Tire failures for logging trucks occur with Tire failures for logging trucks occur with

the Poisson distribution, with µ= 4.0 failures per 10,000 the Poisson distribution, with µ= 4.0 failures per 10,000 miles as the mean rate. miles as the mean rate.

A) If a truck drives 1000 miles per week, what is the A) If a truck drives 1000 miles per week, what is the probability that there will be probability that there will be nono failures during the week. failures during the week.

µ= 0.4 failures per 1,000 miles (see previous slide). µ= 0.4 failures per 1,000 miles (see previous slide). PPXX(0) = e(0) = e - - = = e e -0.4 -0.4 = 0.67 is probability of no failures (X= 0) = 0.67 is probability of no failures (X= 0)

B) B) What is the probability that the truck will have What is the probability that the truck will have at least at least two failures (I.e two or more failures)?two failures (I.e two or more failures)?

P(X P(X 2) = 1 - 2) = 1 - PPXX(0) - P(0) - PXX(1) = 1 - 0.67(1) = 1 - 0.67 - - e e -0.4 -0.4 (0.4) (0.4)11 / 1! = 0.062 / 1! = 0.062

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The Uniform DistributionThe Uniform Distribution• The The UniformUniform distribution is one of the simplest of all distribution is one of the simplest of all

distributions; it is used to describe a situation where the distributions; it is used to describe a situation where the probability for a value of X is the same within an interval a probability for a value of X is the same within an interval a X X b: b: ffXX(x) = 1/(b-a) (x) = 1/(b-a) (the 1/(b-a) gives (the 1/(b-a) gives ffXX(x) dx = 1) (x) dx = 1)

• E(x) = (a + b) / 2 for the Uniform distribution, that is, the E(x) = (a + b) / 2 for the Uniform distribution, that is, the average or mean value of x is halfway between the limits for x.average or mean value of x is halfway between the limits for x.

• V(x) = (b - a)V(x) = (b - a)22 / 12 / 12 (derived from V(x) = E(x (derived from V(x) = E(x 2 2 ) - [E(x)] ) - [E(x)] 2 2 and using and using the formula above for the distribution function, fthe formula above for the distribution function, fXX(x) = 1/(b-a) )(x) = 1/(b-a) )

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The Uniform Distribution--ExampleThe Uniform Distribution--Example• Ex. 5.39, On summer days, T = the time that a suburban Ex. 5.39, On summer days, T = the time that a suburban

commuter train is late, can be modeled as uniformly distributed commuter train is late, can be modeled as uniformly distributed between 0 and 20 minutes:between 0 and 20 minutes:

P(t P(t T T t + dt) = (1/20) dt for 0 t + dt) = (1/20) dt for 0 t t 20, 20,

= 0, otherwise.= 0, otherwise.

A) Find the probability that the train is A) Find the probability that the train is at leastat least 8 minutes late 8 minutes late

P(T P(T 8 ) = 1 - P(T< 8) = 1 - 8 ) = 1 - P(T< 8) = 1 - 0088(1/20)dt = 1 - 8/20 = 3/ 5(1/20)dt = 1 - 8/20 = 3/ 5

B) Find the standard deviation for T:B) Find the standard deviation for T:

V(T) = (20-0)V(T) = (20-0)22/ 12 = 33.3= / 12 = 33.3= TT22, or , or TT = = 33.3 = 5.77;33.3 = 5.77;

(Note: the mean late time will be (0+20)/2 = 10 m) (Note: the mean late time will be (0+20)/2 = 10 m)

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The Exponential DistributionThe Exponential Distribution

• The exponential distribution is useful in modeling waiting time problems, or problems involving time to failure (reliability); if the Poisson distribution is a good model for the probability of an event randomly occurring during a given interval of time, then, with the same assumptions, the exponential distribution is an appropriate model for the probability of the time between two events:– fT(t) = exp (- t), for t 0; fT(t) = 0 for t<0.

– E (t ) = 1/;

– V (t) = 1/ 2

– P(T t) = 1 - exp (- t) is the cumulative distribution function.

• (Note that µ for the Poisson distribution is equal to ) )

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The Exponential Distribution--ExampleThe Exponential Distribution--Example

• Ex. 5.44, text. “The counter service time for unticketed airline Ex. 5.44, text. “The counter service time for unticketed airline passengers follows an exponential distribution, with mean time of passengers follows an exponential distribution, with mean time of 5 minutes = 5 minutes = (1/ or or = 1/(5 m) = 0.2 m = 1/(5 m) = 0.2 m-1-1) )

a) What is P(Ta) What is P(T 2.5 ) (probability that service time will be 2.5 2.5 ) (probability that service time will be 2.5 minutes or less)?minutes or less)?

P(TP(T 2.5 ) = 2.5 ) = 002.5 2.5 0.2 0.2 exp(- 0.2t) dt = 1 - exp(- 0.2*2.5)=0.39exp(- 0.2t) dt = 1 - exp(- 0.2*2.5)=0.39

b) The probability that the service time will be b) The probability that the service time will be longerlonger than 10m than 10m

P(T> 10) = 1 - P(T P(T> 10) = 1 - P(T 10) = exp( -0.2*10) = 0.1365 10) = exp( -0.2*10) = 0.1365 0.14 0.14 Ex. 5.45, text. A) What is the expected number of passengers Ex. 5.45, text. A) What is the expected number of passengers

served per minute?served per minute?

µ = 1*0.2 = 0.2 per minuteµ = 1*0.2 = 0.2 per minute

c) The probability that no passenger is served within 10 m.; c) The probability that no passenger is served within 10 m.;

µ’ = 10*0.2 = 2.0 per 10 minutes give P(X=0) = eµ’ = 10*0.2 = 2.0 per 10 minutes give P(X=0) = e - 2 - 2= 0.1365= 0.1365

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The Normal Probability DistributionThe Normal Probability Distribution

• The Normal probability distribution (also called the “Gaussian” or “Bell-shaped Curve”) is the most important distribution in statistics. It can be shown to occur, generally, if the distribution is the sum or average of many components (“The Central Limit Theorem”). A normal distribution is implicitly assumed for many statistical tests (hypothesis tests, confidence limits, t-tests).

• ffXX(x) = (1/[(x) = (1/[(2(2)]) exp{ -(x- µ))]) exp{ -(x- µ)22/ (2/ (222)},)},

µ is the mean value of X ( E(X) = µ)µ is the mean value of X ( E(X) = µ)

is the standard deviation of X ( V(x) = is the standard deviation of X ( V(x) = 22))

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The Standard Normal Probability DistributionThe Standard Normal Probability Distribution

• The cumulative distribution function, P(X<x), can not be calculated as an explicit function of x, but it can be evaluated numerically. However, to do so for every combination of values for µ and would be a chore, so a more convenient method is to transform X into a universal, “Standard”, variable, Z, as follows:

Z = (X - µ) / Z = (X - µ) / ;;• Note that the mean value of Z, E(Z), is 0, and that the

variance of Z, V(Z), is 1.

• Values of P(Z<z) are given in tables or from software.• Empirical Rule: P(-1 <Z< 1) = 0.68;

P(-1.96 <Z< 1.96) = 0.95

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Z-scores -- ExampleZ-scores -- Example

• The Z-value for a standard normal distribution is often denoted as a “Z-score” (not in this text, however).

• Example, Ex.5.56 (text). The price X of a long-term $1000 bond one year later is normally distributed with mean $980 and SD=$40.

A) P(X > $1000) = ?

P(X > $1000) = 1 - P(X $1000);

For X = $1000, Z = (1000 -980)/ 40 = 0.5 = 0.5;

P(Z 0.5) = 0.1915 + 0.5 = 0.6915 (Table 3, p 800),

so P(X > $1000) = 1 - 0.6915 = 0.3085, or 31 %;• 5.57b. What is the price, x’, such that the probability is 60% that

the value of the bond will exceed x’?

For P(Z z’ ) = 1 - 0.6 = 0.4, z’ = 0.0-0.2053 (Table 3) = -0.2053, or x’ = -0.2053*40 + 980 = $971.79