Probability Essentials Chapter 3

```Special Probability Distributions
Chapter 5
“A throw of the dice will never abolish chance.”
Stéphane Mallarmé, French poet
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Goals for Chapter 5
• Probability Distributions to understand:
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possible outcomes (“choosing” formulas)
Bernoulli Trials
Binomial
Poisson
Uniform
Exponential
Normal
• Standard Normal Random Variable; Z-scores
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Probability from Possible Outcomes
• If we know the number of ways that all possible
events can occur,W, and if we know the number of
ways a specific event, A, can occur, W(A), and if each
way (or possible outcome) is equally likely, then the
probability of the event A is given by
P(A) = W(A) / W
• Example: what is the probability of throwing “7” in
craps (two fair dice): 36 = 6x6 possible outcomes;
(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
and 1, so P(throw=7) = 6/36 = 1/6.
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“Choosing” Formulas for Probability--1
• Use combinatorial (“choosing”) formulas to
calculate number of ways for all outcomes
and for event of interest to occur:
• P(A) = W(A) / W,
where W(A) is the number of ways that
event A can occur, and W is the total number
of ways that all events can occur.
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“Choosing” Formulas for Probability--2
• Example: Ex.5.3, taste test--8 glasses, 4 containing
house brand, 4 national brand. Tester is to identify
four glasses containing house brand; how many
different choices can he make?
W, number of ways to choose 4 things from eight objects is
8! / [4!4!] = 70 (denoted as (84)--”8, choose 4”)
• Ex. 5.4, How many of these choices include 3 correct
(house brand) and 1 incorrect?
W(A) = (43)(41) = 42 = 16.
• What is probability that taster gets 3 correct (out of 4)
by chance?
P(A) = 16 / 70 = 0.228 or about 23%.
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Bernouilli Trials-1
• Count number of successes in a series of similar
events:
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number of heads in n coin tosses;
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number of defective parts in an assembly line;
•
number who vote party-line in the total of
voters at a polling place;
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Bernouilli Trials-2
If the following conditions are met, each event is a
Bernoulli Trial:
1 There are only two possible outcomes for each event-Yes or No; Success or Failure, Test + or -.
2 statistical independence of success.
The probability of success in one event does not depend on
whether a success or failure occurred in previous events
3 probability of success () is constant.
The probability of success in any event remains the same for any
event .
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Examples of non-Bernoulli Trials
• Survey households through the year to determine
employment status of head of household:
Not Bernoulli trials--probability of employment will
show a seasonal variation;
Not Bernoulli trials--more than two outcomes
possible;
• Look at the price of a stock daily over a period of 1 month;
check whether stock price has increased (a success)
Not Bernoulli trials--success of one event not
independent of previous events.
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Bernoulli Trials and the Binomial Distribution
• For a series of n Bernoulli trials, can calculate the
probability of “x” successes (e.g x = 5 heads in 10
tosses) if the order of success events is not critical.
• PX(x; n) = (nx)  x(1- ) n-x
is the probability of x successes in n trials, if  is
the probability of success in an individual trial.
• The coefficient (nx) can be derived as follows: if the order of
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
trials; these are the number of branches in a probability tree
(see board demonstration) and give the total probability for x
successes.
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The Binomial Distribution--Example
• Ex. 5.9-11. 3% discount given to cash customers at motel;
experience indicates that 30% of all customers will take the
discount ( = 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--Expectation
• For the binomial distribution, the expectation value of x,
E(x), the mean or long-term expected value for the number
of successes, is given by
E(x) = n 
• The variance of x is given by
V(x) = n  ( 1 - )
• Previous example (5.9-5.11): n = 20,  = 0.3, so
E(x) = 20  0.3 = 6, the average number of people who
would pay by cash (out of 20);
V(x) = 20  0.3 ( 1 -0.3) = 4.2 = X2 = (2.1) 2;
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The Poisson Distribution
• The Poisson distribution is useful in considering the
frequency of events that occur randomly over time or
over spatial dimensions. (e.g. the frequency of
telephone calls to 911, the number of chocolate bits in
a Toll-house cookie; the frequency that ocean liners
hit icebergs)
• Assumptions 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--Continued
• The Poisson distribution is given by the formula
PX(x) = e - x / (x!), where x, the number of events
counted during the time interval, can have values
x = 0, 1, 2, 3, …., and the quantity  = E(x) = V(x).
(note that  is denoted by  in many other texts; note also that 0!
= 1 and and that anything to the 0 power = 1, so that
PX(0) = e - .)
• Changing interval value changes µ by proportionate
amount
Example; Ex. 5.31, mean rate of tire failures for logging
trucks is 4.0 per 10,000 miles; therefore µ= 0.4 failures
per 1000 miles of driving.
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The Poisson Distribution--Continued
Example: Ex. 5.32--Tire failures for logging trucks occur with
the Poisson distribution, with µ= 4.0 failures per 10,000
miles as the mean rate.
A) If a truck drives 1000 miles per week, what is the
probability that there will be no failures during the
week.
µ= 0.4 failures per 1,000 miles (see previous slide).
PX(0) = e - = e -0.4 = 0.67 is probability of no failures (X= 0)
B) What is the probability that the truck will have at least
two failures (I.e two or more failures)?
P(X  2) = 1 - PX(0) - PX(1) = 1 - 0.67 - e -0.4 (0.4)1 / 1! = 0.062
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The Uniform Distribution
• The Uniform distribution is one of the simplest of all
distributions; it is used to describe a situation where the
probability for a value of X is the same within an
interval a  X  b:
fX(x) = 1/(b-a) (the 1/(b-a) gives fX(x) dx = 1)
• 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.
• V(x) = (b - a)2 / 12 (derived from V(x) = E(x 2 ) - [E(x)] 2
and using the formula above for the distribution function,
fX(x) = 1/(b-a) )
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The Uniform Distribution--Example
• Ex. 5.39, On summer days, T = the time that a suburban
commuter train is late, can be modeled as uniformly
distributed between 0 and 20 minutes:
P(t  T  t + dt) = (1/20) dt for 0  t  20,
= 0, otherwise.
A) Find the probability that the train is at least 8 minutes
late
P(T  8 ) = 1 - P(T< 8) = 1 - 08(1/20)dt = 1 - 8/20 = 3/ 5
B) Find the standard deviation for T:
V(T) = (20-0)2/ 12 = 33.3= T2, or T = 33.3 = 5.77;
(Note: the mean late time will be (0+20)/2 = 10 m)
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The 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--Example
• Ex. 5.44, text. “The counter service time for unticketed airline
passengers follows an exponential distribution, with mean time
of 5 minutes = (1/  or  = 1/(5 m) = 0.2 m-1)
a) What is P(T 2.5 ) (probability that service time will be 2.5
minutes or less)?
P(T 2.5 ) = 02.5 0.2 exp(- 0.2t) dt = 1 - exp(- 0.2*2.5)=0.39
b) The probability that the service time will be longer than 10m
P(T> 10) = 1 - P(T  10) = exp( -0.2*10) = 0.1365 0.14
• Ex. 5.45, text. A) What is the expected number of passengers
served per minute?
µ = 1*0.2 = 0.2 per minute
c) The probability that no passenger is served within 10 m.;
µ’ = 10*0.2 = 2.0 per 10 minutes give P(X=0) = e - 2= 0.1365
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The 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).
• fX(x) = (1/[(2)]) exp{ -(x- µ)2/ (22)},
µ is the mean value of X ( E(X) = µ)
 is the standard deviation of X ( V(x) = 2)
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The 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) / ;
• 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 -- 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
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