Date post: | 23-Dec-2015 |
Category: |
Documents |
Upload: | ambrose-mccormick |
View: | 241 times |
Download: | 0 times |
Chapter 17Chapter 17Probability ModelsProbability Models
Binomial Probability ModelsPoisson Probability Models
Binomial Random Binomial Random VariablesVariables
Binomial Probability Distributions
Binomial Random Binomial Random VariablesVariables
Through 2/25/2014 NC State’s free-throw percentage is 65.1% (315th out 351 in Div. 1).
If in the 2/26/2014 game with UNC, NCSU shoots 11 free-throws, what is the probability that:NCSU makes exactly 8 free-throws?NCSU makes at most 8 free throws?NCSU makes at least 8 free-throws?
““2-outcome” situations are 2-outcome” situations are very commonvery common
Heads/tailsDemocrat/RepublicanMale/FemaleWin/LossSuccess/FailureDefective/Nondefective
Probability Model for this Probability Model for this Common SituationCommon Situation
Common characteristics◦repeated “trials”◦2 outcomes on each trial
Leads to Binomial Experiment
Binomial ExperimentsBinomial Experimentsn identical trials
◦n specified in advance2 outcomes on each trial
◦usually referred to as “success” and “failure”
p “success” probability; q=1-p “failure” probability; remain constant from trial to trial
trials are independent
Binomial Random VariableBinomial Random VariableThe binomial random variable X is the number of “successes” in the n trials
Notation: X has a B(n, p) distribution, where n is the number of trials and p is the success probability on each trial.
Binomial Probability Binomial Probability DistributionDistribution
0 0
trials, success probability on each trial
probability distribution:
( ) , 0,1,2, ,
( ) ( )
( ) (
x n xn x
n nn x n xx
x x
n p
p x C p q x n
E x xp x x p q np
Var x E x npq
P(x) = • px • qn-xn ! (n – x )!x!
Number of outcomes with
exactly x successes
among n trials
Rationale for the Binomial Probability Formula
P(x) = • px • qn-xn ! (n – x )!x!
Number of outcomes with
exactly x successes
among n trials
Probability of x successes
among n trials for any one
particular order
Binomial Probability Formula
Graph of Graph of p(x)p(x); ; xx binomial binomial n=10 p=.5; p(0)+p(1)+ n=10 p=.5; p(0)+p(1)+ …… +p(10)=1+p(10)=1
Think of p(x) as the areaof rectangle above x
p(5)=.246 is the areaof the rectangle above 5
The sum of all theareas is 1
Binomial Probability Histogram: n=100, p=.5
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
Binomial Probability Histogram: n=100, p=.95
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100
Binomial Distribution Binomial Distribution Example: Tennis First Example: Tennis First ServesServes
A tennis player makes 60% of herfirst serves and attempts 4 first-serves.Assume the outcomes of first-serves
are independent.What is the probability that exactly 3
of her first serves are successful?
3 14 3
4;success=successful first serve; .60
(3) (.60) (.40) 4(.216)(.40) .3456
n p
p C
15
Binomial Distribution ExampleBinomial Distribution Example
• Shanille O’Keal is a WNBA player who makes 25% of her 3-point attempts.
• Assume the outcomes of 3-point shots are independent.1. If Shanille attempts 7 3-point shots in a game, what is
the expected number of successful 3-point attempts?2. Shanille’s cousin Shaquille O’Neal makes 10% of his 3-
point attempts. If they each take 12 3-point shots, who has the smaller probability of making 4 or fewer 3-point shots?
Shaq: 12, .10; ( 4) .996n p P X ( ) 7(.25) 1.75E X np
Shanille: 12, .25; ( 4) .842n p P X Shanille has the smaller probability.
Using binomial tables; Using binomial tables; n=20, p=.3n=20, p=.3
P(x 5) = .4164P(x > 8) = 1- P(x 8)=
1- .8867=.1133P(x < 9) = ?P(x 10) = ?P(3 x 7)=P(x 7) - P(x 2)
.7723 - .0355 = .7368
9, 10, 11, … , 20
8, 7, 6, … , 0 =P(x 8)
1- P(x 9) = 1- .9520
Binomial n = 20, p = .3 Binomial n = 20, p = .3 (cont.)(cont.)P(2 < x 9) = P(x 9) - P(x 2)
= .9520 - .0355 = .9165P(x = 8) = P(x 8) - P(x 7)
= .8867 - .7723 = .1144
Color blindness
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population.
We can model this situation with a B(n = 25, p = 0.08) distribution.
What is the probability that five individuals or fewer in the sample are color blind?
Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)”
P(x ≤ 5) = BINOMDIST(5, 25, .08, 1) = 0.9877
What is the probability that more than five will be color blind?
P(x > 5) = 1 P(x ≤ 5) =1 0.9877 = 0.0123
What is the probability that exactly five will be color blind?
P(x = 5) = BINOMDIST(5, 25, .08, 0) = 0.0329
0%
5%
10%
15%
20%
25%
30%
0 2 4 6 8
10
12
14
16
18
20
22
24
Number of color blind individuals (x)
P(X
= x
)
Probability distribution and histogram for
the number of color blind individuals
among 25 Caucasian males.
x P(X = x) P(X <= x) 0 12.44% 12.44%1 27.04% 39.47%2 28.21% 67.68%3 18.81% 86.49%4 9.00% 95.49%5 3.29% 98.77%6 0.95% 99.72%7 0.23% 99.95%8 0.04% 99.99%9 0.01% 100.00%
10 0.00% 100.00%11 0.00% 100.00%12 0.00% 100.00%13 0.00% 100.00%14 0.00% 100.00%15 0.00% 100.00%16 0.00% 100.00%17 0.00% 100.00%18 0.00% 100.00%19 0.00% 100.00%20 0.00% 100.00%21 0.00% 100.00%22 0.00% 100.00%23 0.00% 100.00%24 0.00% 100.00%25 0.00% 100.00%
B(n = 25, p = 0.08)
What are the mean and standard deviation of the
count of color blind individuals in the SRS of 25
Caucasian American males?
µ = np = 25*0.08 = 2
σ = √np(1 p) = √(25*0.08*0.92) = 1.36
p = .08n = 10
p = .08n = 75
µ = 10*0.08 = 0.8 µ = 75*0.08 = 6
σ = √(10*0.08*0.92) = 0.86 σ = √(75*0.08*0.92) = 2.35
What if we take an SRS of size 10? Of size 75?
Recall Free-throw Recall Free-throw questionquestion
Through 2/25/14 NC State’s free-throw percentage was 65.1% (315th in Div. 1).
If in the 2/26/14 game with UNC, NCSU shoots 11 free-throws, what is the probability that:
1. NCSU makes exactly 8 free-throws?
2. NCSU makes at most 8 free throws?
3. NCSU makes at least 8 free-throws?
1. n=11; X=# of made free-throws; p=.651
p(8)= 11C8 (.651)8(.349)3
=.2262. P(x ≤ 8)=.798
3. P(x ≥ 8)=1-P(x ≤7)=1-.5717 = .4283
22
Poisson Probability Poisson Probability ModelsModels
The Poisson experiment typically models situations where rare events occur over a fixed amount of time or within a specified region
Examples◦The number of cellphone calls per
minute arriving at a cellphone tower.◦The number of customers per hour
using an ATM◦The number of concussions per game
experienced by the participants.
24
◦Properties of the Poisson experiment1) The number of successes (events) that
occur in a certain time interval is independent of the number of successes that occur in another time interval.
2) The probability of a success in a certain time interval is
the same for all time intervals of the same size, proportional to the length of the interval.
3) The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.
Poisson Poisson ExperimentExperiment
25
The Poisson Random Variable◦The Poisson random variable X is the
number of successes that occur during a given time interval or in a specific region
Probability Distribution of the Poisson Random Variable.
( ) 0,1,2...!
( ) ( )
where 0is the average number of occurences
in an interval of time or in a specified region.
xep x x
xE X V X
Poisson Prob Dist Poisson Prob Dist =1=1
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 1 2 3 4 5
1 0
1 1
1 2
1( 0) (0) .3679
0!
1( 1) (1) .3679
1!
1( 2) (2) .1839
2!
eP X p
eP X p
eP X p
Poisson Prob Dist Poisson Prob Dist =5=5
00.020.040.060.080.1
0.120.140.160.180.2
0 1 2 3 4 5 6 7 8 9 10
0067
0337
28
Example ◦Cars arrive at a tollbooth at a rate of
360 cars per hour.◦What is the probability that only two
cars will arrive during a specified one-minute period?
The probability distribution of arriving cars for any one-minute period is Poisson with = 360/60 = 6 cars per minute. Let X denote the number of arrivals during a one-minute period. 0446.
!26e
)2X(P26
29
◦Example (cont.)◦What is the probability that at least
four cars will arrive during a one-minute period?
◦P(X>=4) = 1 - P(X<=3) = 1 - .151 = .849