Post on 31-Mar-2015
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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
Classic binomial experiment: Classic binomial experiment: tossing atossing acoin a pre-specified number coin a pre-specified number of timesof times
Toss a coin 10 timesResult of each toss: head or tail
(designate one of the outcomes as a success, the other as a failure; makes no difference)
P(head) and P(tail) are the same on each toss
trials are independent◦ if you obtained 9 heads in a row, P(head) and
P(tail) on toss 10 are same as P(head) and P(tail) on any other toss (not due for a tail on toss 10)
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.
ExamplesExamples
a. Yes; n=10; success=“major repairs within 3 months”; p=.05
b. No; n not specified in advancec. No; p changesd. Yes; n=1500; success=“chip is
defective”; p=.10
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
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Binomial Probability Histogram: n=100, p=.95
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0.1
0.11
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70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100
ExampleExample
A production line produces motor housings, 5% of which have cosmetic defects. A quality control manager randomly selects 4 housings from the production line. Let x=the number of housings that have a cosmetic defect. Tabulate the probability distribution for x.
SolutionSolution(i) D=defective, G=goodoutcome x P(outcome)GGGG 0 (.95)(.95)(.95)(.95)DGGG 1 (.05)(.95)(.95)(.95)GDGG 1 (.95)(.05)(.95)(.95) : : :DDDD 4 (.05)4
SolutionSolution
0 44 0
1 34 1
2 24 2
3 14 3
44 4
( ) is a binomial random variable
( ) , 0,1,2, ,
4, .05 ( .95)
(0) (.05) (.95) .815
(1) (.05) (.95) .171475
(2) (.05) (.95) .01354
(3) (.05) (.95) .00048
(4) (.05) (.9
x n xn x
ii x
p x C p q x n
n p q
p C
p C
p C
p C
p C
05) .00000625
SolutionSolution
x 0 1 2 3 4p(x) .815
.171475 .01354 .00048 .00000625
Example (cont.)Example (cont.)
x 0 1 2 3 4p(x) .815
.171475 .01354 .00048 .00000625
What is the probability that at least 2 of the housings will have a cosmetic defect?
P(x p(2)+p(3)+p(4)=.01402625
Example (cont.)Example (cont.)
What is the probability that at most 1 housing will not have a cosmetic defect? (at most 1 failure=at least 3 successes)
P(x )=p(3) + p(4) = .00048+.00000625 = .00048625
x 0 1 2 3 4p(x) .815 .171475 .01354 .00048 .00000625
Using binomial tables; Using binomial tables; n=20, p=.3n=20, p=.3
P(x 5) = .416P(x > 8) = 1- P(x 8)=
1- .887=.113P(x < 9) = ?P(x 10) = ?P(3 x 7)=P(x 7) - P(x 2)
.772 - .035 = .737
9, 10, 11, … , 20
8, 7, 6, … , 0 =P(x 8)
1- P(x 9) = 1- .952
Binomial n = 20, p = .3 Binomial n = 20, p = .3 (cont.)(cont.)P(2 < x 9) = P(x 9) - P(x 2)
= .952 - .035 = .917P(x = 8) = P(x 8) - P(x 7)
= .887 - .772 = .115
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
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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 expected value and standard
deviation of the count X of color blind individuals
in the SRS of 25 Caucasian American males?
E(X) = np = 25*0.08 = 2
SD(X) = √np(1 p) = √(25*0.08*0.92) = 1.36
p = .08n = 10
p = .08n = 75
E(X) = 10*0.08 = 0.8 E(X) = 75*0.08 = 6
SD(X) = √(10*0.08*0.92) = 0.86 SD(X) = (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
Recall from beginning of Recall from beginning of Lecture Unit 4: Hardee’s vs Lecture Unit 4: Hardee’s vs
The ColonelThe ColonelOut of 100 taste-testers, 63
preferred Hardee’s fried chicken, 37 preferred KFC
Evidence that Hardee’s is better? A landslide?
What if there is no difference in the chicken? (p=1/2, flip a fair coin)
Is 63 heads out of 100 tosses that unusual?
Use binomial rv to analyzeUse binomial rv to analyze
n=100 taste testersx=# who prefer Hardees chickenp=probability a taste tester
chooses HardeesIf p=.5, P(x 63) = .0061 (since
the probability is so small, p is probably NOT .5; p is probably greater than .5, that is, Hardee’s chicken is probably better).
Recall: Mothers Recall: Mothers Identify NewbornsIdentify Newborns
After spending 1 hour with their newborns, blindfolded and nose-covered mothers were asked to choose their child from 3 sleeping babies by feeling the backs of the babies’ hands
22 of 32 women (69%) selected their own newborn
“far better than 33% one would expect…”Is it possible the mothers are guessing?Can we quantify “far better”?
Use binomial rv to Use binomial rv to analyzeanalyze
n=32 mothersx=# who correctly identify their own babyp= probability a mother chooses her own
babyIf p=.33, P(x 22)=.000044 (since the
probability is so small, p is probably NOT .33; p is probably greater than .33, that is, mothers are probably not guessing.