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Discrete probability distributions

The Binomial Distribution

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Revision

Combinations the number of waysxelements

can be selected from nelements.

35)123)(1234(

1234567

!3!4

!7

)!47(!4

!7

1and1NB)!(!

!

47

0

C

CCxnx

nC

nnnxn

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Binomial Probability Distribution

The binomial distribution has the following characteristics:

An outcome of an experiment is classified into one of twomutually exclusive categor ies, such as a success or failure.

The data collected are the results of counts in a series of

trials. The probability of success stays the same for each trial.

The trials are independent.

For example, tossing an unfaircoin three times. H is labeled success and T is labeled failure.

The data collected are number of H in the three tosses.

The probability of H stays the same for each toss.

The results of the tosses are independent.

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Binomial Probability Distribution

To construct a binomial distribution, let

n = the total number of trials

p = the probability of success

q =1-pbe the probability of failure

x= number of successes in ntrials

The formula for the binomial probability distribution

is:

P(x) = nCxpxqn-x

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EXAMPLEx = number of patients who will experiencenausea following treatment with Phe-Mycin

2 4-2 2 24!p(2) P(x=2)= (0.1) (0.9) =6(0.1) (0.9) =0.04862!(4-2)!

Find the probability that 2 of the 4 patientstreated will experience nausea.

n = 4, p = 0.1, q = 1 p = 1 - 0.1 = 0.9

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Binomial Probability Distribution

The formula for the binomial probability distribution is:

P(x) = nCxpxqn-x

TTT, TTH, THT, THH,

HTT, HTH, HHT, HHH.

X=number of heads

The coin is fair, i.e., P(head) = 1/2.

P(x=0) = 3C00.50

(1- 0.5)3-0

=3!/(0!3!) (1) (1/8)=1/8 P(x=1) = 3C10.51(1- 0.5)3-1 =3!/(1!2!) (1) (1/8)= 3/8 P(x=2) = 3C20.52(1- 0.5)3-2 =3!/(2!1!) (1) (1/8)= 3/8 P(x=3) = 3C30.53(1- 0.5)3-3 =3!/(3!0!) (1) (1/8)= 1/8

When the coin is not fair, simple counting rule will not work.

x)!x!(n

n!Cxn

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Mean & Standard Deviation of

the Binomial Distribution

np

npq

n= total number of trials

p= probability of success

q= probability of failure

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EXAMPLE

The Alabama Department of Labor

reports that 20% of the workforce in

Mobile is unemployed. From a sample of14 workers, calculate the following

probabilities:

Exactly three are unemployed. At least three are unemployed.

At least one are unemployed.

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EXAMPLE

The probability of exactly 3:

The probability of at least 3 is:

The probability of at least one being unemployed:

2501.0)0859.0)(0080.0)(364(

)20.01()20.0()3( 113314

CP

551.000....172.250.

)80(.)20(....)80(.)20(.)3( 0141414113

314

CCxP

.956.0441

.20)(1(.20)C1

P(0)11)P(x140

014

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EXAMPLE

Since p = 0.2 and n = 14.

Hence, the mean is:

=np= 14(0.2) = 2.8.

The standard deviation is:

)8.0)(2.0)(14(npq

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Shape of a Binomial Distribution

Symmetric if p = .5

Skewed to right if p < .5

Skewed to left if p > .5

SYMMETRIC

SKEWED LEFT

SKEWED RIGHT