chapter1(2) edit

8/11/2019 Chapter1(2) Edit

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A probability function is a function which assignsprobabilities to the values of a random variable.

Individual probability values may be denoted by thesymbol P(X=x), in the discrete case, which indicatesthat the random variable can have various specificvalues.

All the probabilities must be between 0 and 1;0 P(X=x) 1.

The sum of the probabilities of the outcomes must be1.

P(X=x)=1

It may also be denoted by the symbol f(x), in thecontinuous , which indicates that a mathematicalfunction is involved.

1.4 Probability Distributions


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ContinuousProbability

Distributions

Binomial

Poisson

ProbabilityDistributions

DiscreteProbability

Distributions

Normal


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Definition :

An experiment in which satisfied the following characteristic is calleda binomial experiment :

1. The random experiment consists of n identical trials.2. Each trial can result in one of two outcomes, which we denote by

success, S or failure, F .3. The trials are independent.4. The probability of success is constant from trial to trial, we denote the

probability of success by p and the probability of failure is equal to( 1 - p ) = q .

Examples:1. No. of getting a head in tossing a coin 10 times.2. No. of getting a six in tossing 7 dice.3. A firm bidding for contracts will either get a contract or not

3


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Definition:A binomial experiment consist of n identical trial with probabilityof success, p in each trial. The probability of x success in n trialsis given by

x = 0, 1, 2, ......, n

The M ean and Vari ance of X If X ~ B(n, p), then

Mean :Variance :Std Deviation :

where n is the total number of trials, p is the probability ofsuccess and q is the probability of failure.

4

( ) n x n x x P X x C p q

( ) E X np 2 ( ) (1 )V X np p npq

npq


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5

Given that ~ (12, 0.4), finda) ( 2)

b) ( 3)

c) ( 4)d) (2 5)

e) E( )

f) Var( )

X b

P X

P X

P X

P X

X

X


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6

12 2 102

12 3 93

12 4 84

a) ( 2) (0.4) (0.6) 0.0639

b) ( 3) (0.4) (0.6)

0.1419

c) ( 4) (0.4) (0.6)

0.2128

d) (2 5) ( 2) ( 3) ( 4)

P X C

P X C

P X C

P X P X P X P X

0.0639 0.1419 0.2128

=0.4185


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e) ( )

= 12(0.4) =4.8

f) ( ) = 12(0.4)(0.6)

= 2.88

E X np

Var X npq

7


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When the sample is relatively large, tables of Binomialare often used. Since the probabilities provided in thetables are in the cumulative form , the followingguidelines can be used:

P X k


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Definition:

A random variable X has a Poisson distribution and it isreferred to as a Poisson random variable if and only if its

probability distribution is given by

is the long run mean number of eventsfor the specific time or space dimension of interest.A random variable X having a Poisson distribution canalso be written as

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( ) for 0,1, 2,3,...!

xe P X x x x

(Greek lambda)

~ ( )o X P

with ( ) and ( ) E X Var X


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Poisson distribution is the probability distribution

of the number of successes in a given space*.*space can be dimensions, place or time or combination ofthem

Examples:1. No. of cars passing a toll booth in one hour.2. No. defects in a square meter of fabric3. No. of network error experienced in a day.


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Given that , find

13

~ (4.8)o X P

a) ( 0)

b) ( 9)

c) ( 1)

P X

P X

P X


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14

4.8 0

4.8 9

4.8a) ( 0) 0.00820!

4.8 b) ( 9) 0.03079!

c) 1 ( 0) 1 0.0082

= 0.9918

e P X

e P X

P X


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Suppose that the number of errors in a piece of software has aPoisson distribution with parameter . Find

a) the probability that a piece of software has no errors.

b) the probability that there are three or more errors in piece ofsoftware .

c) the mean and variance in the number of errors.

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3


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16

3 0

3

3 0 3 1 3 2

3

3a) ( 0)0!

0.050

b) ( 3) 1 ( 0) ( 1) ( 2)

3 3 3 1

0! 1! 2!

1 3 9 11 1 2

1 0.423 0.577

e P X

e

P X P X P X P X

e e e

e


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Example :


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Definition

19

2

2

2

12

2

A continuous random variable is said to have a

normal distribution with parameters and ,

where and 0, if the pdf of is

1( ) 2

If ~ ( , ) then and

x

X

X

f x e x

X N E X

2 V X


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ell haped

SymmetricalMean, Median and Mode

are Equal

Location is determined by themean, Spread is determined by thestandard deviation,

The random variable has aninfinite theoretical range:+ to

Mean = Median = Mode

X

f X)


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By varying the parameters and , we obtain differentnormal distributions

Many Normal Distributions


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Determine the probability or area for the portions of the Normal distribution described.

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a) (0 0.45)

b) ( 2.02 0)c) ( 0.87)

d) ( 2.1 3.11)

e) (1.5 2.55)

P Z

P Z P Z

P Z

P Z


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a) (0 0.45) ( 0.45) ( 0) 0.67364 0.5

= 0.1736

b) ( 2.02 0) (0 2.02)

= ( 2.02) ( 0)

P Z P Z P Z

P Z P Z

P Z P Z

0.97831 0.5

= 0.47831

c) ( 0.87) 0.80785 P Z


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d)

e)

28


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31

a) ( 0.6745) 0.25

b) ( 0.3585) 0.36

c) ( 2.1201) 0.983

d) ( 1.2265) 0.89

P Z

P Z

P Z

P Z


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Suppose X is a normal distribution N (25,25). Find

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a) (24 35)

b) ( 20)

P X

P X


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24 25 35 25

a) (24 35) 5 5 ( 0.2 2)

= ( 2) ( 0.2)

= ( 2) ( 0.2)

=0.

P X P Z P Z

P Z P Z

P Z P Z

97725 0.42074 0.55651

20 25 b) ( 20)

5 ( 1)

( 1) 0.84134

P X P Z

P Z

P Z


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When the number of observations or trials n in a binomial

experiment is relatively large, the normal probabilitydistribution can be used to approximate binomial probabilities.A convenient rule is that such approximation is acceptablewhen

Definition

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30, and both 5 and 5.n np nq

Given a random variable ~ ( , ), if 30 a nd both 5

and 5, then ~ ( , )

with

X b n p n np

nq X N np npq

X np Z

npq


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The continuous correction factor needs to bemade when a continuous curve is being used to

approximate discrete probability distributions.0.5 is added or subtracted as a continuouscorrection factor according to the form of theprobability statement as follows:

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.

.

.

.

.

a) ( ) ( 0.5 0.5)

b) ( ) ( 0.5)

c) ( ) ( 0.5)

d) ( ) ( 0.5)e) ( ) ( 0.5)

. continuous correction factor

c c

c c

c c

c c

c c

P X x P x X x

P X x P X x

P X x P X x

P X x P X x P X x P X x

c c


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Example

In a certain country, 45% of registered voters are male. If300 registered voters from that country are selected atrandom, find the probability that at least 155 are males.

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Solutions

38

.

is the number of male voters.~ (300,0.45)

( 155) ( 155 0.5) ( 154.5)

300(0.45) 135 5

300(0.55) 165 5

154.5 300(0.45) 154.5 135300(0.45)(0.55) 74.25

c c

X X b

P X P X P X

np

nq

P Z P Z

( 2.26)

0.01191

P Z


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Suppose that 5% of the population over 70 yearsold has disease A. Suppose a random sample of 9600 people over 70 is taken. What is theprobability that fewer than 500 of them havedisease A?


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Example

A grocery store has an ATM machine inside. An average

of 5 customers per hour comes to use the machine. Whatis the probability that more than 30 customers come touse the machine between 8.00 am and 5.00 pm?

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Solutions

.

is the number of customers come to use the ATM machine in 9 hours.

~ (45)

45 10

~ (45,45)

( 30) ( 30 0.5) ( 30.5)

30.5 45 ( 2.16)45

0.98461

o

c c

X

X P

X N

P X P X P X

P Z P Z

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