Lecture Discrete and continuous.pdf

7/29/2019 Lecture Discrete and continuous.pdf

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1

Random Variables

isa number that you dont know yet

is variable defined by the probabilities of eachpossible value in the population.

A random variable

Types ofRandom Variables

Discrete Random Variable Whole Number (0, 1, 2, 3 etc.)

Countable, Finite Number of Values Jump from one value to the next and cannot take any

values in between.

Continuous Random Variables Whole or Fractional Number

Obtained by Measuring

Infinite Number of Values in Interval Too Many to List unlike Discrete Variable


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Discrete Random VariableExamples

Experiment Random

Variable

Possible

Values

Children of One

Gender in Family

# Girls 0, 1, 2, ..., 10?

Answer 33 Questions # Correct 0, 1, 2, ..., 33

Count Cars at Toll

Between 11:00 & 1:00

# Cars

Arriving

0, 1, 2, ...,

Open Check in Lines # Open 0, 1, 2, ..., 8

Washington State PopulationSurvey and Random Variables

A telephone survey ofhouseholds throughoutWashington State.

But some households dont havephones.

number of telephones,x P(x)

0 0.03500

1 0.70553

2 0.21769

3 0.02966

4 0.00775

5 0.00332

6 0.00088

7 0.00002

8 0.00000

9 0.00015

Total 1.00000

0.04

0.71

0.22

0.03 0.01 0.000.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7 8 9

Number of Telephone Lines (x)

P(x)


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Definitions: Binomial

Binomial: Suppose that n independent experiments, or

trials, are performed, where n is a fixed number, and that

each experiment results in a success with probabilityp

and a failure with probability 1-p. The total number of

successes, X, is a binomial random variable with

parameters n and p.

We write:X ~ Bin (n, p) {reads: X is distributedbinomially with parameters n and p}

And the probability that X=r(i.e., that there areexactly rsuccesses) is:

rnrn

r

pprXP

)1()(

Binomial example

Take the example of 5 coin tosses. Whats

the probability that exactly 3 heads in 5coin tosses will occur?


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P(getting exactly 3 heads ) = (1/2)3 (1-1/2)2

= (1/2)3 (1-1/2)2

=10 ()5 =0.3125=31.25%

5

3

rnrn

r

pprXP

)1()(

SOLUTION

!3!5(/!5 3)-

x0 3 4 52

Binomial distributionfunction:X= the number of heads tossed in 5 cointosses

number of heads

p(x)

number of heads

0.3125 -

0.03125 -0.15625 -


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Definitions: Bernouilli

Bernouilli trial: If there is only 1 trial with

probability of successp and probability of

failure 1-p, this is called a Bernouilli

distribution. (special case of the binomial

with n=1)

Probability of success:

Probability of failure:

pppXP

111

1

1

)1()1(

pppXP

1)1()0( 010

1

0

Binomial distribution: example In tossing a coin 20 times, whats the

probability of getting exactly 10 heads?

176.)5(.)5(. 101020

10


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Binomial distribution: example If 20 consumers are asked whether they like

to buy a certain product , whats theprobability that 2 or fewer of themdislike?(Assume that there is a 50% chancea consumer buys the product )

4

472018220

2

572019120

1

72020020

0

108.1

108.1105.9190)5(.

!2!18

!20)5(.)5(.

109.1105.920)5(.!1!19

!20)5(.)5(.

105.9)5(.!0!20

!20)5(.)5(.

x

xxx

xxx

x

Example: Poisson distribution

Suppose that food poisoning case in certain restaurant has an

incidence of 1 in 1000 person-years. Assuming that members of

the population are affected independently, find the probability of kcases in a population of 10,000 (followed over 1 year) for k=0,1,2.

The expected value (mean) = = .001*10,000 = 1010 new cases expected in this population per year

00227.!2

)10()2(

000454.!1

)10()1(

0000454.!0

)10()0(

)10(2

)10(1

)10(0

eXP

eXP

eXP


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more on Poisson

Poisson Process (rates)

Note that the Poisson parameter can be given asthe mean number of events that occur in a definedtime period OR, equivalently, can be given as arate, such as =2/month (2 events per 1 month) thatmust be multiplied by t=time (called a PoissonProcess)

X ~ Poisson ()

!

)()(

k

etkXP

tk

E(X) = tVar(X) = t

Example

For example, if closing of savings accountsin a certain bank are occurring at a rate of

about 2 per month, then whats theprobability that exactly 4 cases will occur inthe next 3 months?

X ~ Poisson (=2/month)

%4.13!4

6

!4

)3*2(months)3in4P(X

)6(4)3*2(4

ee

Exactly 6 cases?

%16!6

6

!6

)3*2(months)3in6P(X

)6(6)3*2(6

ee


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Continuous Probability Distributions

A continuous random variable is a variable thatcan assume any value in an interval

thickness of an item

time required to complete a task

temperature of a solution

height, in inches

These can potentially take on any

value, depending only on the ability to measure

accurately.

The Uniform Distribution

The uniform distribution is a probability

distribution that has equal probabilities for all

possible outcomes of the random variable

xmin xmaxx

f(x)Total area under the

uniform probability

density function is 1.0


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The Continuous Uniform Distribution:

otherwise0

bxaifab

1

where

f(x) = value of the density function at any x value

a = minimum value of x

b = maximum value of x

The Uniform Distribution(continued)

f(x) =

Uniform Distribution Example

Example: Uniform probability distributionover the range 2 x 6:

2 6

.25

f(x) = = .25 for 2 x 66 - 21

x

f(x)

42

62

2

ba

1.33312

2)-(6

12

a)-(b

222


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The Normal ProbabilityDensity Function

The formula for the normal probability densityfunction is

Where e = the mathematical constant approximated by 2.71828 = the mathematical constant approximated by 3.14159 = the population mean = the population standard deviationx = any value of the continuous variable, < x <

22 /2)-( x

2e

2

1=f( x)

Cumulative Normal Distribution

For a normal random variable X with mean and

variance 2 , i.e., X~N(, 2), the cumulative

distribution function is

)xP(X)F(x 00

x0 x0

)xP(X 0

f(x)


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xba

xba

xba

Finding Normal Probabilities(continued)

F( a)F( b)=b)


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Example

If X is distributed normally with mean of 100and standard deviation of 50, the Z value for

X = 200 is

This says that X = 200 is two standard

deviations (2 increments of 50 units) above

the mean of 100.

2.050

100200

XZ

The Standardized Normal Table

Z0 2.00

.4772

Example:

P(Z < 2.00) = 0.5000+0.4772=.9772

.5000


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The Standardized Normal Table

Z0-2.00

Example:

P(Z < -2.00) = 1 0.9772

= 0.0228

Fornegative Z-values, use the fact that thedistribution is symmetric to find the needed

probability:

Z0 2.00

.9772

.0228

.9772.0228

(continued)

Finding Normal Probabilities

Suppose X is normal with mean 8.0 and

standard deviation 5.0

Find P(X < 8.6)

X

8.6

8.0


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Suppose X is normal with mean 8.0 and

standard deviation 5.0. Find P(X < 8.6)

Z0.120X8.68

= 8 = 10

= 0 = 1

(continued)

Finding Normal Probabilities

0.125.0

8.08.6

XZ

P(X < 8.6) P(Z < 0.12)

EXAMPLE 2

The daily water usage per person inTetuan, Zamboanga City follows the a normaldistribution with a mean of 20 gallons and a

standard deviation of 5 gallons. About 68percent of those living in Tetuan will use howmany gallons of water?

About 68% of the daily water usage will liebetween 15 and 25 gallons.


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EXAMPLE 2

What is the probability that a person

from Tetuan selected at random will

use at least 20 to less than 24 gallons

per day?

00.05

2020

Xz

80.05

2024

Xz

- 5

0 . 4

0 . 3

0 . 2

0 . 1

. 0

x

f

(

x

r a l i t r b u i o n : = 0 ,

-4 -3 -2 -1 0 1 2 3 4

P(0 z


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EXAMPLE 2 continued

What percent of the population usewater at least 18 to less than 26 gallons

per day?

40.05

2018

Xz

20.15

2026

Xz

Example 2 continued

The area associated with a z-value of0.40 is

.1554.

The area associated with a z-value of 1.20 is

.3849.

Adding these areas, the result is .5403.

We conclude that 54.03 percent of the

residents use at least 18 but less than 26

gallons of water per day.

Lecture Discrete and continuous.pdf

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