cis2460 statistics tutorial part 4

8/3/2019 CIS2460 Statistics Tutorial Part 4

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Statistical Tutorial(Chapter 5)

Compiled and Presented by

Alana Cordick

Histograms

Example:

0.04128

0.0277

0.04136

0.04125

0.24714

0.15453

0.371102

0.10301

Relative FrequencyFrequencyArrivals /

Party

Histogram of Party Size

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

1 2 3 4 5 6 7 8

# in Party

Relative

Frequency

Statistical Concepts Part 4

Continuous Distributions

Continuous Distributions

Continuous random variables can be used to describe randomphenomena in which the variable can take on any value in someinterval

In this section, the distributions studied are: Normal

Uniform

Exponential

Gamma

Erlang

Weibull

Triangular

Lognormal

Beta distribution


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Normal (Gaussian)Distribution Continuous

AKA Bell Curve

Most common continuous distribution found in nature (as seenthrough the Central Limit Theorem to come)

Distribution:

Mean:

Variance:

Denoted by:X ~ N(,2)

= pp x

xxf ,

2

1exp

2

1)(

2

pp

02f

Normal (Gaussian)Distribution

Evaluating the distribution:

Use numerical methods (no closed form)

Independent of and , using the standard normaldistribution:

Z ~ N(0,1)

Transformation of variables: let Z = (X - ) /,

=z

tdtez

2/2

2

1)(where,

( )

)()(

2

1

)(

/)(

/)(2/

2

==

=

==

x

x

xz

dzz

dze

xZPxXPxF

Normal (Gaussian)

Distribution

Standard Normal Distribution

Has a = 0 and = 1 i.e.

Special properties:

.

f(-x)=f(+x); the pdf is symmetric about

The maximum value of the pdf occurs atx =

The mean and mode are equal

Linearity

~X N(0, 1)

0)(limand,0)(lim ==

xfxf xx

Normal Distribution

Linearity Any linear combination of normally distributed

random variable is normally distributed

Y = aiX

i

i=1

n

Normally distributed

In General

~X+ Y N(X + Y,X2 + Y2)~X N(X, X2)If thenand ~Y N(Y, Y2)

~X Y N(X Y,X2 + Y2)


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Normal Distribution Evaluating the distribution:

Use numerical methods (no closed form)

Independent of and , using the standard normal

distribution:

Z ~ N(0,1)

Transformation of variables: let Z = (X - ) /,

=z

tdtez

2/2

2

1)(where,

( )

)()(

2

1

)(

/)(

/)(2/

2

==

=

==

x

x

xz

dzz

dze

x

ZPxXPxF

Uniform Distribution

A random variable X is uniformly

distributed on the interval (a,b) if its pdf is

given by:

And a cdf of:

=

otherwise,0

,1

)(bxa

abxf

=

bx

bxaab

ax

ax

xF

,1

,

,0

)( p

p


P(x1 < X < x2) is proportional to the length

of the interval for all x1 and x2 such that:

a


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Applications:

Random numbers, uniformly distributed

between zero and 1, provide the means to

generate a truly random event

Used to generate samples of random variates

from all other distributions


Example

A bus arrives every 20 minutes at a specified

stop beginning at 6:40 a.m. and continuing until

8:40 a.m. A certain passenger does not know

the schedule, but arrives randomly (uniformly

distributed) between 7:00 a.m. and 7:30 a.m.every morning. What is the probability that the

passenger wait more than 5 minutes for a bus?


Example

First concentrate on the bus arrival times at the stop:

6:40

7:00

7:20

7:40

We are only interested in its arrival between 7:00 and

7:30


Example

So we know the bus will arrive at 7:00 and 7:20

What ranges of time will the passenger nothave to wait more than 5 minutes?

7:00

7:15 7:20

Therefore, we want to find the probability thatthe passenger arrives during the other times

P(0 < X < 15) + P(20 < X < 30)


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Example

Since X is a uniform random variable (0, 30),

we have the following:

6

5

30

201

30

15

030

020

030

030

030

015

)20()30()15(

=

+=

+

=

+= FFF

There is an 5/6 =

83% chance thepassenger will

have to wait more

than 5 minutes

ab

axxF

=)(

Exponential Distribution

A random variable X is said to be

exponentially distributed with parameter >

0 if its pdf is given by:

With cdf:

=

otherwise,0

0,)(

xexf

x

==

0,1

00,)(

0xedte

x

xF x xt

p


Application

Model interarrival times (time between arrivals) whenarrivals are completely random

= arrivals / hour

Model service times

= services / minute

Model the lifetime of a component that failscatastrophically (i.e. light bulb)

= failure rate


is the value where the pdf intersects the y

axis


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Memoryless property

For all s>= 0 and t >=0

P(X > s + t | X > s) = P(X > t)

In other words, if you know a component hassurvived s hours so far, the same distribution of

the remaining amount of time that it survives isthe same as the original distribution

It does not remember the it already has been usedfor s amount of time


Example:

Suppose the life of an industrial lamp is

exponentially distributed with failure rate =

1/3 (one failure every 3000 hours on the avg.)

Determine the probability the lamp will last

longer than its mean life

368.0)1(1)3(1)3(1)3(13/3=====>

eeFXPXP

=

0,1

00,)(

xe

xxF

x

p


Example:

The probability that an exponential random

variable is greater than its mean is always 0.368 Regardless of the value of

The probability the industrial lamp will last

between 2000 and 3000 hours is:

145.0487.0632.0)1()1(

)1()1()2()3()32(

3/21

3/23/3

===

==

ee

eeFFXP


Example:

The probability that the lamp will last for another 1000

hours given that it is operating after 2500 hours

717.0

)1(1

)1(1

)1(

)5.2|15.2(

)5.2|5.3(

3/1

3/1

==

=

=

>=

>+>=

>>

e

e

XP

XP

XXP

XXP


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Gamma Distribution

A function used in defining the gammadistribution is the gamma function, which isdefined for all > 0 as:

Gamma function is a generalization of thefactorial notion to all positive #s (not justintegers)

)!1()( =

Gamma Distribution

A random variable X is gamma distributedwith parameters and if its pdf is givenby:

Where is a shape parameter and is a scaleparameter

>

=

otherwise

xexxf

x

,0

0,)()()(

1

Gamma Distribution

Mean:

Variance:

Cdf of X:

1)( =XE

21)( =XV

0,)(

0,0

)(1

)(1

>

=

xdtet

x

xFt

x

Gamma Distribution

Gamma and exponential distributions are

related when is an integer

If the random variable, X, is the sum ofindependent, exponentially distributedrandom

variables, each with parameter , then X has a

gamma distribution with parameters and

Independent if: X= X1+ X2 + + X


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Weibull Distribution

A random variableXhas a Weibulldistribution if its pdf has the form

3 parameters:

Location parameter: ,

Scale parameter: , (> 0)

Shape parameter. , (> 0)

=

otherwise,0

,exp)(

1

xxx

xf


Example: = 0 and = 1

When = 1,

X~ exp(= 1/)


Mean:

Variance:

Cdf:

++= 1

1)(

vXE

+

+=

2

2 1112)(

XV

=

otherwise

xexxf

x

,0

0,)()()(

1

Erlang Distribution

Mean:

Variance:

Cdf:

1111)( =++=

kkkXE L

( ) ( ) ( ) 22221111

)( kkkk

XV =++= L

( )

>=

=

0,0

0,!

1)(

1

0

x

xi

xke

xF

k

i

ixk

Erlang Distribution

Application

When a customer must complete a series ofk

stations The next customer cannot enter start the first station

until the previous customer has completed all k

stations

Erlang Distribution

Example A college professor of electrical engineering is leaving

home for the summer, but would like to have a lightburning at all times to discourage burglars. Theprofessor rigs up a device that will hold two light bulbs.The device will switch the current to the second bulb ifthe first bulb fails. The box in which the light bulbs arepackages says, Average life 1000 hours, exponentiallydistributed. The professor will be gone 90 days (2160hours).

What is the probability that a light will be burning when thesummer is over and the professor returns?


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Erlang Distribution

Example

Reliability function the probability that the systemwill operate at least x hours

R(x) = 1 F(x)

= k= 2

k= 1 / 1000, so = 1/2000 per hour

Determine F(2160):[ ]

( )

636.0

!

16.21

!

)2160)(2000/1)(2(1)2160(

1

0

16.2

1

0

)2160)(2000/1)(2(

=

=

=

=

=

i

i

i

i

ie

i

eF

Erlang Distribution

Example

Now:

R(x) = 1 F(x)

R(x) = 1 0.636 = 0.364

Therefore, there is a 36% chance that a light

will be burning when the professor returns

Summary

Normal

Exponential

Weibull

= pp x

xxf ,

2

1exp

2

1)(

2

=

otherwise,0

0,)(

xe

xf

x

==

0,1

00,)(

0xedte

x

xF x xt

p

eeeFXPXP


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Triangular Distribution

A random variable X has a triangular distribution if

its pdf is given by

( )( )( )

( )( )( )

cbawhere

otherwise,0

,2

1

,2

)(

0 if its pdfis given by:

Range: (0, 1)

( )( )

( )( ) ( )( )21

21

21

21

11

,

otherwise,0

10,,

1

)(

21

+

=

=0} is a Poisson

process with mean rate if:

1. Arrivals occur one at a time

2. {N(t), t>=0} has stationary increments (completely

random, without rush or slack periods)

3. {N(t), t>=0} has independent increments


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Poisson Distribution

Properties

Equal mean and variance:E[N(t)] = V[N(t)] = t (= as before)

Since stationary increments, the number of arrivals in

time s to t, such that s < t, is also Poisson-distributedwith mean (t-s)

,...2,1,0and0for,!

)(])([ ===

ntn

tentNP

nt

( )[ ]

[ ] [ ])()()()()(

,...2,1,0,!

])()([

)(

sNtNVstsNtNE

nn

stensNtNP

nst

==

=

==

Interarrival Times

Consider the interarrival times of a Poisson process (A1, A2, ), whereAi is the elapsed time between arrival i and arrival i+1

The 1st arrival occurs after time t iff there are no arrivals in the interval[0,t], hence:

P{A1 > t} = P{N(t) = 0} = e-t

Since: P{A1


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Nonstationary Poisson Process

(NSPP)

Poisson Process without the stationary increments,characterized by (t), the arrival rate at time t

Useful when arrival rates vary i.e. restaurant

The expected number of arrivals by time t:

Relating stationary Poisson process n(t) with rate =1 andNSPPN(t) with rate (t):

Let arrival times of a stationary process with rate = 1be t1, t2, , and arrival times of a NSPP with rate (t)be T1, T2, , we know: ti = (Ti) & Ti =

1(ti)

=t(s)ds(t)

0


(NSPP)

Example: Suppose arrivals to a Post Office haverates 2 per hour from 8 am until 12 pm, and then0.5 per hour until 4 pm

Let t = 0 correspond to 8 am, NSPPN(t) has ratefunction:

Expected number of arrivals by time t:

=

84,5.0

40,2

)(p

p

t

t

t

( )

+=+=+

=

84,62

5.0*45.0*4*25.02

40,2

)( 4

0 4p

p

tt

tdsds

tt

t t


(NSPP)

Hence, the probability distribution of the number

of arrivals between 11 am and 2 pm.

11 am = 3rd hour

2 pm = 6th hour

P[N(6) N(3) = k]

= P[N((6)) N((3)) = k]

= P[N(9) N(6) = k]

= e(9-6)(9-6)k/k!

= e3(3)k/k!

Empirical Distributions

A distribution whose parameters are the observed values ina sample of data.

May be discrete or continuous

May be used when it is impossible or unnecessary to establish thata random variable has any particular parametric distribution.

Advantage: no assumption beyond the observed values in thesample.

Disadvantage: sample might not cover the entire range of possiblevalues.


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References

Discrete-Event System Simulation, 4th Ed.

Banks, Carson, Nelson, Nicol

Published by Prentice-Hall, 2005

cis2460 statistics tutorial part 4

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