chapter 4 continuous random variable and probability...

Chapter 4 Continuous Random Variable and ProbabilityDistributions

Seungchul Baek

STAT 355 Introduction to Probability and Statistics for Scientists andEngineers

Seungchul Baek STAT 355 Introduction to Probability and Statistics for Scientists and Engineers 1

Continuous Random Variable

A continuous random variable is one with an interval (either finite orinfinite) of real numbers for its range.

Examples

Let X = length in meter.Let X = temperature in ¶F.Let X = time in seconds


o -

- in sense of

the length ofinterval

.

Continuous Random Variable

Because the number of possible values of X is uncountably infinite, theprobability mass function (pmf) is no longer suitable.

For a continuous random variable, P(X = x) = 0, the reason for thatwill become clear shortly.

For a continuous random variable, we are interested in probabilities ofintervals, such as P(a Æ X Æ b), where a and b are real numbers.

We will introduce the probability density function (pdf) to calculateprobabilities, such as P(a Æ X Æ b).


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Probability Density Function

Every continuous random variable X has a probability densityfunction (pdf), denoted by fX (x).

Probability density function fX (x) is a function such that

fX (x) Ø 0 for any x œ Rs Œ

≠Œ fX (x)dx = 1

P(a Æ X Æ b) =s b

a fX (x)dx , which represents the area under fX (x)from a to b for any b > a.

If x0 is a specific value, then P(X = x0) = 0. We assign 0 to areaunder a point.


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Cumulative Distribution Function

Here is a pictorial illustration of pdf:

Let x0 be a specific value of interest, the cumulative distributionfunction (CDF) is defined via

FX (x0) = P(X Æ x0) =⁄ x0

≠ŒfX (x)dx .


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of art- X.

area

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x discrete F×lxo7=PHEKo)¥P=

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Cumulative Distribution Function

If x1 and x2 are specific values, then

P(x1 Æ X Æ x2) =⁄ x2

x1fX (x)dx

= FX (x2) ≠ FX (x1).

From last property of a pdf, we have

P(x1 Æ X Æ x2) = P(x1 < X < x2)= P(x1 Æ X < x2)= P(x1 < X Æ x2)


÷::Plx=r

②=PlF•IxpT

Example: Electric Current

Let the continuous random variable X denote the current measured in athin copper wire in milliamperes. Assume that the range of X (measured inmA) is [0, 20], and assume that the probability density function of X isfX (x) = 0.05 for 0 Æ x Æ 20. What is the probability that a currentmeasurement is less than 10 milliamperes?


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.o 57 0

[found a = f."

o.osold = 0.0544! 0,5?

= fo' 8. ofda -- aos do

= 0.5P ( o#Elo)

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Example

Suppose that Y has the pdf

fY (y) =I

3y2, 0 < y < 1

0, otherwise.

Find the CDF of Y .

Calculate P(Y < 0.3)

Calculate P(0.3 Æ Y Æ 0.8)


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"I g.'gurdy

--I'? !

- ↳in-- PIYE 's)

= f.VzEdt#= Fy 19=0.3)-

= [t3) It

%;Er±÷!a⇒, .⇐. ⇒ our= "

= 0, of 3 - o , 33 = 0 ,

485

Example 4.5

“Time headway” in tra�c flow is the elapsed time between the time that onecar finishes passing a fixed point and the instant that the next car begins topass that point. Let X denote the time headway for two randomly chosenconsecutive cars on a freeway during a period of heavy flow. The pdf of X is

fX (x) =I

0.15e≠0.15(x≠0.5), x Ø 0.5

0, otherwise.

What is the probability that headway time is at most 5 seconds?


Valid

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"I /!,f in , DX= I

0

PIKE 5) = go.si?.is-e-I5" da

÷← o.iseSo .

? e-""' da

Mean of a Continuous Random Variable

The mean (expectation) and variance can also be defined for a continuousrandom variable. Integration replaces summation in the calculation ofexpectation for a discrete r.v.

Recall that for a discrete random variable Y . The mean of Y is defined as

E (Y ) = µY =ÿ

all yy · pY (y).

Definition: For a continuous random variable X . The mean of X isdefined as

E (X ) = µX =⁄ Œ

≠ŒxfX (x)dx


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Mean of a Continuous Random Variable

Theorem: Let X be a continuous random variable with pdf fX (x). Supposethat g is a real-valued function. Then, g(X ) is a random variable and

E [g(X )] =⁄ Œ

≠Œg(x)fX (x)dx .


discrete-

Efglx ) ) =a ,9124 Px

-

Variance of a Continuous Random Variable

Definition: The variance of X , denoted as var(X ) or ‡2, is

‡2 = var(X ) = E [(X ≠ µ)2] =⁄ Œ

≠Œ(x ≠ µ)2

fX (x) dx .

The population standard deviation of X is

‡ =Ô

‡2,

the positive square root of the variance.

The computing formula for variance is the same as the discrete case,i.e.,

var(X ) = E (X 2) ≠ [E (X )]2.


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Example: Electric Current

Recall that the pdf of X is

fX (x) =I

0.05, 0 Æ x Æ 200, otherwise.

Compute E (X ) and var(X ).


-

El x ) =kfcxidK= 12%10.05)dK

=o.o5[the] too)

=④

Eff) = 1202540.05> DX

Varix) - Ekg - {Eggs= "5LIzx3] ?

= 533.33 - Co2 =(203-03)--1133.33=33.33

Relationship Between fY (y) and FY (y)

Definition: A r.v. Y is continuous if its cdf FY (y) is continuous for≠Œ < y < Œ.

Definition: Suppose Y is a continuous r.v. with cdf FY (y). Then pdf of Y

isfY (y) = d

dyFY (y).

Corollary: From the Fundamental Theorem of Calculus, We see that

FY (y) =⁄ y

≠Œf (t)dt.


Idef

¥Fyc8)=fcy#

Percentiles

Definition: Let p be a number in [0, 1]. The 100p-th percentile of thedistribution of a continuous random variable Y , denoted by ÷(p), is definedby

p = F{÷(p)} =⁄ ÷(p)

≠Œf (y)dy

The median of a continuous distribution, denoted m, is the 50-thpercentile, i.e.,

12 = F (m) =

⁄ m

≠Œf (y)dy


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I

① ② eta

•

Fcm

Uniform DistributionDefinition: A continuous random variable X is said to have a uniformdistribution on the interval [a, b] if the pdf of X is

fX (x) =I 1

b≠a , a Æ x Æ b

0, otherwise.

Notation: X ≥ U(a, b).

Mean and variance

E (X ) = a + b

2

var(X ) = (b ≠ a)2

12

If [a, b] = [0, 1], it is said that X follows the standard uniformdistribution.Seungchul Baek STAT 355 Introduction to Probability and Statistics for Scientists and Engineers 16

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Example 4.9

The distribution of the amount of gravel (in tons) sold by a particularconstruction supply company in a given week is a continuous randomvariable with pdf

fX (x) =I3

2(1 ≠ x2), 0 Æ x Æ 1

0, otherwise.

What is the cdf of X?

What is the median?


Fires f! fetid t= So

"

Izu- t' Idf = ? ft - ft']!

I= Zz (x - 13×3)

⑨mine:÷:"

Fcm) = 3⇒m - Ism3) = Izm3-3m

.

Example 4.10 and 4.12

The pdf of X is

fX (x) =I3

2(1 ≠ x2), 0 Æ x Æ 1

0, otherwise.

Compute E (X ) and var(X ).


E (x )

t

f a f ex> d k= fo

'

x I a- isda-

- II tie- tax a ] ! -- Ift -ta)= I

D= EH7 - ee#⇒Idef

=g x 2few d kI::::::.i.÷.t - t

÷ o. 0593-

Normal Distribution

Most widely used distribution.

Central Limit Theorem (Chapter 5): Whatever the distribution therandom variable follows, if we repeat the random experiment over andover, the average result over the replicates follows normal distributionalmost all the time when the number of the replicates goes to large.

Other names: “Gaussian distribution”,“bell-shaped distribution” or“bell-shaped curve.”


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Density of Normal Distribution

A random variable X with the pdf

f (x) = 1Ô2fi‡

e≠ (x≠µ)2

2‡2 , ≠Œ < x < Œ

is a normal random variable with parameters µ and ‡, where≠Œ < µ < Œ, and ‡ > 0. Also,

E (X ) = µ, var(X ) = ‡2.

We use X ≥ N (µ, ‡2) to denote the distribution. If X ≥ N(0, 1), it iscalled the standard normal distribution.

Now our objective is to calculate probabilities (of intervals) for anormal random variable through R or normal probability table.


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Density of Normal Distribution

The plot of the pdfs of normal distributions with di�erent parametervalues:

CDF: The cdf of a normal random variable does not exist in closedform. Probabilities involving normal random variables and normalquantiles can be computed numerically


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Characteristics of Normal pdf

Bell-shaped curve.

≠Œ < x < Œ, i.e., the range of X is the whole real line.

µ determines the location of a distribution and has the highest value off (x) at µ.

The curve is symmetric about µ.

‡ determines the dispersion of a distribution.

Inflection points at µ ± ‡.


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Example: Strip of Wire

Assume that the current measurements in a strip of wire follow a normaldistribution with a mean of 10 milliamperes and a variance of 4(milliamperes)2. What is the probability that a measurement exceeds 13milliamperes?


O O

XNN ( 10,4)

PIX > 13 ) = I - P ( XE 13 )-# (x- 1012

= I - L! e da

-

0.0668'

=. Shove solve numerically=

Empirical RuleFor any normal random variable X , we have

P(µ ≠ ‡ < X < µ + ‡) = 0.6827P(µ ≠ 2‡ < X < µ + 2‡) = 0.9543P(µ ≠ 3‡ < X < µ + 3‡) = 0.9973

These are summarized in the following plot:


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Earthquakes in a Town

Since 1900, the magnitude of earthquakes that measures 0.1 or higher onthe Richter Scale in a certain location in California is distributedapproximately normally, with µ = 6.2 and ‡ = 0.5, according to dataobtained from the United States Geological Survey.

Approximately what percent of the earthquakes are above 5.7 on theRichter Scale?

What is the approximate probability that an earthquake is above 6.7?


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so%→x% .

Standard Normal Distribution

If X is a normal random variable with E (X ) = µ and var(X ) = ‡2, therandom variable

Z = X ≠ µ

‡

is a normal random variable with E (Z ) = 0 and var(Z ) = 1. That is, Z iscalled the standard normal random variable.

Creating a new random variable by this transformation is referred to asstandardizing.

Z is traditionally used as the symbol for a standard normal randomvariable.

�(z) is commonly used to stand for the cdf of Z .


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O D =D

② n Nfo , 't ⇒ standard

←

¥7 plz) : the pdf of -2 . d¥=fE).

Standard Normal Probability Table


Standard Normal Probability Table

With the help of normal probability table, we can calculate the probabilitiesfor nonstandard normal distribution through standardizing.

Suppose X ≥ N (10, 4), we want to calculate P(X > 13).

P(X < 7)?

P(X > 7)?

P(5 < X < 7)?


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(M "taste

= I - 0.768=79332 .

= Pl a f-Pleas)-

Peso .

÷.

=p ( Z 71.5)

2- ) = I - PIZEI.NL= p ( -2.54 Z s - 1.5 ) = I - 0.9332

= plzH#E÷ :&:b -

-o

-out

.

Example: Steel Bolt

The thickness of a certain steel bolt that continuously feeds amanufacturing process is normally distributed with a mean of 10.0 mm andstandard deviation of 0.3 mm. Manufacturing becomes concerned about theprocess if the bolts get thicker than 10.5 mm or thinner than 9.5 mm.

Find the probability that the thickness of a randomly selected bolt isgreater than 10.5 or smaller than 9.5


= I - pfzcjhtplzc-l.de÷ 0.095 C

O°- §PIXIES- or xca

= PIX > 10.5) t P ( X 29.5)= FEES) t PC XL 9.5)= I - pl ' a

" + pl'it÷s

Inverse Normal ProbabilitiesSometimes we want to answer a question which is the reverse situation. Weknow the probability, and want to find the corresponding value of Y .


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Example: Steel Bolt

What is the cuto� value that approximately 2.5% of the boltsproduced will have thicknesses less than this value?

What is the cuto� value that approximately 1% of the bolts producedwill have thicknesses greater than this value?


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qq.at#Z=-t%f2C-.96)--o.o2SplZsBf=o.ozs-←A =P - 1.96)"g2-

-

ME > z )/9

-X-lo s - 1.96×0.3

I - p (2-783)=0.01 XL- 1.96×0-3-110

Pl o.gg =T⇒ pfx#0.699 ) 10.99

Example: Volume

The fill volume of an automatic filling machine used for filling cans ofcarbonated beverage is normally distributed with a mean of 12.4 fluidounces and a standard deviation of 0.1 fluid ounce.

What is the probability that a randomly chosen can will containbetween 12.3 and 12.5 ounces?

2.5% of the cans will contain less than ounces.


✓ l2.④

MY?f!-

-"* ⇐ Pf c-i.gg""

<⇒ p ( X L 12.2041=0.025

Approximating the Binomial Distribution

Suppose that X ≥ B(n, p). It can be approximated by a normal distributionvia

P(X Æ x) ¥ �A

x + 0.5 ≠ npÔ

npq

B

,

where � is the cdf of the standard normal distribution. In practice, theapproximation is adequate provided that both np Ø 10 and nq Ø 10.

What is the constant 0.5?

Why do we need a condition like np Ø 10 and nq Ø 10?


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Example 4.20

Suppose that 25% of all students at a large public university financial aid.Let X be the number of students in a random sample of size 50 who receivefinancial aid, so that p = 0.25.

What is the approximate probability that at most 10 students receiveaid? How about the exact probability?

What is the approximate probability that between 5 and 15 studentsreceive aid? How about the exact probability?


*NB (50,0125)

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-- 840.98-7-Ff=

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Exponential Distribution

We have discussed Poisson distribution in the previous chapter whichfor example can model the number of car accidents for a given lengthof time t.

The waiting time between accidents is another random variable thatis often of interest. We can use an exponential distribution to modelsuch a waiting period.


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We define

X = the waiting time between two car accidents

andN = the number of accidents during time of length x

We know that if the mean number of accidents is ⁄ per base unit, thenthe random variable N ≥ Poisson(⁄x).


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0 A

←o_0

Exponential DistributionTo model the waiting time, suppose there is no accident during thetime of length x . Now,

P(X > x) = P(N = 0) = e≠⁄x (⁄x)0

0! = e≠⁄x ,

which means that there is no events in [0, x ].

By the complement rule, it follows that

FX (x) = P(X Æ x) = 1 ≠ P(X > x) = 1 ≠ e≠⁄x .

By di�erentiating the CDF of X , the pdf of X is

fX (x) =I

⁄e≠⁄x , x Ø 0

0, otherwise,

where ⁄ is “mean number of events per unit interval.”Seungchul Baek STAT 355 Introduction to Probability and Statistics for Scientists and Engineers 37

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The plot of pdfs of exponential distributions with di�erent values of ⁄ isshown below. The shorthand notation for X following exponentialdistribution is given by X ≥ expon(⁄)


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Mean and Variance

Suppose that X ≥ expon(⁄), then

E (X ) = 1⁄

var(X ) = 1⁄2


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Summary

Suppose that X ≥ expon(⁄), then

pdf: fX (x) = ⁄e≠⁄x , for x Ø 0

dexp(x, ⁄) in Rcdf: FX (x) = P(X Æ x) = 1 ≠ e

≠⁄x , for x Ø 0

pexp(x, ⁄) in RNote: P(X > x) = 1 ≠ FX (x) = e

≠⁄x , for x Ø 0

Mean: E (X ) = 1⁄

Variance: var(X ) = 1⁄2


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'

.

Example: Computer Usage

Let X denote the time in hours from the start of the interval until the firstlog-on. Then, X has an exponential distribution with 1 log-on per hour. Weare interested in the probability that X exceeds 6 minutes. (Hint: Because

⁄ is given in log-ons per hour, we express all time units in hours. That is, 6

minutes =0.1 hour)


petanhour

I↳ = hour

P ( X> o . 1) = I - pix co - y

= I - J!- '

e-"da

- I -Ff÷ o

. 904 .

Example: Accidents

The time between accidents at a factory follows an exponential distributionwith a historical average of 1 accident every 900 days.

What is the probability that there will be more than 1200 days betweenthe next two accidents?

What is the probability that there will be less than 900 days betweenthe next two accidents?


unit interval

f -- I 9ood⑦P'( X 7 12%-0 ) = PIX > 1.33)

= I - P ( X El . 33)

= I - ( I - e- ' ' 33)

• atWr

-

plxc 9%) -- pixel,= I - e-

'

÷ o.

632.

Exponential or Poisson Distribution?

We model the number of industrial accidents occurring in one year.

We model the length of time between two industrial accidents(assuming an accident occurring is a Poisson event).

We model the time between radioactive particles passing by a counter(assuming a particle passing by is a Poisson event).

We model the number of radioactive particles passing by a counter inone hour


⇐

Example: Radioactive Particles

The arrival of radioactive particles at a counter is Poisson events. Thenumber of particles in an interval of time follows a Poisson distribution.Suppose, on average, we have 2 particles per millisecond. What is theprobability that no particles will pass the counter in the next 3 milliseconds?


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PIN > 3) =L - PlwE3 )=L - fo3ze-2ddk÷ 0.002

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--

✓ I.IMI X - Poisson ( 2)=

3ms Tn Poisson ( 2×3=6) .

i. pH -

-o) = 6°E÷÷ 0,002 .

Example: Machine Failures

If the number of machine failures in a given interval of time follows aPoisson distribution with an average of 1 failure per 1000 hours, whatis the probability that there will be no failures during the next 2000hours?

What is the probability that the time until the next failure is more than2000 hours?


X N Poisson (1×2--2)

PlX=o )④ n Poisson l'd 100ohm" '

= e-Z÷o④=

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Memoryless Property

An even more interesting property of an exponential random variable isconcerned with conditional probabilities.

The exponential distribution is often used in reliability studies as themodel for the time until failure of a device. The memoryless propertyof the exponential distribution implies that the device randomly wearsout, i.e., P(X > t + �t|X > t) remains the same for any t.

However, the lifetime L of a device that su�ers slow mechanical wear,such as bearing wear, is better modeled by a distribution s.t.P(X > t + �t|X > t) increases with t, such as the Weibulldistribution (later).


X n exponly

✓ PIX >tto.to/X7t)--Plx7ot7plX7ttotfx7ty=plx> ot)E

Memoryless Property

Show P(X > t + �t|X > t) = P(X > �t).


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time

yEt pH>t;Y

=P ( x > t tot )

p¥

↳ FtpexetThe

calf of It l - pl X Et)

Fieri - e-"

=

=

= P(X7#

Understanding Memoryless Property

To understand this property of exponential distribution, let us assumeX models the life time of a light bulb.

The memoryless property tells you that given the fact that the lightbulb still “survives” at time t, the probability it will last longer thanadditional �t amount of time (the conditional probability) equals tothe probability that it will last longer than �t amount of time from thebeginning (the unconditional probability).

The exponential distribution is the only continuous distribution withthe “memoryless” property.


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Gamma Function

Definition: The gamma function is defined as

�(–) =⁄ Œ

0y

–≠1e

≠ydy ,

where – > 0.

Facts about the gamma function:

�(1) = 1.

�(–) = (– ≠ 1)�(– ≠ 1), for any – > 1.

�(n) = (n ≠ 1)!, n is an integer.


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Gamma Distribution

A random variable Y has a gamma distribution if its pdf is

fY (y) =

Y]

[

1�(–)—– y

–≠1e

≠y/—, y Ø 00, otherwise

where the shape parameter – > 0 and the scale parameter — > 0.

Remark: The kernel of a pdf fY (y) is the part that depends on y . Thegamma pdf consists of a kernel and a “normalizing constant.” Thisconstant is free of y , but it forces the pdf to integrate to 1 over thesupport.

⁄ Œ

0

1�(–)—–

y–≠1

e≠y/—

dy = 1 =∆⁄ Œ

0y

–≠1e

≠y/—dy = �(–)—–


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Gamma Distribution


Mean and Variance

If Y ≥ Gamma(–.—),E (Y ) = –—

var(Y ) = –—2


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og =p÷ )! yd e-Hedy

← =p¥!y"""e-Hedy-

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‰2 Distribution

For any integer ‹ Ø 1, a random variable Y has a ‰2 distribution with ‹degrees of freedom if Y is a gamma random variable with – = ‹/2 and— = 2.

E (Y ) = –— = ‹

22 = ‹

var(Y ) = –—2 = ‹

222 = 2‹


Toni,.ie?n:Y

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Example 4.21

Suppose that a random variable Y follows an exponential distribution with⁄ = 1/6.

What is the probability that Y is at most 10?

What is the probability that Y is between 5 and 10?


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to":O.

-

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-= P LY Elo ) - PHE5 )

= ¥640 ) - ( l - e-4×51=0.246

Weibull Distribution

The Weibull distribution is often used to model the time until failure ofmany di�erent physical systems.

The random variable X with probability density function

fX (x) = —

”

3x

”

4—≠1e

≠(x/”)—, for x Ø 0

is a Weibull random variable with scale parameter ” > 0 and shapeparameter — > 0. The shorthand notation is X ≥ Weibull(—, ”).


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-

when p -- I, fix , =Fft

no aging eftet .= Ig e- ÷run

Weibull DistributionBy changing the values of — and ”, the Weibull pdf can assume manyshapes. Because of this flexibility (and for other reasons), the Weibulldistribution is very popular among engineers in reliability applications.


C- expon

k

cdf of Weibull Distribution

If X has a Weibull distribution with parameters ” and —, then thecumulative distribution function of X is

FX (x) = P(X Æ x) = 1 ≠ e≠(x/”)—

, for x Ø 0

It follows thatP(X > x) = e

≠(x/”)—.


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"

I"

e-'Etat

•i. is:* ÷..- ⇒ .

Example: Bearing Wear

The time to failure (in hours) of a bearing in a mechanical shaft issatisfactorily modeled as a Weibull random variable with — = 1/2 and” = 5000 hours. Determine the probability that a bearing lasts at least 6000hours.


- itsFai -- I - e

-

-

-

P (XE 6000)

= i - e- fs÷jk

÷ o.

665

=

Example: Battery Lifetime

The lifetime of a rechargeable battery under constant usage conditions,denoted by T (measured in hours), follows a Weibull distribution withparameters — = 2 and ” = 10.

What is the probability that a battery is still functional at time t = 20?

What is the probability that a battery is still functional at time t = 20given that the battery is functional at time t = 10?


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Summary

Suppose that X ≥ Weibull(—, ”),

pdf: fX (x) = —”

! x”

"—≠1e

≠(x/”)—, for x Ø 0

dweibull(x , —, ”) in R

CDF: FX (x) = 1 ≠ e≠(x/”)—

, for x Ø 0

pweibull(x , —, ”) in R

Mean: E (X ) = ”�11 + 1

—

2

Variance: var(X ) = ”25�

11 + 2

—

2≠

1�

11 + 1

—

2226

where � is called gamma function and �(n) = (n ≠ 1)! if n is apositive integer.


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chapter 4 continuous random variable and probability...

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