math f113 -chapter-4.pdf

BITS Pilani Pilani Campus

BITS Pilani presentation

Dr RAKHEE Department of Mathematics


MATH F111 & AAOC C111 Probability and Statistics


Chapter 4

Continuous Distribution

BITS Pilani, Pilani Campus

Definition: A random variable is continuous if it can assume any real numbers and some interval (or intervals) of real numbers and the probability that it assume any specific value is 0 (zero).

Continuous Random Variables


Definition: Let X be a continuous random variable. A function f(x) is called continuous density (probability density function i.e. pdf ) iff integral converges.

CONTINUOUS DENSITY (Probability density function )

=

1)(.2

0f(x)1.

dxxf


1.P[X = a] =0. 2. P[X = b] =0. 3. P[a X b] = P[ a X < b] = P[a < X b] = P[a < X < b] 4. P[a X b] = where a and b are real numbers. Area under the curve of f between x = a to x = b. TOTAL AREA IS ONE.

Probability for Interval

;)(b

a

dxxf


Let X be the continuous r.v. with density f(x). The cumulative distribution function (cdf) for X, denoted by F(X) , is defined by F(X) = P ( X x ) , all x =

CUMULATIVE DISTRIBUTION FUNCTION

x

dttf )(


1. If X is a Continuous random variable, then F(x) is also a continuous function for all x.

2. The F(x) is nondecreasing function.

Note:

PROBABILITY by using cdf F(x) P(a X b) = F(b) F(a). FIND f(x) from F: f(x) = dF(x)/dx = F(x)

for all x.


0)()(lim

)(lim)(

===

=

dttfdttf

xFF

x

x

x

1)()(lim

)(lim)(

+

+

+

===

=+

dttfdttf

xFF

x

x

x


Consider the density of X f (x ) = x / 6 ; 2 x 4 , = 0 ; e.w. (a) Find the cumulative distribution function. (b) Use cdf F, find P[2.5X 3], P[1 X3.5].

EXERCISE 4.1, 9 PAGE 140


(a) cumulative density function

4;1)(

42);4(121

261

60)()(

2;0)(

2

2

2

2

2

=


Use cdf F, find P[2.5X 3], P[1X3.5].

[ ]22917.0

)45.2()43(121

)5.2()3(]35.2[

22

=

=

= FFxP

[ ]6875.0

0)45.3(121

)1()5.3(]5.31[

2

=

=

= FFxP


The Cumulative distribution function (c.d.f.) of a continuous random variable X is

Determine and . Hence find the probability density function (p.d.f.).

Example

( )


In parts (a) and (b) proposed cumulative distribution functions are given. In each case, find the density that would be associated with each, and decide whether it really does define a valid continuous density. If it does not, explain what property fails. (a) Consider the function F defined by F(x) = 0 ; x < -1, = x + 1 ; -1 x 0, = 1 ; x > 0.

EXERCISE 4.1.14, PAGE 141


Clearly f(x) 0 for all x. But F is cdf as

>

1 .

. continue


Clearly f(x) 0 for all x. But F is not cdf as

0 02 0 1/ 2

( )1/ 2 1/ 2 10 1

xx xdFf x

xdxx


(5) (Continuous uniform distribution) A random variable X is said to be uniformly distributed over an interval (a, c) if its density is given by

cxaac

xf >

,01


Secondly,

11 =

dxac

c

a(b) Sketch the graph of the uniform density.

X=a X=c

f(x)

x

1/(c-a)

( ii ) Shade the area in the graph of part (b) that represents P[X (a + c)/2].

X= a X= c

f(x)

x (a+c)/2

(c ) Find the probability pictured in part: ii

(e) Let (l, m) and (d, f) be subintervals of (a, c) of equal length. What is the relationship between P[l X m] and P[d X f]

Sol: Probability is same on equal length of interval.

+

=

=

+

2

5.012

ca

a

dtac

caXP

(10) Find the general expression for the cumulative uniform distribution for a random variable X over (a, c)

cxxF

cxaacax

dsac

xXPxF

axxFx

=


MEAN ,VARIANCE, S.D. & MGF (m.g.f.) 1. Mean = E [X] =

where X is the c.r.v. with density f(x).

EXPECTATION & DISTRIBUTION PARAMETERS

dxxxf )(


2. Let X be the c.r.v. with density f(x). Let H(x) be a random variable. The expected value of H(x) is defined as:

provided is finite.

( )[ ]

= dxxfxHXHE )()(

dxxfxH )(|)(|


3. Var X = E[X2] (E[X])2 = 2 , where

Variance is shape parameter in the sense that a random variable with small variance will have a compact density; one with a large variance will have a density that is rather spread out or flat.

continue

= dxxfXE )(x][ 22


4. m.g.f. = Moment Generating Function = E[ etX] = mX (t) =

continue

dxxfetx )(


Let X denote the length in minutes of a long distance telephone conversation. The density for X is given by (a)Find the moment generation function

mX(t).

Section 4.2, page 141, 17

0,101)( 10 >=

xexf

x


(b) Use mX(t) to find the average length of such call.

(c) Find variance and standard deviation for X.


Assume that the increase in demand for electric power in millions of kilowatt hours over the next 2 years in particular area is a random variable whose density is given by f(x) = (1/64 ) x3 ; 0 < x < 4 , = 0 ; e.w. . (a) Verify that this is a valid density.

Exercise 4.2.24 page 143


(b) Find the expression for the cumulative distribution function F for X, and use it to find the probability that the demand will be at most 2 million kilowatt hours.

(c) If the area only has the capacity to generate an additional 3 million kilowatt hours, what is the probability that demand will exceed supply?

(d) Find the average increase in demand.

continue


MGF of uniform distribution random variable on (a, c) :

0,e1

1][

tc

=

=

tace

t

dxac

eeE

ta

c

a

txtX


Use definition to find mean and variance can be found.

12)()(

2)(

2acXVar

caXE

=

+=


If a pair of coils were placed around a homing pigeon and magnetic field was applied that reverses the earths field, it is thought that the bird would become disoriented. Under these circumstances it is just as likely to fly in one direction as in any other.

Section 4.1, page no 139, 6


Home (0)

Pigeon

is uniformly distributed over the interval [0, 2]. a) Find the density for .


(b) Sketch the graph of the density. (c) Shade the area corresponding to the

probability that a bird will orient within /4 radians of home, and find this area using plane geometry.

(d) Find the probability that a bird will orient within /4 radians of home by integrating the density over the appropriate region(s).


(e) If 10 birds are released independently and at least 7 orient within /4 radians of home, would you suspect that perhaps the coil are not disorienting the birds to the extent expected? Explain, based on the probability of this occurring.

(f) Find mean, variance for .


Example : A random variable X with density

0a,b,x

,)bx(a

a1)x(f 22>


Cauchy distribution with density


dxx1

x1dxx1x1

02

0

2

++

+

=

Multiply and divide by 2, we get

+

+

+

=0

20

2 |1|ln21|1|ln

21 xx

which does not exist, as )ln(


A random variable X with density function = 0 , for x 0, is said to have a Gamma Distribution with parameters and ,for x > 0, > 0, > 0.

GAMMA DISTRIBUTION

/1

)(1)( xexxf

=


Definition : The function defined by is called the Gamma function. Properties : 1. (1) = 1 2. () = ( -1) ( -1 ); where is the +ve

real number. 3. (n) = (n-1)! (Factorial of n-1), n = 1,2,3. 4. (1/2)=

Gamma Function

,0,0;)( 10

>>=

zdzze z


1. (1) = 1 Proof: By definition of Gamma function, we

have

1

,1,0;)1(

0

0

0

==

=>=

dze

zdzze

z

z


2. () = ( -1) ( -1 ); where is the +ve real number.

Proof:

by integrating by parts, we have

,0,0;)( 10

>>=

zdzze z

[ ] dzzeze zz 1)1(0

01 )1()(

+=


01lim)!1(

)2)(1(lim

)1(lim

lim)(lim

3

2

11

=

=

=

=

zz

zz

zz

zz

z

z

e

ez

ez

ezze


Thus, Since,

[ ])1()1(

)1()( 1)1(0

01

=

+=

dzzeze zz

)!1()(.3 =

)!1()1()...2)(1(...)1()1()(

====


==

0

21

21 dzez z


A random variable X with density function is said to have a Gamma Distribution with parameters and ,for x > 0, > 0, > 0.

GAMMA DISTRIBUTION

( )

>>>

=

wiseother

xexxf

x

,0

0,0,0,)(

1)(

/1


To check the necessary and sufficient condition of pdf:

0xallfor0)x(f >

=

===

=

0

11

0

/1

)(1

,

)(1)(,

dtet

txanddtdxtxLet

dxexdxxfFurther

t

x


pdfaisxfHence

ezdxxf z

)(

1)(

)(0

1 =

=


Let X be a Gamma r.v. with parameters and . Then

MEAN, VARIANCE AND MGF OF GAMMA FUNCTION

1)exp()(

1

0

1 =

dxxx

1. m.g.f. = mX(t)=(1- t)- , t < 1/ 2. E[X] = Mean = X = 3. Var(X) = 2 = 2


Proof:

)1(

)1()1(

)(1

)(1

0

) 1(1

/1

tdzdxand

tzxxtzlet

dxex

dxexe

xt

xtx

=

=

=

=

=

][)(, txX eEtmdefBy =


=

=

0

1

1

0

)1()(1

)1()

1(

)(1

dzezt

tdze

tz

z

z


==

=

=

=

0

1

0

)()1(

)(][

t

tX

t

tmdtdXE


20

2

22

)1(

))((][

+=

==t

X tmdtdXE

222

22

)()1(][][)(

=+=

== XEXEXVar


and both play a role in determining the mean and the variance of the random variable.

Graphs of Gamma Densities are not symmetric and are located entirely to the right of the vertical axis.


0

0.05

0.1

0.15

0.2

0.25

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Gamma(1, 4)

Gamma(2, 3)

Gamma(20, 0.5)

f(x)

x


For > 1, the maximum value of the density occurs at the point x = ( 1).


0

0.05

0.1

0.15

0.2

0.25

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Gamma(1, 4)

Gamma(2, 3)

Gamma(20, 0.5)

f(x)

x


Exercise 4.3 (Page No: 143)

Let X be a gamma random variable with = 3 and = 4

(i) What is the expression for the density for X

0,128

1

0,4!21

)(1)(

42

423

1

>=

>=

=

xex

xexexxf

x

xx


(ii) What is the moment generating function for X

41,)41()1()( 3


In Gamma Distribution, put = 1, we get

Exponential Distribution

>>

=

elsewhere

xexf

x

,0

0,0,1)(


0

0.05

0.1

0.15

0.2

0.25

0.3

1 3 5 7 9 11 13 15 17 19

x

= 3

f(x)


For the above function Mean = = , var (X) = 2,

mX(t)=(1- t)-1 , t < 1/

Cont


The c.d.f. of exponential distribution with Parameter is given by

0 x1

111)(

0

0

0

>==

=

=

xxs

xs

x s

ee

edsexF


The c.d.f. of exponential distribution

>

==

Here, we have that the first occurrence of the event will take place after time w only if number of occurrences in the time interval [0,w] is zero. Let X be the number of occurrences of the event in this time interval [0,w].

Proof:


X is a Poisson random variable with parameter w. Thus,

ww

eweXPwWP

=

=

==>

!0)(

]0[][0

wewWPwF =>= 1][1)(


Since, in the continuous case, the derivative of the cumulative distribution function is the density

wewfwF == )()(

1 =

This is exactly density for an exponential random variable with


A particular nuclear plant releases a detectable amount of radioactive gases twice a month on average. Find the probability that at least 3 monthswill elapse before the release of first detectible emission. What is the average time one must wait to observe the first emission?

Section 4.3, page no 144,34


0,2)(, 2 >= xexfTherefore x

Let X be time elapsed before the release of the first detectable emission a then X is an exponential distribution with = 1/2

Solution :

,)1(1)3(1]3[ 66 ===> eeFXP


A computer centre maintains a telephone consulting service to trouble shoot for its users. The service is available for 9:00 to 5:00 each working day. Past experience shows number of calls received per day is a Poisson distribution with parameter 50. For a given day find the probability that first call will be received (i) before 10:00 a.m. (ii) after 3:00 p.m.

Example


A random variable X with density function is said to have a Gamma Distribution with parameters and ,for x > 0, > 0, > 0.

( )

>>>

=

wiseother

xexxf

x

,0

0,0,0,)(

1)(

/1

GAMMA RANDOM VARIABLE


If a random variable X has a gamma distribution with parameters = 2 and = /2, then X is said to have a chi-square (2) distribution with degrees of Freedom and denoted by , is a positive integer.

Chi-square distribution

2


0,2

2

1)( 2/1

2

2/>

=

xexxf x

E[ ] = , Var[ ] = 2

= 2 and = /2,

2

2


1for =

x

f(x)

( )

>

=

wiseother

xexxf

x

,0

,0,2)2/1(

1)(

2/1)2/1(

2/1

Chi square distribution


2=for

Chi square distribution

0,21)( 2/ >= xexf x

0,2)

2(

1)( 2/1

2

2/>

=

xexxf x


x

f(x)

2=forChi square distribution

0,21)( 2/ >= xexf x


x

f(x)

-2.08E-16

0.05

0.1

0 5 10 15 20 25

for = 10


We do not have explicit formula for CDF F of X2 . Instead values are tabulated on page no. 695-696 as below (F occurs in margin here, and related value of r.v. inside the table):


If F is CDF for Chi square random variable Having 5 degrees of freedom F(1.61) = 0.1, F(4.35) = 0.5

P[X2 < t]

F 0.100 0.250 0.500

5 1.61 2.67 4.35 6 2.20 3.45 5.35

7 2.83 4.25 6.35


For 0 < r < 1, we denote by , for a chi-square r.v. with degrees of freedom, a unique number such that P[X2 ] = r

Probability to the right of is r.

2r

2r

Another notation

2r

for exampleP[X210 ] = 0.05 2

05.0


x

f(x)

0

0.05

0.1

0 5 10 15 20 25

P[X210 ] = 0.05 2

05.0

Probability to the right of is r. 2r

0.05

= 1 - P[X210 ] = 0.95 2

95.0


P[X2 t] F 0.90 0.95

10 16.0 18.3 11 17.3 19.7 21.9

12 18.5 21.0 23.3

0.975

20.5


x

f(x)

-2.08E-16

0.05

0.1

0 5 10 15 20 25

P[X210 ] = 0.05 2

05.0

Probability to the right of is r. 2r

3.18205. =

0.05


38. Consider a chi squared random variable with 15 degrees of freedom. (i) What is the mean of ?215X

Sol: Chi square distribution is Gamma with = /2 & = 2, Mean = = =15 and 2 = 2 = 30


otherwise 0

0,2)2/15(

1)(

,2/,2var

2/1)2/15(2/15

2

=

>

=

==

xexxf

hencewithiablerandomis

x

Sol:

(ii) What is the expression for the density for ?215X


(iii) What is the expression for the moment Generating function for

215X

Sol: 2/1,)21()1()( 2/15


P[X2 < t] F 0.005 0.010 0.900

15 4.60 5.23 22.3 16 5.14 5.81 23.5

17 5.70 6.41 24.8

.025

6.26

10.0900.01]3.22[1]3.22[ 215

215

==


875.0025.0900.0)26.6()3.22(]3.2226.6[ 215

===

FFXP

]3.2226.6[ Find (v) 215 XP

P[X2 < t] F 0.005 0.010 0.900

15 4.60 5.23 22.3 16 5.14 5.81 23.5

17 5.70 6.41 24.8

.025

6.26

7.56

6.91


Sol. P[X215 > 2r] = r.

freedom of degree 150.25205.0 for=

P[X215 > 20.05] = 0.05.

P[X2 < t] F 0.005 0.010 0.950

15 4.60 5.23 25.0

.025

6.26

(vi) Find 20.05, & 20.01 for 15 degree of freedom

P[X215 20.05] = 0.95.


6.302 01.0 =

P[X215 > 2r] = r. P[X215 > 20.01] = 0.01.

P[X2 < t] F 0.005 0.010 0.990

15 4.60 5.23 30.6

.025

6.26

P[X215 20.01] = 0.99.


A random variable X with density f(x) is said to have normal distribution with parameters and > 0, where f(x) is given by:

4.4 The Normal Distribution

( )

.0);,(,,21)(

2

2

2 >=

xexfx


=

-

1f(x)dx (ii)

),(-x 0)()( xfi

=

==

=

-

21

-

21

2

-

21

2

22

2

21

21

let

121

dzedxe

dzdxzx

dxeproveTo

zx

x


=

0

21

-

21 22

212

21 dzedze

zz

=

=

==

+

0

)(21

0

0

21

0

21

0

21

0

21

22

22

22

dxdye

dyedxeII

Idzedze

yx

yx

zz

BITS Pilani, Pilani Campus 2

lim

lim

2/

0

2/

0

0

)(212/

0

2

2

=

=

=

Rw

R

R r

R

ddwe

rdrde

+

0

)(21

0

22

dxdyeyx


2I .. =ei

12

2

212

21

0

21

-

21 22

==

=

dzedze

zz


Mean and Standard deviation for Normal distribution

Theorem: Let X be a normal random variable with parameters and . Then is the mean of X and is its standard deviation.


The density Curves


Cumulative Distribution Function


Standard Normal Distribution

Theorem: Let X be normal with mean and standard deviation . The variable

is standard normal. Z has mean 0 and standard deviation 1.

=XZ


),(,21)( 2

2

=

zezfz

Standard Normal Distribution

dze

dzzfzZPzF

z z

z

Z

=

==

2

2

21

)()()(

The probability density function is given by

The corresponding distribution function is given by


( )

,21)()(

2

2

2 dsexXPxFx s

==

dzzzsLet =+==

ds s

( )zFdzexF Z

xz

=

=

=

-xZP 21)( 2

2


Moment Generating Function

21

21][)(

2

2

2

2

dze

dzeeeEtm

zzt

zztZt

Z

=

==


=+

+=

2)(

2222

2

222222 tzttttzzzzt

Let z - t = w dz = dw

222

222

21)(

twt

Z edweetm ==

21)( 2

)(

2

22

dzeetmtzt

Z

=



Let Z be normally distributed with parameters = 0 and = 1, then the moment generating function for Z is

2

2

)(t

Z etm =


Let X be normally distributed with parameters and , then the moment generating function for X is given by


2

2)(

)Zt(Xt

22

22

][

][e]E[e )(

+

+

=

==

==

tt

ttZtt

X

e

eeeEe

Etm


Mean and Variance for Normal r.v.

2

22222

22

02

22

0

E[X]-]E[X)(

)(]E[X and

)(][

=

+==

+==

===

=

=

XVar

tmdtd

tmdtdXEMean

tX

tX


=

aFcF ZZc)XP(a

),N( is X If(i)

)(1)(F )( Z zFzii Z=

3.5z if 1)(F -3.5z if 0)(F )(

Z

Z

zziii


Using table, find the values of (i) P[Z 1.31] (ii) P[Z < 1.31] (iii) P[Z = 1.31] (iv) P[Z > 1.31] (v) P[-1.305 Z 1.43] (vi) z .10 (vii) z .90 (viii) The point z such that P(-z Z z) = 0.9

Example


dzedzzfzZPzFz zz

Z

=== 2

2

21)()()(

z

1.4

0.00 0.01 0.02 0.03 0.04

1.3 .9032 .9049 .9066 .9082 .9099

.9192 .9207 .9222 .9236 .9251

(i) P[Z 1.31] = .9049 (ii) P[Z < 1.31] = .9049

(i) P[Z 1.31] (ii) P[Z < 1.31]


dzedzzfzZPzFz zz

Z

=== 2

2

21)()()(

z

1.4

0.00 0.01 0.02 0.03 0.04

1.3 .9032 .9049 .9066 .9082 .9099

.9192 .9207 .9222 .9236 .9251

(iv) P[Z > 1.31] = 1 - P[Z 1.31] = 1- 0.9049

(iii) P[Z = 1.31] (iv) P[Z > 1.31]

(iii) P[Z = 1.31] = 0


z

1.4

0.00 0.01 0.02 0.03 0.04

1.3 .9032 .9049 .9066 .9082 .9099

.9192 .9207 .9222 .9236 .9251

(v) P[-1.305 Z 1.43]

P[-1.305 Z 1.43] = F(1.43) - F(-1.305) = 0.9236 (1 F(1.305)) = 0.9236 1+(0.9032 + 0.9049)/2) = 0.82765


(vi) z .10 (vii) z .90

(vi) P(Z zr ) = r P(Z z.10 ) = 0.10 P(Z < z.10 ) = 0.90 z.10 = 1.28

z

1.4

0.05 0.06 0.07 0.08 0.09

1.2 .9015 .8997


(vii) z .90 = ? (viii) The point z such that P(-z Z z) = 0.9 P(-z Z z) = F(z) - F(-z) =2F(z) 1 = 0.9 i.e. F(z) = 0.95

z

1.4

0.04 0.05

1.6 0.9505 0.9495

z = 1.645


X is normally distributed and the mean of X is 12 and standard deviation is 4. (i) Find P(X 10), P(X 10), P(0 X 12) (ii) Find r, where P(X > r) = 0.24 (iii) Find a and b, where P(a < X< b) = 0.50 and P(X > b) = 0.25

Example


If X is normal random variable with P(X 35) = 0.07 & P(X 63) = 0.89 Find mean & standard deviation of X.

Example


Most galaxies take the form of a flatten disc, with the major part of the light coming from this very thin fundamental plane. The degree of flattening differs from galaxy to galaxy. In the Milky Way Galaxy most gases are concentrated near the center of the fundamental plane. Let X denote the perpendicular distance from this center to a gaseous mass. X is normally distributed with mean 0 and standard deviation 100 parsecs. (A parsec is equal to approximately 19.2 trillion miles.)

Section 4.4, page no 145, 41


(a) Sketch a graph of the density for X. Indicate on this graph the probability that a gaseous mass is located within 200 parsecs of the center of the fundamental plane. Find this probability.

(b) Approximately what percentage of the gaseous masses are located more than 250 parsecs from the center of the plane?

(c) What distance has the property that 20% of the gaseous masses are at least this far from the fundamental plane?

(d) What is the moment generating function for X?


The positive random variable Y is said to have a log-normal distribution, if logeY is normally distributed. That is,

logeY ~ N(,)=X

Log-Normal Distribution

& are not mean & standard deviations of Log-normal random variable.

PresenterPresentation Notes


Let X be normal with mean and variance 2. Let G denote the cumulative distribution for Y = eX and let F denote the cumulative distribution for X. Show that (i) G(y) = F(lny) (ii) G(y) =F(lny)/y (iii) Find density of Y

Section 4.4, Problem 45/ p 146


.1)(ln

ln)(ln)(ln)(

yyF

dyydyF

dyydF

dyydG

=

==

a) G(y) = P(Y y) = P(eX y) = P(X ln y) = F(ln y) b)


( )

.0);,(,

,21)ln()(

ln2 2

2

>

==

x

dxeyXPyGy x

( )

dwew

yG

wx

y w

=

===

0

2ln

x

2

2

121)(

,w

dwdx ,ew ,ln

c) Now X ~ N(,). Therefore,


( )

.other wise 0

0,);,(,2

1)(2

2

2ln

=

>=

yey

ygy

Hence, the density for Y is given by

( )

otherwise 0

0y , 121 2

2

2ln

=

>

=

y

eydy

dG


Let Y denote the diameter in millimeters of styrofoam pellets used in packing. Assume that Y has a log-normal distribution with parameter = 0.8, = 0.1. (i) Find the probability that a randomly

selected pellet has a diameter that exceeds 2.7 mm

(ii) Find two values of Y such that probability is approximately 0.95 between these values?

Problem 46


P(Y > 2.7) = P(exp(X) > 2.7) = P(X > ln(2.7))

(i) Find the probability that a randomly selected pellet has a diameter that exceeds 2.7 mm

0.02680.9732-1 )93.1(1)93.1(

1.08.)7.2ln(-XP

===>=

>

FZP


P[y1 < Y < y2] = 0.95

(ii)

95.01.0

8.ln1.0

8.ln

95.01.0

8.ln-X1.0

8.lnP

21

21

=


0.95 0.025 0.025

z0.025 z0.975


025.02

975.01

21

1.08.0ln

1.08.0ln

95.01.0

8.0ln1.0

8.0lnP

zy

zy

yZy

=

=

=


707.2,96.11.0

8.0ln

829.1,96.11.0

8.0ln

2025.02

1975.01

===

===

yzy

yzy


Thank You

BITS Pilani presentationMATH F111 & AAOC C111Probability and StatisticsSlide Number 3Continuous Random VariablesCONTINUOUS DENSITY (Probability density function ) Probability for IntervalCUMULATIVE DISTRIBUTION FUNCTION Slide Number 8Slide Number 9EXERCISE 4.1, 9 PAGE 140Slide Number 11Slide Number 12Slide Number 13EXERCISE 4.1.14, PAGE 141Slide Number 15. continueSlide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22EXPECTATION & DISTRIBUTION PARAMETERSSlide Number 24continuecontinueSlide Number 27Slide Number 28Exercise 4.2.24 page 143continueSlide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39GAMMA DISTRIBUTIONGamma FunctionSlide Number 42Slide Number 43Slide Number 44Slide Number 45Slide Number 46GAMMA DISTRIBUTIONSlide Number 48Slide Number 49MEAN, VARIANCE AND MGF OF GAMMA FUNCTIONSlide Number 51Slide Number 52Slide Number 53Slide Number 54Slide Number 55Slide Number 56Slide Number 57Slide Number 58Slide Number 59Slide Number 60Slide Number 61Exponential DistributionSlide Number 63ContSlide Number 65Slide Number 66Slide Number 67Slide Number 68Slide Number 69Slide Number 70Slide Number 71Slide Number 72Slide Number 73Slide Number 74Slide Number 75Slide Number 76Slide Number 77Slide Number 78Slide Number 79Slide Number 80Slide Number 81Slide Number 82Slide Number 83Slide Number 84Slide Number 85Slide Number 86Slide Number 87Slide Number 88Slide Number 89Slide Number 90Slide Number 91Slide Number 92Slide Number 934.4 The Normal DistributionSlide Number 95Slide Number 96Slide Number 97Slide Number 98Slide Number 99Slide Number 100Slide Number 101Slide Number 102Standard Normal DistributionSlide Number 104Slide Number 105Slide Number 106Slide Number 107Slide Number 108Moment Generating FunctionSlide Number 110Slide Number 111Slide Number 112Slide Number 113Slide Number 114Slide Number 115Slide Number 116Slide Number 117Slide Number 118Example Slide Number 120Slide Number 121Slide Number 122Log-Normal DistributionSlide Number 124Section 4.4, Problem 45/ p 146a) G(y) = P(Y y) = P(eX y) = P(X ln y) = F(ln y)b)Slide Number 127Hence, the density for Y is given by Problem 46(i) Find the probability that a randomly selected pellet has a diameter that exceeds 2.7 mmSlide Number 131Slide Number 132Slide Number 133Slide Number 134Slide Number 135

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