1 Schaum’s Outline Probability and Statistics Chapter 4 Presented by Carol Dahl Special...

1

Schaum’s Outline

Probability and Statistics

Chapter 4

Presented by Carol Dahl

Special Probability Distributions

4-2

Chapter 4 Outline

Binomial Distribution

Normal Distribution

Poisson Distribution

Relations Between Distributions

Binomial and Normal

Binomial and Poisson

Poisson and Normal

Central Limit Theorem

4-3

Outline

Multinomial Distribution

Hypergeometric Distribution

Uniform Distribution

2 Distribution

t-Distribution

F-Distribution

Cauchy

Exponential

Lognormal

4-4Introduction

Special Probability Distributions

give probabilities for random variables

discrete and continuous

help us make inferences

4-5Distributions Help Make Inferences

Powerful tools - uncertainty

prediction

confidence intervals Q = ßo + ß1P + ß2Y hypothesis tests

World Metal Production 1999 Million metric tons

0 5

10 15 20 25 30 35

Aluminum Lead Copper

4-6Binomial Distribution

You own ten draglines for mining coal

4-7 Binomial (Bernoulli) Distribution

Probability associated with no repairs

P(no repairs) = p

P(repairs) = (1-p) = q

If breakdown between machines independent

Bernoulli Trial

n trials = 10

x number with no repairs out of n draglines

Binomial distribution

P(X=x)=n pxqn-x=n!/(x!(n-x)!)pxqn-x

x

4-8Binomial Example

Dragline P(repairs) = 0.2

Binomial P(X=x) = ( n ) px(1-p)n-x

( x

)

P(X=2) = (10) 0.22 (1-0.2)10-2 = 10! 0.22*0.88

2 2!(10-2)!

= 0.302

Excel insert, function, statistical, binomdist

=binomdist(x,n,p,cumulative)

(true or false)

= binomdist(2,10,0.2,false)

4-9 Probabilities of Dragline Repairs

Number of Probability of repairs Repair

0 0.1071 0.2682 0.3023 0.2014 0.0885 0.0266 0.0067 0.0018 0.0009 0.000

10 0.000

4-10Probabilities of Dragline Repairs

Probability of Repair

0.000.050.100.150.200.250.300.35

0 1 2 3 4 5 6 7 8 9 10

4-11Properties of Binomial Distribution

Discrete

Suppose n = 10, P = 0.2

Mean =np = 10*0.2 = 8

Variance 2=npq = 10*0.2*0.8 = 1.6

Standard deviation = (1.6 )0.5 = 1.265

Coefficient of skewness

α3=(q-p)/ = (0.8 – 0.2)/ 1.265 = 0.474

Coefficient of kurtosis

α4=3+(1-6pq)/npq = 3 + (1-6*0.8*0.2)/1.6 = 3.025

4-12Functions Relating to Binomial

Moment generating function M(t)=(q+pet)n

M(t) = E(etX) = xetXP(X)

E(X) = xXP(X) = M'(0)

M'(t)=n(q+pet)n-1pet

M'(0)=n(q+pe0)n-1pe0 = n((1-p)+p)n-1p =np

E(X2) = xX2P(X) = M''(0)

E(X3) = xX3P(X) = M'''(0)

Replacing t by iω with i imaginary number

(-1)0.5 then we get another useful function

Characteristic function φ(ω)=(q+peiω)n

4-13Law of Large Numbers

for Bernouilly Trials

Estimate p by sampling p = x/n

By increasing number of trials

can get as close as we want to true mean

lim P (|X/n – p|>ε) = 0 n->

4-14Standard Normal (0,1)

Most important continuous distribution (Gaussian)

f(Z)= 1 exp(-Z2/2) dZ (22)0.5

4-15Standard Normal (0,1) Example

Building a hydro – Three Gorges –18,000 MW

4-16

Standard Normal (0,1)

Want to know how much rainfall deviated from normal

Z

Z ~ N(0,1) with Z measured in inches

Five things you might want to know

P(Z < a) P(Z > b)

P(Z < -c) P(Z > -d)

P(e < Z < f)

4-17Standard Normal (0,1)

f(Z) = 1 exp(-Z2/2) dZ (22)0.5 P(Z < a) P(Z > b)P(Z < -c) P(Z > -d)P(e < Z < f)

4-18Standard Normal (0,1)

P(Z < a) = 1 aexp(-Z2/2) dZ (2)0.5 -

P(Z > b) = 1 exp(-Z2/2) dZ (2)0.5 b

P(Z < -c) = 1 -cexp(-Z2/2) dZ (2)0.5 -

P(Z > -d) = 1 exp(-Z2/2) dZ (2)0.5 -d

P(e < Z < f) = 1 f exp(-Z2/2) dZ (2)0.5 e

P(e < Z < f) = P(Z <f) – P(Z < e)

4-19Standard Normal (0,1)

But difficult to integrateuse Tables, Excel, other computer packages

Normal Tables – Schaums GHJP(0<Z<z)= 1 z exp(-Z2/2) dZ (2)0.5 0

4-20Standard Normal (0,1) - Table

Table: P(0<Z<z) P(0<Z< 2) = 0.477

P(Z<2) = 0.5 + 0.477

Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-21Standard Normal (0,1) - Table

Table: P(0<Z<z)

(Z>1.52)=1–P(Z< 1.52)1 – (0.5 + 0.436) = 0.564

Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-22Standard Normal (0,1) - Table

Table: P(0<Z<z)(Z>-1.58)=P(Z<1.58)

= 0.5 + 0.443

Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-23Standard Normal (0,1) - Table

Table: P(0<Z<z)

(Z<-2.56)=P(Z>2.56)=

=1- P(Z<2.56) = (1–(0.5+0.495)=0.005

Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-24Standard Normal (0,1) - Table

Table: P(0<Z<z)=> = (-0.54<Z<2.02)

=P(Z<2.02)-(Z<-0.54) = ?

Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-25Standard Normal (0,1) - Table

ExcelP(- <Z<z) = normsdist(z,true)P(Z<2.0) = normsdist(2) = 0.977P(Z < -2) = normsdist(-2) = 0.023

4-26Standard Normal (0,1) - Table

ExcelP(- <Z<z) = normsdist(z,true)P(Z>2.0) = 1- normsdist(2) = 0.023P(Z > -2) = 1- normsdist(-2) = 0.97

4-27

Inverse Normal

Normal: P(Z<1) =

Deviation in rainfall < 1 inch above normal what % of the time?

P(Z<-1.5) =

Deviation in rainfall < 1.5 inch below normal

what % of the time?

Inverse Normal

P(Z<z) = 0.05

Rain fall deviates < what amount 5% of the time

4-28Standard Normal (0,1) – Inverse

Table: P(0<Z<z)=>

P(Z<a) = 0.698 = 0.5 + 0.198

a = 0.52 Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-29Standard Normal (0,1) - Inverse

Table: P(0<Z<z)=>P(Z>a) = 0.476P(Z>a) = 1 – P(Z<a)

P(Z<a) = 0.524 = 0.5 + 0.024

a = 0.06Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-30Standard Normal (0,1) - Inverse

Table: P(0<Z<z)=> P(Z<a) = 0.288= P(Z>-a)

P(Z<-a) = 1 – P(Z>-a) = 0.712 = 0.5 + 0.212

-a = 0.56 -> a=?

Z 0.00 0.02 0.04 0.06 0.080.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-31Standard Normal (0,1) - Inverse

Table: P(0<Z<z)=>

P(Z>a) = 0.846P(Z>a) =P(Z<-a) P(Z<-a) = 0.5 + 0.346- a = 1.02 a = -1.02 Z 0.00 0.02 0.04 0.06 0.08

0.0 0.000 0.008 0.016 0.024 0.0320.5 0.191 0.198 0.205 0.212 0.2191.0 0.341 0.346 0.351 0.355 0.3601.5 0.433 0.436 0.438 0.441 0.4432.0 0.477 0.478 0.479 0.480 0.4812.5 0.494 0.494 0.494 0.495 0.4953.0 0.499 0.499 0.499 0.499 0.499

4-32Normal Example Inverse Excel

P (Z<a) = 0.698 =normsinv(0.698) = 0.519

P (Z<a) = 0.288 =normsinv(0.288) = -0.559

P (Z>a) = 0.846 =-normsinv(0.846) = -1.019

P (Z>a) = 0.210 =normsinv(1-0.210) = 0.806

P(-a<Z<a)= 0.5 =normsinv(0.75) = 0.67449

4-33Properties of Normal Distribution

Mean Variance 2

Standard deviation

Coefficient of skewness 3=0

Coefficient of kurtosis 4=3

Moment generating function M(t)=e t+(^2*t^2/2)

Characteristic function ()=ei -( ^2 ^2/2)

4-34

Relation between DistributionsBinomial => Normal n gets large

0

0 . 1

0 . 2

0 . 3

0 . 4

0 . 5

0 1 2 3 4 5

B i n o m i a l ( 5 , 0 . 2 )

00 . 0 5

0 . 10 . 1 5

0 . 20 . 2 5

0 . 30 . 3 5

0 1 2 3 4 5 6 7 8 9 1 0

B i n o m i a l ( 1 0 , 0 . 2 )

B i n o m i a l ( 5 0 , 0 . 2 )

0

0 . 0 5

0 . 1

0 . 1 5

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48

4-35Poisson Distribution

Discrete infinite distribution with pdf

f(x)=P(X=x)=xe- /x! x=0,1,2,3,...

= mean

decay radioactive particles

demands for services

demands for repairs

4-36Poisson Distribution

Example: X number of well workovers/month

X ~ poisson mean = 5

P(X = 3) = 53e-5 = 0.140 5!

4-37 Poisson Distribution in Excel

Excel

= poisson(x,(mean),cumulative) true or false

P(X = 3)

= poisson(3,5,false) = 0.14

P(X< 3)

= poisson(3,5,true) = 0.265

p(X > 8)

= 1 - poisson(7,5,true) = 0.13

4-38Properties of Poisson

Mean =

Variance 2 =

Standard deviation = 1/2

Coefficient of skewness 3= -1/2

Coefficient of kurtosis 4= 3 + 1/

Moment generating function M(t)=e (e^t-1)

Characteristic function ()=e (e^(i)-1)

4-39

Relations between Distributions

Poisson and Binomial close

when n large & p small

Poisson and Normal are close when n gets large

To standardize Poisson

Z = (X - )/ 0.5

4-40

random variables

X1, X2, …  independent identically distributed

finite mean and variance 2.

then X = (X1 + X2 +…+Xn)/n

goes to a N(2/n) as n ->

Central Limit Theorem

4-41Multinomial Distribution

Example: You work for a gas company

likelihood a family will buy

gas furnace is 1/2 (p1)

electric furnace is 1/3 (p2)

fuel oil furnace is 1/6 (p3)

10 furnaces (n) replaced in next heating season probability that

5 (x1) gas?

4 (x2) electric

1 (x3) fuel oil?

4-42

Multinomial Distribution

Generalization of binomial

A1, A2, A3,…Ak are events

occur with probabilities p1, p2,…pk

If X1, X2, …Xk are random variables

number of times that A1, A2,…Ak occur n trials

X1 + X2+…+Xk = n then

P(X1 = n1, X2 = n2,…, nk) = n! p1n1 p2

n2…pknk

n1! n2!…nk!

Where n1 + n2+…,+ nk = n

4-43

Work out probability for

5 (x1) gas?

4 (x2) electric

1 (x3) fuel oil?

P(x1=5, x2=4, x3=1) =

Multinomial Distribution

081.0)61

()31

()21

(!1!4!5

!10 145

4-44

Example:

Box of drilling bits contains

5 “X” bit

4 “+” bit

3 “button” bit

6 bits selected at random from box

no replacement

Find probability: 3 are “X” bits,

2 are “+” bits and

1 is “button” bits

Hypergeometric Distribution

4-45

Probability

P(choose x1 from n1, x2 from n2, etc

Hypergeometric Distribution

k21

k21

k

k

2

2

1

1

x...xx

n...nnx

n...

x

n

x

n

195.0

!6!6!12

!1!2!2!2!2!3!3!4!5

123

3451

3

2

4

3

5

4-46

Example:

Box contains 6 assays of copper

4 assays of gold

choose an essay at random

no replacement

5 trials

X = number of copper essays chosen

P(X=3)

Use hypergeometric

Hypergeometric Distribution

Excel

4-47Hypergeometric

Distribution Excel

We are choosing from

two categories - copper (s) and gold (not s)

X - number of copper assays chosen

number in s = ns = 6, number in not s = nns= 4

total population = ns+nns = 6 + 4 = 10

sample size = n = 5

=hypgeomdist(x,n,ns,ns+nns) = P(X<x).

P(X=3) = hypgeomdist(3,5,6,10) -

hypgeomdist(2,5,6,10) = 0.476

4-48

Random variable X

uniformly distributed in a<=x<=b

if density functions:

1/(b-a) a<=x<=b

f(x) = 0 otherwise

Uniform Distribution

4-49

Failure rate on bits X ~ f(x) = 1/2 2 < X < 4 years

P (X > 3) = 34(1/2)dX = 0.5X| 3

4

= 0.5*4 – 0.5*3 = 0.50

half of bits last more than 3 years

P (2<X< 2.5) = 22.5(1/2)dX = 0.5X| 2

2.5

= 0.5*2.5 – 0.5*2 = 0.25

P(2.7<X<3.3) = ?

Uniform DistributionExample

4-50Distributions for Econometric Inference

Y = ßo + ß1

X1 + ß2X2 +

estimate ßs

Y^ = bo + b1X1 + b2X2

assumptions about distribution of

mean and variance

gives us distributions of Y^, bo, b1, b1

mean and variance

4-51Distributions derived from Normal

2

21

N(0,1)2

4-52Distributions derived from Normal

2

2n

N(0,1)2

+ . . . .

2

+

n

1i

4-532 Distribution – Tests on Variance

^

Y = (N-1)s2/2 ~ 2(N-1) 0 < Y <

Want to knowP(Y < b)P(Y> a)P(a<Y<b)

4-542 Distribution Probability from Table

P(Y2 > 0.103) = 0.95P(Y2 > 5.991) = 0.05P(Y4 < 9.488) = 1 - P(Y4 > 9.488) =

1 – 0.05 = 0.95 P(0.297<Y4<9.448) = ?P(C>c) 0.99 0.95 0.05 0.01

Df c= c= c= c=1 0.000 0.004 3.841 6.6352 0.020 0.103 5.991 9.2103 0.115 0.352 7.815 11.3454 0.297 0.711 9.488 13.2775 0.554 1.145 11.070 15.086

4-552 Distribution Inverse Probability from Table

P(Y3 > c) = 0.95

c = 0.352

P(Y4 < c) = 0.99

1 - P(Y4 > c) = 1 – 0.99

= 0.01 => c = 13.277P(C>c) 0.99 0.95 0.05 0.01Df c= c= c= c=

1 0.000 0.004 3.841 6.6352 0.020 0.103 5.991 9.2103 0.115 0.352 7.815 11.3454 0.297 0.711 9.488 13.2775 0.554 1.145 11.070 15.086

4-562 Distribution Probability from Excel

P(Y5 > c) = chidist(c,5) =

P(Y5 > 2) = chidist(2,5) = 0.849

P(Y6 < c) = 1 – P(Y6 > c) = 1- chidist(c,6)

P(Y6<4) = 1 - P(Y6 > 4)

= 1- chidist(4,6) = 0.323

4-57

2 Distribution Inverse Probability from Excel

P(Y3 > c) = => c = chiinv(,3)

P(Y3 > c) = 0.05 => c = chiinv(0.05,3) = 7.815

P(Y6 < c) =

P(Y6 > c) = 1 - => c = chiinv(1- ,6)

P(Y6<c) = 0.1 => c = chiinv(0.90,6) = 2.204

4-58Distributions for Econometric Inference

Y = ßo + ßX1 + ß2X2 + Y^ = bo + b1X1 + b2X2

assumptions about distribution of

= mean E() = 0, variance 2

gives us distributions of Y^, bo, b1, b2

Each has mean

E(Y^), E(bo), E(b1), E(b2)

Each has variance

2, 2bo, 2

b1, 2b2

2 tests on variance

part of other distributions

4-59

t-Distribution

= N(0.1)

df

2/df

=tdf

4-60

t-Distribution k degrees of freedom

4-61

Properties:

   When df >30 approximates Standard Normal

   Symmetrical with mean 0 and variance

df > 2

2dfdf

t-Distribution Properties

4-62

Used in Econometrics to:

Make inferences on means

Similar to Normal but tables set up differently

df bigger the larger the sample

if n large use normal tables

t-Distribution Uses

4-63

t-Distribution Probabilities Tables

P(t>)df 0.100 0.050 0.025

1 3.078 6.314 12.7062 1.886 2.920 4.3033 1.638 2.353 3.1824 1.533 2.132 2.776

Inf 1.282 1.645 1.960

P(t2 > 1.886) = 0.10

P(t4 < 2.132) =

1 - P(t4 > 2.132) = 1 – 0.05 = 0.95

P(t2>-1.886) =

P(t2<1.886) =

1- (t2>1.886)=

1 – 0.10 = 0.90

P(t4 < -2.132) =

P(t4 > 2.132) = 0.05 P(1.638<t3<3.182)=?

4-64t-Distribution Inverse from Tables

P(t>)df 0.100 0.050 0.025

1 3.078 6.314 12.7062 1.886 2.920 4.3033 1.638 2.353 3.1824 1.533 2.132 2.776

Inf 1.282 1.645 1.960

P(t3 > c) = 0.025 c=3.182P(t4< c) = 0.90

1 - P(t4 > c) = 0.90 P(t4 > c) = 0.10

c = 1.533P(t3 > -c) = 0.975 P(t3 < c) = 0.975 = 1 - P(t3 > c) = 0.975 P(t3 > c) = 0.025 c = 3.182P(t1< -c) = 0.05

= P(t1> c) c = 6.314P(c1<t3<c2) = 0.90 ?

4-65

t-Distribution from Excel P(tdf> c) =

= tdist(c,df,1)

= tdist(c,df,2)/2

P(t10> 2.10)

= tdist(2.10,10,1)

= 0.031

P(t20< 1.86)

= 1 - P(t20>1.86)

= 1 - dist(1.86,20,1)

= 0.961

4-66

t-Distribution from Excel

P(t15>-1.56) =

P(t15<1.56) =

1- (t15>1.56) =

1 – tdist(1.56,15,1) =

0.93

P(t12 < -2.132) =

P(t12 > 2.132) =

tdist(2.132,12,1)=

0.027

P(c1<t3<c2)= 0.99 ?

4-67

t-Distribution Inverse from Excel P(tdf > c) =

c=tinv(*2,df1)

P(t10 > c) = 0.15

c = tinv(0.15*2,10)

= 1.093

P(t25< c) = 0.90

1 - P(t25> c) = 0.90

P(t25>c) = 0.10

c= tinv(0.10*2,25)

c = 1.316

4-68

t-Distribution Inverse from Excel P(t3 > -c) = 0.975

P(t3 < c) = 0.975

= 1 - P(t3 > c)

P(t3 > c) = 0.025

c = tinv(0.025*2,3)

c = 3.182

P(t1< -c) = 0.05

= P(t1> c)

c =tinv(0.05*2,1)=6.314

P(c1<t17<c2) = 0.90 ?

4-69

F-Distribution

df1

df2

2df

1df

2

2

4-70

F-Distribution

4-71

Used in Econometrics to: 

   Test between

two variances

several means

subset ’s not simultaneously = 0

Comes from 2 so 0<F

F-Distribution

4-72

2,1 kk tF

If k2 (denominator df) is large

F-Distribution Relation to Other Distributions

2

2k

1

2k

k,k

k

k~F

2

1

21

1

2k

k,k k~F 1

21

4-73

Properties:k1 = numerator df; k2 = denominator df

Skewed to right approaches N(0,1) as k1 and k2 get larger

Mean = k2 > 2

Variance = k2 > 4

22

2

kk

)4()2(

)2(2

22

21

2122

kkk

kkk

F-Distribution

4-74F-Distribution Probabilities from Tables

P(F1,2>18.513) = 0.05

P(F3,6<4.757) = 1 - P(F3,6>4.757)

= 1 – 0.05 = 0.95

DFDen. 1 3 5 7

2 18.513 19.164 19.296 19.3533 10.128 9.277 9.013 8.8874 7.709 6.591 6.256 6.0945 6.608 5.409 5.050 4.8766 5.987 4.757 4.387 4.207

Df NumeratorP(Fdfn,dfd>c)= 0.05

4-75F-Distribution Inverse from Tables

P(F1,2>c) = 0.05=> c = 18.513

P(F3,6<c) = 0.95 = 1 - P(F3,6>c)

P(F3,6>c) = 0.05 => c = 4.757

DFDen. 1 3 5 7

2 18.513 19.164 19.296 19.3533 10.128 9.277 9.013 8.8874 7.709 6.591 6.256 6.0945 6.608 5.409 5.050 4.8766 5.987 4.757 4.387 4.207

Df NumeratorP(Fdfn,dfd>c)= 0.05

4-76F-Distribution Probabilities from Excel

P(F2,1>26) = fdist(f,df1,dfd2)

= fdist(26,2,1) = 0.137

P(F6,3<5.8) = 1 - P(F6,3>5.8) = 1-fdist(5.8,6,3)

= 1 - 0.0392 = 0.9608

4-77F-Distribution Inverses from Excel

P(F6,8>c) = 0.07

c = finv(,df1,df2) = finv(0.07,6,8) = 3.12

P(F9,1<c) = 0.96 = 1 - P(F9,1>c)

P(F9,1>c) = 0.04

c = finv(0.04,9,1) = 376.06

4-78

Cauchy Distribution af(x) = π (x2 + a2) a>0, - <x<

symmetric around 0

no moment generating but characteristic function

like normal but fatter tails

no variance

can have bimodal

t = 1 degree of freedom is a Cauchy

4-79Exponential

f(x) = e-(x/) x>0

excel =expondist(x,,true)

  applications

queuing theory

continuous

related to Poisson (same lambda)

Poisson = number of repairs

exponential time between repairs

life of light bulbs

4-80Exponential

graph the exponential function for = 1, 3, 10

4-81Exponential Example

Example: When drilling oil well

breakage or lost tool down hole

fishing for it

expensive

no luck fishing

deviate well

Suppose fish time t ~ exponential =1/3.

longer than 7 hours better to deviate

percent of time better to deviate?

  =1-expondist(7,1/3,true)=0.096972

4-82Integrate Exponential Functions

Make review mineral examples to show how to integrate

ex, e-x, eax

4-83Integrate Exponential Functions Rules

add exampleGeneral integration rule:

∫ ekx dx = ekx / k for example ∫ ex dx = ex

Integration by substitution rule:

∫ x ex2 dx

Substitute u = x2 du = 2x dx or dx = du/2x

Then

∫ x eu (du/2x) = ½ ∫ eu du = ½ eu

4-84Integrate Exponential Functions Rules

Integrate by parts ∫ x ex dx

Let f (x) = x and g\ = ex, then f\(x)= 1 and g(x) = ex

∫ x ex dx = f (x) . g (x) - ∫ g (x) . f\ (x) dx

= x ex - ∫ ex dx = x ex – ex

To check take the derivative for this expression (x ex - ex) it will give (x ex )

4-85Integrate Exponential Functions

0

)()( dxexdxxfxxE x

0 0)( duvdxex x

integrate by parts:

v = x u = e-λx

dv = dx du = -λe-λxdx

4-85

4-86

00

dxexedvuuv xx

01

0

1 xxxx exedxexe

)(00

lim 1011 xEeebeb

bb

4-86

Integrate Exponential Functions

4-87

If = 1 find P(x < 3)

30

3

0

3

0

)1()1( xxx edxedxe

950213.01049787.003 ee

4-87Integrate Exponential Functions

4-88Log Normal

X is lognormally distributed if

Y = Ln(X) is normally distributed

Uses - failure rates - mineral deposits

4-89Log Normal

m = mean s = sigma = standard deviation

Source:http://mathworld.wolfram.com/LogNormalDistribution.html

4-90Log Normal

4-91Log Normal in Excel

lognormal distribution is =lognormdist(x, mean, std. dev.)

=lognormdist(1.5,0.5,0.5)=0.425019

4-92

Chapter 4 Sum Up

Special Distributions Powerful tools

Binomial Distribution

Normal Distribution

Poisson Distribution

Relations Between Distributions

Binomial and Normal

Binomial and Poisson

Poisson and Normal

4-93

Chapter 4 Sum Up

Central Limit Theorem

Multinomial Distribution

Hypergeometric Distribution

Uniform Distribution

Distributions from Normal

2 Distribution

t-Distribution

F-Distribution

Y = ßo + ß1X1 + ß2X2 +

4-94

Chapter 4 Sum Up

Cauchy

Exponential

Lognormal

4-95

End of Chapter 4