1 schaum’s outline probability and statistics chapter 4 presented by carol dahl special...
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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