chptr2 prob and stats

Maritime Safety and Environmental Engineering Probability and Statistical Theory _________________________________________________________________________________________

_________________________________________________________________________________________ SESS6002 1 2006/2007

Chapter 2 PROBABILITY AND STATISTICAL THEORY Dr. James Blake

2.1 AIM OF CHAPTER The hard bit ! Revision of probability and elementary set theory, Venn diagrams etc. More advanced treatment of probabilistic variables - probability density functions and the like. Common distributions, Normal(Gaussian), Log-normal, and Rayleigh. Extreme value distributions... Some statistics and the use of confidence limits.


_________________________________________________________________________________________ SESS6002 2 2006/2007

TABLE OF CONTENTS

TABLE OF CONTENTS ................................................................................................................ 2

2.1 INTRODUCTION ..................................................................................................................... 3 2.2 Probability Concepts .................................................................................................................. 3 2.3 Probabilistic Formulation .......................................................................................................... 5 2.3.1 The Basics ............................................................................................................................... 5 2.4 Review of Set Theory ................................................................................................................ 5 2.4.1 Some Theorems: ..................................................................................................................... 7 2.4.2 Operational Rules ................................................................................................................... 8 2.4.3 Minor Digression: ................................................................................................................... 9 2.5 Conditional Probability: .......................................................................................................... 11 2.6 Bayes' Theorem ....................................................................................................................... 13 2.6.1 Another digression ................................................................................................................ 14 2.7 Permutations and Combinations .............................................................................................. 15 2.8 Analytical Models .................................................................................................................... 16 2.8.1Discrete Random Variables and discrete distributions. ......................................................... 21 2.8.4 Some More Statistics ............................................................................................................ 22 2.8.4 Statistical Hypothesis testing: ............................................................................................... 23 2.9 Conclusion ............................................................................................................................... 23 Bibliography .................................................................................................................................. 23 Example Sheet 1 Probability ......................................................................... 24


_________________________________________________________________________________________ SESS6002 3 2006/2007

2.1 INTRODUCTION This lecture is intended as a revision of basic probability, set and statistical theory for use in quantitative risk analysis techniques. As such it is no intended to cover all the details of probability theory or statistics. For this due reference should be made to relevant engineering mathematics textbooks. However, what is intended is a review of the basics which are of relevance to risk analysis and other quantitative techniques of use in marine safety and environmental engineering analyses.

2.2 Probability Concepts Uncertainties can be roughly categorised into four groups, namely physical, statistical, modelling, and human actions. Probability concepts only deal with the first three groups. Because of these uncertainties, the realistic description of an engineering system should require a probabilistic representation. Example 1: Ultimate bending strength Mult of a mild steel beam. For a particular beam Mult is a single but unknown value. What is known is only a likely range of values within which Mult is likely to be. The uncertainty (or variability) results from physical uncertainties of the material yield stress σy and its plastic modulus Zp.

Z = M pyult σ


_________________________________________________________________________________________ SESS6002 4 2006/2007

Figure 1 Variability of Ultimate Bending Strength Example 2: Fatigue life of a structural component under a given stress range S. Here the fatigue life is represented by the number of stress cycles which the structural component can endure before fracture Figure 2 Fatigue life of structural component


_________________________________________________________________________________________ SESS6002 5 2006/2007

Uncertainties arise from random occurrences of extreme events, such as storms, earthquakes, floods etc.

2.3 Probabilistic Formulation Any probabilistic formulation requires: Definition and identification of events; and Calculation of probabilities of pertinent events. Definition of events requires the use of set theory. A sample space S consists of all possible outcomes also known as sample points. A sample space can either be discrete (countable) or continuous. One example for a discrete sample space is the defects in a fabricated structural member. The ultimate bending strength of a steel beam should be modelled by a continuous sample space... 2.3.1 The Basics Probability: Classical Definition (Ref. 1) If a single trial of a chance situation can have one of N different outcomes which are exhaustive, mutually exclusive, and equally likely, and if f of these N possibilities are favourable to a specified event A, then the probability of the event A, written P(A), is defined as f/N The law of averages, Venn argued that the terms probability and chance presuppose a series of trials, some of which exhibit an attribute of interest. The probability of interest is then defined as the numerical fraction which represents the proportion between the two events in the long run.

In 1933 a Russian mathematician, A.N. Kolmogorov published an axiomatic structure of probability theory. The advantage of such a structure is that it allows the development of a general logical structure which can be applied to any specific situation which satisfies its assumptions.

2.4 Review of Set Theory A set is a precisely specified collection of objects, called elements of the set

limit number of heads tossedP(H) =

n n⎛ ⎞⎜ ⎟→∞ ⎝ ⎠


_________________________________________________________________________________________ SESS6002 6 2006/2007

A set can be specified either by listing elements or by describing them If all the elements of a set B are members of a set A then B is called a subset of A That is:

In any discussion involving sets, there is some all-inclusive collection of objects to which the discussion is limited; this collection is called the universe and it is conventionally designated Ω For any subset A, A ⊂ Ω, the complement of A, written A , (or not A) is the set of all elements of the universe that are not elements of the set A. The complement of Ω clearly contains no elements and is called the null or empty set, φ Relationships between sets are more clearly illustrated by the use of Venn diagrams. The universe is designated by a rectangle and the sets under consideration by circles drawn within the rectangle. Figure 3 Venn Diagrams The set obtained by combining two sets is referred to as the union of two sets. eg. a set C containing all the elements included in sets A and B is defined as:

The set D which contains elements which are members of both A and B is defined as the intersection of A and B

A B⊂

C = A B∪


_________________________________________________________________________________________ SESS6002 7 2006/2007

Two sets which have no elements in common are called disjoint and would be written as

How does this help with understanding probability ? Set Vocabulary Probability Vocabulary Universe Sample Space Subset Event Disjoint Mutually Exclusive Empty Set Impossible Event Probability: Axiomatic Definition For any event A in a sample space Ω, the probability of A denoted P(A) is a real number such that Axiom 1 P(A) ≥ 0 for every event A in Ω Axiom 2 P(Ω) = 1 Axiom 3 If A1 , A2 , A3 , ... are disjoint events then P(A1 ∪ A2 ∪ A3 ∪...) = P(A1 ) + P(A2 ) +P(A3 ) + ... 2.4.1 Some Theorems: The probability of an impossible event is zero The probability of the complement of any event A is given by P(A) = 1 - P(A). The probability of any event is a number between 0 and 1 for every event A

B A = D ∩

φ = B A∩

1 P(A) 0 ≤≤


_________________________________________________________________________________________ SESS6002 8 2006/2007

If events A and B are such that B ⊂ A , that is B is a subset of A, then P(B) ≤ P(A) For any two events A and B

2.4.2 Operational Rules The rules for intersection ∩ and union ∪ are similar to that for addition and multiplication

respectively. Example 3: Failure of a mild steel strut Consider the failure of a mild steel strut subjected to an axial compression. If B = buckling and Y = plastic yield, then F = B ∪ Y denotes failure and S = F = B ∩ Y no failure. Example 4: Failure of a truss frame

P ( A B ) = P ( A ) + P ( B ) - P ( A B )∪ ∩

and + x

A B = B A A B = B A

(A B) C = A (B C ) (A B) C = A (B C)

(A B) C = (A C) (B C) (A B) C = (A C) (B C)

A B = A B

∪≡ ∩≡

∪ ∪ ∩ ∩

∪ ∪ ∪ ∪ ∩ ∩ ∩ ∩

∪ ∩ ∩ ∪ ∩ ∩ ∪ ∪ ∩ ∪

∪ ∩ A B = A B A A = ∩ ∪ ∪ Ω


_________________________________________________________________________________________ SESS6002 9 2006/2007

Figure 4 Failure of a truss frame If events E1, E2 and E3 represent component failures of members 1, 2 and 3 respectively, then

The following exceptions should be remembered:

2.4.3 Minor Digression: Subjective probability: How can the probability of say a liquefied natural gas catastrophe be a long-term relative frequency? as the situation is unique and technology will improve. To understand consider two opponents betting on the occurrence of some event A. If A occurs you receive a payoff whereas if it doesn't your opponent receives a payoff. An elementary gambling situation (EGS) is an agreement in which you pay u1 units if the event A does not occur and you receive u2 units if A does occur. The amounts u1 and u2 are called the stakes in the gamble; the ratio u2 : u1 is called the odds; and the fraction that your stake bears to the total is called the betting quotient q

Obviously want u2 as large as possible in comparison to u1. but you have to take account of both your opponent and the likelihood of A occurring. ie. you want q as close to zero as possible while opponent wants it as close to 1. Attitude to bet ranges from displeasure to pleasure as q moves from 1 to 0. When both sides have same attitude, that is indifference, for such a value of q the bet is called a fair bet . In terms

failure of truss

no failure of truss

1 2 3

1 2 3 1 2 3

= E E E

= = E E E E E E

∪ ∪

∪ ∪ ∩ ∩

2A A = A 2 A A A = A A∪ ≠ ∩ ≠

u + u

u = q21

1


_________________________________________________________________________________________ SESS6002 10 2006/2007

of stakes over N trials would expect to break even. That is (Win N[P(A)] times and lose N[1-P(A)] times ):

or after rearranging

another definition: Your probability P(A) of the event A is the fraction q of total stakes you are willing to wager on the occurrence of A in what you regard as a fair bet in an EGS. and another in terms of betting odds: If you consider u2 : u1 fair betting odds for betting on the occurrence of A in an EGS, then your probability P(A) of the event P(A) is q. which can be arranged as:

or fair odds are the ratio of failure to success probability. u2 : u1 is odds in favour, u1 : u2 is odds against All arguments involving assets are based on tacit assumption that the utility (value) of, say, 100 monetary units is 100 times the utility of 1 unit. In practice this is not always so and utility depends heavily on the individual and circumstance. Cases illustrated in Figure 5 (from Ref. 1)

u ]. ) A ( P - 1 [ N = u . ] ) A ( P [ N 12

q = u + u

u = ) A ( P21

1

( )

P(A) P(A) -1 =

uu

1

2


_________________________________________________________________________________________ SESS6002 11 2006/2007

Figure 5 Modes of utility dependent on character of individual Case (a) represents fair or linear utility; here the utility of any monetary unit is directly proportional to the amount. Case (b) represents a reckless gambler who otends to overestimate the positive utilities and underestimate the negative ones. Case (c) represents the converse of a cautious gambler. For a poor person (case d) the values of both positive and negative utilities are overestimated, while a rich person (case e) can afford to underestimate both of them. It has been determined experimentally that there is a common attitude towards utilities represented by case (f): each one of us is a gambler to a certain point, after which wetend to be overly cautious. The threshold depending on the individual. The winning attitude, winning is not everything it is the only thing. The mere fact of winning, no matter how small the amount, yields a certain fixed positive utility, and the mere fact of losing even a negligible amount results in a definite negative utility. Finally case h of a desperate attitude when a certain fixed amount of money is urgently required: no lesser amount will do, while a larger amount is inconsequential.

2.5 Conditional Probability: For any two events A and B, the conditional probability of A given that B has occurred is written P(A|B) and is defined as:

provided that P(B)>0 Example (Ref1.) Probability of clear day in January is 0.45, and the probability of warm, clear day is 0.1. Observe day is clear what is probability it will be warm as well ? W is event "the day is warm" and C as "the day is clear", hence P(W ∩ C) = 0.1 and P(C)=0.45. P(W|C) = 0.1/0.45 = 0.22 General multiplication rule of probability obtained by rearranging definition of conditional probability:

providing P(B)>0. Sometimes called the law of compound probability

) B ( P

] B A ( P = ) B A ( P ∩_

P ( A B ) = P ( A B ) P ( B )∩


_________________________________________________________________________________________ SESS6002 12 2006/2007

If, two events A and B, represented on a Venn diagram as Figure 6 Law of compound probability then A ( A B ) ( A B ) = ∩ ∪ ∩ as these two components are mutually exclusive then can write

which is known as the law of inflating probabilities. Example: Probability of train arriving on time is 90% if there is no snow and 10% if there is snow. According to weather forecast there is a 40% chance of snow. What is the probability of arrival on time ? Let A be event of arrival on time and S the event of snow.

hence P(A) = (0.40)(0.10) + (0.60)(0.9) = 0.04 + 0.54 = 0.58 There is a 58% chance the train will arrive on time.

hence

P ( A ) = P(A B) + P(A B )

P(A) = P(B) P(A B ) + P( B ) P(A B )

∩ ∩

P(A) = P(A S) + P(A S ) = P(S) P(A S ) + P( S ) P(A S )∩ ∩


_________________________________________________________________________________________ SESS6002 13 2006/2007

The concept of conditional probability allows us to formulate models for practical problems in which results at some stage may be influenced by previous occurrences For two events, A is independent of B if the conditional probability of A(given B) is the same as the unconditional probability of A. P(A|B)=P(A) and hence also P(B|A)=P(B) Multiplication rule for independent events: If A and B are independent then

Have to be careful when extending the idea to more than two events. Independence of three or more events. The events A1, A2, ... Ak are independent if and only if the probability of the intersection of any combination of them is equal to the product of the probabilities of the single events. Independence of events is a useful and powerful concept as it simplifies calculations involving complex probabilities. Most real-world probabilistic models can only be analysed if independence is assumed. Therefore, there is a temptation to abuse system and hence need for random sampling techniques. Martin Gardener (1982) noted that "most people find it difficult to believe that the probability of an independent event is not somehow influenced by its proximity to other independent events of the same sort. eg. coin tossing with long sequence of heads. Rare events ?

2.6 Bayes' Theorem If B1 , B2 , ..., Bk are mutually exclusive events, and A is any event with P(A)>0, then for any Bi

Baye's rule is sometimes interpreted as the probability that Bi was the cause of A. In such a context, the unconditional probabilities P(Bi ) are called prior probabilities and the conditional probabilities P(A │ Bi ) posterior probabilities. Example 5.

P(B) P(A) = B) P(A∩

iii

i

1 1 2 2 k k

P( ) P(A B )P( A) BBP( A ) = = B P(A) P(A)

P( ) P(A B )B =P( ) P(A| ) + P( ) P(A| ) + ... + P( ) P(A| )B B B B B B

∩


_________________________________________________________________________________________ SESS6002 14 2006/2007

In 30 years, the likelihoods of different earthquale in tensities within one area are listed as below:- P(I) = 0.14 P(II) = 0.30 P(III) = 0.20; P(IV) = 0.15 P(V) = 0.10 P(VI) = 0.05 P(VII) = 0.05 P(VIII) = 0.01 P(IX) = P(X) = P(XI) = 0.0 The conditional probabilities of major structural damage depend on earthquake intensities, namely P(F│≤ III) = 0.0 P(F ⏐ IV) = 0.001 P(F ⏐ V) = 0.01 P(F ⏐ VI) = 0.05 P(F ⏐ VII) = 0.10 P(F ⏐ VIII) = 0.20 P(F ⏐ IX) = 0.40 P(F ⏐ X) = 0.75 P(F ⏐ ≥ XI) = 1.0 Hence in 30 years, the total probability of structural damage due to an earthquake is predicted as

Example 6. Following example 5, if a structure collapses due to an earthquake occuring within 30 years, the probability that the collapse due to an earthquake of intensity VI is found to be

Similarly the total probability caused by earthquakes with intensity greater than VI can be estimated

2.6.1 Another digression When asking sensitive questions it is unreasonable to expect truthful answers. The purpose behind asking the questions is to estimate the proportion of a population that belong in the 'yes'

0.011 = P(XII) x XII) | P(F + ... + P(I) x I) | P(F = P(F)

0.23 = 0.011

0.05 x 0.05 = P(F)

P(VI) x VI) | P(F = F) | P(VI

0.864 =

P(F)P(XII) x XII) | P(F + ... + P(VI) x VI) | P(F = F) | VI P( ≥


_________________________________________________________________________________________ SESS6002 15 2006/2007

response category. For example, polling about political preferences. This is done by selecting a 'random' sample of n people and estimating p by the proportion of affirmative replies in the sample p'. If you can't rely on truthful responses the estimate is obviously inaccurate (biased) so use randomised response technique.

2.7 Permutations and Combinations If there are n elements they can be laid out in a row in any order. Each such arrangement is called a permutation. Theorem 1: The number of permutations of n different elements taken all at a time is:

Theorem 2: If n given elements can be divided into c classes such that elements belonging to the same class are alike while things belonging to different classes are different, then the number of permutations of these things taken all at a time is:

A permutation of n elements taken k at a time is a permutation containing only k of the given elements. Two such permutations consisting of the same k elements in different order, are different, by definition. eg. there are 6 permutations of three letters a, b, c taken 2 at a time: ab, ac, bc, ba, ca, cb. A permutation of n elements taken k at a time with repetition, is as above but with individual elements chosen from all n at every stage. eg. there are 9 permutations with repetition of three letters a, b, c taken 2 at a time: as above plus aa, bb, cc Theorem 3: The number of different permutations of n different elements taken k at a time without repetition is:

and with repetition is

x x x xn ! = 1 2 3 ..... n

x x x x1 2 3 c

1 2 3 c

n ! ! ! ! ... !n n n n

where + + + ... + = nn n n n

! k) - (n

! n = ) 1 + k - n ( ... ) 3 - n ( ) 2 - n ( ) 1 - n ( n


_________________________________________________________________________________________ SESS6002 16 2006/2007

The number of combination of n different elements taken k at a time, without repetitions is the number of sets that can be made up from the n given elements each set containing k different elements and no two sets containing exactly the same k elements. The number of combinations of n different elements taken k at a time with repetitions is the number of sets that can be made up of k elements chosen from the n elements, each element being used as often as desired: eg. there are 3 combinations of three letters a, b, c taken 2 letters at a time without repetition, namely ab, ac, bc and 6 such combinations with repetitions ab, ac, bc, aa, bb, cc. Theorem 4: The number of different combinations of n different elements, k at a time, without repetitions, is:

The factorial function is defined such that 0 ! = 1 and further values are computed from (n + 1) ! = ( n + 1 ). n ! and for large n the function is very large and a convenient approximation is Stirling's asymptotic formula:

2.8 Analytical Models Random variables are introduced to map events into numerical systems in order to describe these events analytically. Distribution functions provide pre-defined rules for assigning probabilities to values of random variables. A function which assigns a number to each outcome of a random experiment is called a random

nk

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

k

1 - k + n

is repetition withnscombinatio ofnumber the and

1.2...k

1)+k-2)...(n-1)(n-(n n = ! ) k - n ( ! K

! n = kn

nnn ! 2 n (e = 2.718...)

eπ ⎛ ⎞

⎜ ⎟⎝ ⎠

;


_________________________________________________________________________________________ SESS6002 17 2006/2007

variable. The probability distribution or sometimes the distribution of the random variable X is given by a list of the values of the random variable with corresponding probabilities which sum to 1. Mathematically a probability distribution function f(x) has a total area equal to one, and the cumulative distribution function F(x) is defined as:

If we perform a random experiment and the event corresponding to a number a occurs then the event X = a and its probability is P(a). similarly the probability we obtain a value in a given interval is P(a<X<b). Importantly as P(-∞ < X < ∞ ) = 1 then:

The probability of X occurring in the interval [a,b] can be found to be,

The statistical properties of random variable X are commonly specified by its mean, standard deviation, skewness, and kurtosis, and they are defined as follows

hence

x

-

F(x) = f(y) dy

dF(x)f(x) =

dx

∞∫

)a X P( - 1 = )a > X P( ≤

F(a)- F(b) = f(x)dx - f(x)dx =

f(x)dx = ) b X a P(

a

-

b

-

b

a

∫∫

∫

∞∞

≤≤


_________________________________________________________________________________________ SESS6002 18 2006/2007

The non-dimensional parameter skewness characterises the degree of asymmetry of a distribution around its mean. The kurtosis is also a non-dimensional quantity and measures the relative peaked-ness or flatness of a distribution. For a Gaussian variable, the skewness and kurtosis are equal to zero. The coefficient of variation, COV, is often used to indicate the variability of random variable X.

Most engineering problems involve multiple random variables. The following equations for multiple random variables are relevant.

X-

= E(X) = x f(x) dxµ∞

∞∫

( )Var 22X X

-

= (X) = x - f(x) dxµσ∞

∞∫

Skewness3

X

X-

x - = f(x) dxµ

σ

∞

∞

⎛ ⎞⎜ ⎟⎝ ⎠∫

Kurtosis4

X

X-

x - = f(x) dx - 3µ

σ

∞

∞

⎛ ⎞⎜ ⎟⎝ ⎠∫

µσ

X

X = COV


_________________________________________________________________________________________ SESS6002 19 2006/2007

A linear function of several Gaussian variables is also a Gaussian variable. The moments of a linear function of a random variable can be written as follows, for example Y = α X

where α is a constant. The equations given below can also be derived,

here Var(X1,X2) is the covariance of X1 and X2. The above equations can be generalised for a general linear function.

The statistical moments of a general non-linear function involving several random variables can also be approximated by using Taylor series. The estimation depends on the point at which the

Y-

X

= E(Y) = f(x) dx

= E(X) =

αµ

α α µ

∞

∞

⋅∫

( )

( )

22Y X X

-

22 2 2XX

= (x) dx f

= E X - =

α α µσ

µα α σ

∞

∞

⋅

⎡ ⎤⎣ ⎦

∫

Var

1 2

1 2

1 21 2

1 2Y X X

2 2 2 2 21 2Y 1 2 1 2X X

Y = + X X

= +

= + + 2 ( , )X X

α α

µ µ µα α

σ α σ α σ α α

( )Var

n

i0 ii=1

n

i0 iYi=1

n n2

i jY i ji=1 j=1

Y = + X

= +

, X X

=

α α

µ µα α

σ α α

∑

∑

∑∑


_________________________________________________________________________________________ SESS6002 20 2006/2007

non-linear function is linearised. Details on this matter will be discussed later. A natural choice for such a point is the mean value point, but this may not give the most accurate prediction. If the non-linear function is linearised at X* = ( X1* , ... , Xn* ), the following equations can be derived,

Example Estimate the first and second moments of a function Y = X1X2

2. Here X1 and X2 are statistically independent. The function is to be linearised at the mean value point. Common Distribution Functions 1. Normal (Gaussian) distribution

2. Log-normal distribution

( ) ( ) X - X XXg

+ )Xg( )Xg( *ii

i

*n

1=i

*

∂

∂≈ ∑

( ) ( )X - X

Xg + )Xg( *ii

i

*n

1=i

*Y µµ

∂

∂≈ ∑

( ) ( ) ( )Var

* *n n2

i jYi ji=1 j=1

g g X X , X XX X

σ∂ ∂

≈∂ ∂∑∑

exp2x

X X12

X XX-

x - x - 1F(x) = - dy = 2

µ µ

π σ σσ∞

⎡ ⎤⎛ ⎞ ⎛ ⎞Φ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦∫

exp2

X12

XX

x - 1f(x) = - , - < x <2

µ

π σσ

⎡ ⎤⎛ ⎞∞ ∞⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦


_________________________________________________________________________________________ SESS6002 21 2006/2007

The median value of X corresponds to a probability of 50% and is equal to exp(λ). The following equations can be established for a log-normal distribution function,

where xm is the median value 3. Rayleigh distribution

2.8.1Discrete Random Variables and discrete distributions. A random variable is said to be discrete if it has the following properties: 1: The number of values for which X has a probability different form 0 if finite or at most

countable infinite. 2: If an interval a < X ≤ b does not contain such a value then the probability of within that

interval is zero The probability function of a discrete random variable is defined as :

ln lnexp

2

12

1 x - x - f(x) = - = , 0 < x 2 x

λ λξ ξπ ξ

⎡ ⎤⎛ ⎞ ⎛ ⎞Φ <∞⎢ ⎥⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

ln Var ln2 = E( x ) , = ( X )λ ξ

⎟⎟⎠

⎞⎜⎜⎝

⎛ + 1 =

x = - =

2

22

m2

21

µσξ

ξµλ

ln

lnln

exp2x xf(x) = - , 0 < x <

m 2 m⎡ ⎤

∞⎢ ⎥⎣ ⎦

when

otherwise

jj x = (j = 1, 2, .... )p xf ( x ) =

0

⎧⎪⎨⎪⎩


_________________________________________________________________________________________ SESS6002 22 2006/2007

2.8.4 Some More Statistics Confidence Intervals We know that from a sample we cannot draw conclusions about the corresponding population that are 100% certain. So we have to be more modest and modify our problem as follows: Chose a probability γ close to 1 (for example 95%, 99% etc ). Then determine two quantities Θ1 and Θ2 such that the probability that Θ1 and Θ2 include the exact unknown value of the parameter θ is equal to γ The idea is to replace the impossible requirement with a preassigned probability close to 1. If we know such functions Θ1 and Θ2 and a sample is given can compute numerical value of θ1 of Θ1 and a numerical value of θ2 of Θ2 . The interval between end points is called a confidence interval with lower and upper confidence limits and the number γ the confidence level. Determination of a confidence interval for the mean µ of a normal distribution with known variance σ2: 1: Choose confidence level 2: Determine corresponding c 3: Compute mean of the sample x' 4: Compute k = c σ/∫n. The confidence interval for µ is:

Determination of a confidence interval for the mean µ of a normal distribution with unknown variance σ2 : 1: Choose a confidence level 2: Determine the solution c of the equation:

from the table of the t-distribution due to Student with n-1 degrees of freedom ( n is sample size)

3: Compute the mean x' and the variance s2 of the sample 4: Compute k = c σ/∫n to give a confidence interval as before. Determination of a confidence interval for the variance σ2 of a normal distribution whose mean need not be known: 1: Choose a confidence level 2: Determine solutions c1 and c2 of the equations

( ) k + x k - x CONF ʹ′≤≤ʹ′ µ

) + 1 ( = ) c ( F 21 γ


_________________________________________________________________________________________ SESS6002 23 2006/2007

from the table of the chi-square distribution with n - 1 degrees of freedom (n = size of

sample ) 3: Compute (n-1)s2 where s2 is the variance of the sample 4: Compute k1 = (n-1)s2/c1 and k2 = (n-1)s2/c2 . The confidence interval is:

2.8.4 Statistical Hypothesis testing: Define statistics as the art of learning from experience. The actual occurrence of an event to which a certain statistical hypothesis attributes a small probability is an argument against that hypothesis, and the smaller the probability, the stronger is the argument...if a statistical hypothesis survives several opportunities of refutation, we may consider it corroborated to a certain extent.

2.9 Conclusion A review has been made of some of the basics of probability and statistics. The next lecture will progress from these to discuss how lessons can be drawn for decision making in uncertainty and risk analysis techniques.

Bibliography E. Kreyszig, Advanced Engineering Mathematics, 5th Edition, John Wiley & Sons,1983 S. Kotz and D.F. Stroup, Educated Guessing: How to cope in an uncertain world, Marcel Dekker Inc., 1983. QA 273 KOT

) + 1 ( = ) c ( F ,) - 1 ( = ) c ( F 21

221

1 γγ

( ) k k CONF 12

2 ≤≤σ


_________________________________________________________________________________________ SESS6002 24 2006/2007

Example Sheet 1 Probability 1. Let A be the event that you have enough money in your bank account and let B be the event that you postdate your check. if P(A) = 0.8, P(B│A) = 0.1 and P(B│A*) = 1., compute P(A│B) and interpret the numerical result. (Ref. Kotz and Stroup) 2. A survey of 400 students reported that 200 smoked regularly, 150 drank coffee regularly, 90 both smoked and drank coffee regularly, and 180 neither smoked nor drank coffee. Are you suspicious of the results of this survey? 3. For a twin-screw vessel to lose all forward power (event L), either both power drive systems A and B have to fail or control of both is lost from the bridge, event C. Then

Show that

4. The Department of Wildlife needs to estimate the number of deer within a wildlife reserve. Deer do not generally cooperate by standing still to be counted, so a more subtle technique has been developed. Suppose that the reserve is known to contain either 1, 2, 3, 4, or 5 deer with equal probability. We choose a deer at random, fire a big tranquillizer dart at it, while it is knocked out it is marked with a tag and then let go. The next day we observe a deer from the reserve and note that it is not tagged. What are the posterior probabilities that the reserve contains 1, 2, 3, 4, and 5 deer respectively. 5. Suppose in the example given for drilling for oil or selling the rights the prior probabilities are P(oil)=0.3 and P(no oil)=0.7. Show that the Bayesian decision for the no-data problem is to sell the drilling rights. What value for P(oil) would cause us to be indifferent to whether we drill or sell the drilling rights. 6. A motor drives an electric generator. During a 30-day period, the motor needs repair with probability 5% and the generator needs repair with probability 6%. What is the probability that during a given period, the entire apparatus will need repair. 7. A box of 100 gaskets contains 10 gaskets with type A defects, 5 gaskets with type B defects and 2 gaskets with both types of defects. Find the probability that a gasket to be drawn has a type B defect under the condition that it has a type A defect. 8. Another wildlife reserve has a small number of deer A. You know that A is 1, , or 3, and you assign probabilities of 0.2, 0.2, and 0.6 respectively to these values. You catch a deer at random, tag it and let it go. The next day you again catch a deer, note whether it is tagged and let it go. You must then decide whether the reserve has three deer, winning a prize, if your decision is correct. What is the best method in deciding ?

( ) C ) B A ( P = P(L) ∪∩

) C A P( - ) C B P( + ) C A P( = P(L) ∪∪∪∪

chptr2 prob and stats

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