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Preparatory Program in Mathematics

Module-II

Amit Vatsa

[email protected]

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Introduction to Probability Theory

• Probability: Quantitative measure of chance.

– 0.0 Absolutely no chance

– 1.0 Absolute certainty

• Some terminology

– Experiment Tossing a die

– Event (Experimental outcome) 1,2,3,4,5,6

• Elementary Event 1,2,3,4,5,6

• Composite Event < 3, even, odd – A composite event includes more than one elementary outcomes.

– Sample Space: A complete or exhaustive listing (set) of elementary events.{1,2,3,4,5,6}

– All elements in the sample space are mutually exclusive and collectively exhaustive.

(Composite events need not be mutually exclusive).

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Example

1. Distributing 3 distinguishable balls in 2 cells. How many

possible outcomes? What is the sample space?

Let a, b, c be the 3 balls. Possible outcomes are:

{a b c| -}

{a b| c}{a c| b}

{b c| a}

{a | b c}

{b | c a}

{c | a b}

{- | a b c}

Event : 1

st

cell is not empty?Event: 1st cell contains two balls?

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Example

1. Distributing 3 indistinguishable balls in 2 cells. How many

possible outcomes? What is the sample space?

Possible outcomes are:

{- | ***}

{*| **}{**| *}

{***| -}

Event : 1st cell is not empty?

Event: 1st cell contains two balls?

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Example

• What is the sample space for two coin toss?

• What is the sample space for two dice throws?• In a family of two children

– Sample space of number of girls in the family

– Are all outcomes equally likely??

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• Assumption: All elements in the sample space are equally likelyto occur???

– How valid is this?

• Throwing a die: sample space {1,2,3,4,5,6}

• What is the chance of each elementary event occurring?

• Can we now determine probabilities of composite events? How?

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Composite Events

• Event A: outcome of the toss is an even number.

– Which elementary outcomes are favorable to event A? 2,4,6.

– Since each of these has a 1 in 6 chance, P(A) = 3/6 = 0.5

• Event B: outcome is divisible by 3

– Favorable outcomes: 3, 6. P(B) = 2/6 = 0.333.

• Complementary Event: The event in question does not occur.

– Complement of event A is designated A‟.

– What is P(A‟) ? Clearly 1-P(A). Why?

• Sum of probabilities of events that are mutually exclusive and

col lectively exhaustive is always 1.0.

• Odds: ratio of favorable to unfavorable outcomes.

– What are the odds that the outcome is divisible by 3? 2:4

• If the odds of an event are a:b , then its probability is a/(a+b).

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Joint or Non-exclusive Events

• Two events are non-exclusive if they have one or more elementrary outcomes

common between them, i.e. they can occur together.

– A: outcome is even.

– B: outcome is divisible by 3.

– Are they exclusive or non-exclusive? Why?

• A Venn Diagram is useful

1

23

4 56Sample space

A B

Another example: Draw one card from a deck of 52 cards.

Event A: A Heart is drawn

Event B: A King is drawn

Are they exclusive?

A Black King is drawn

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Probability of Union of Events

• P(A or B) or P(AB)

• If A and B are mutually exclusive events, P(A or B) = P(A)+P(B)

• A – Outcome of die toss is less then 3

• B – Outcome is divisible by 3

1

234 56

AB

1

2

3

4

56P(A or B) = 2/6+2/6 = 4/6 = 0.667

What if they are non-exclusive?

A – Even number

B – Divisible by 3

P(AB) = P(A)+P(B)-P(AB) General Rule of addition

P(AB) = 3/6+2/6-1/6 = 4/6

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Example

• Two dice are thrown

– Probability both of them show 6?

– Probability none of them show 6?

– Probability at least one of them show 6?

– Probability exactly one of them show 6?

• Cell and ball problems

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Independent Events

• Draw a card from a deck.

• Put it aside.• Now draw another card.

• What is the probability that first card is a King?

• What is the probability that second card is a King?

• Does it depend of the first draw? Why?

• What if the first card is put back in the deck before the second

draw? Does it still depend on the first draw? Why?

• Two events are independent if the occurrence of non-occurrence

of one event has no effect on the probability of the other event.

• They are dependent when the occurrence of one event does affect

the probability of the other event.

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Probability of Intersection of Events

• P(AB) = ?

• Toss a coin twice – A : Head on first toss P(A) = 0.5

– B : Head on second toss P(B) = 0.5

• AB : Head on both tosses (Joint event)

• P(AB) = P(A).P(B) (Joint probability)• This is true only when the two events are independent .

• What if A and B are dependent?

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Conditional Probability

• Event A : A King on the first draw.

• Event B : A King on the second draw.

• P(A) = 4/52

• What about of P(B)? It depends on whether first draw was a King or not.

– If A occurred, probability of B = 3/51

– If A did not occur, probability of B = 4/51.

• Notice that there is a condition attached to the probability in each case. – P(B/A) = 3/51

– P(B/A‟) = 4/51.

• Why is the probability different in the two cases? Dependence!

• Test of Independence

– Events A and B are independent if P(A/B) = P(A/B‟) = P(A) and

P(B/A) = P(B/A‟) = P(B).

– Here P(A) and P(B) are unconditional probabilities of the two events.

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New Sample Space

Joint Probability

• For two events A and B, what is the probability that both events

take place. P(A and B) or P(AB)

• If A and B are independent events, P(A and B) = P(A).P(B)

• What if they are dependent?

• View this as a two step process:

– A occurs. – Given that A has occurred, B occurs.

• P(A and B) = P(A). P(B/A).

• A die toss example:

– A : outcome is less than 5 – B : outcome is a multiple of 3

• P(A) = 4/6, P(B) = 2/6, P(B/A) = ?

26

4

1

53

A B

1/4

P(A and B) = (4/6).(1/4) = 1/6 P(B/A) = P(A and B) / P(A)

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B

Total Probability Formula

• Consider dependent events A and B.

– P(B) = P(AB) + P(A B)

– = P(B/A)P(A) + P(B/A)P(A)

• Consider a mutually exclusive and collectively exhaustive set of

events: A1, A2, A3,…An

• Let B be some event in the same sample space.

• P(B) = P(BA1)+P(BA2)…..+P(BAn)

• = P(A1)P(B/A1)+P(A2)P(B/A2)….+P(An)P(B/An)

A1

A2

A4

A3

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Unconditional (Marginal) Probability of B

• Revisit the Card experiment

– A: King on first draw

– B: King on second draw

• We know the following probabilities: – P(A) = 4/52 P(A‟) = 1-P(A) = 48/52

– P(B/A) = 3/51 P(B/A‟) = 4/51

• Given these, can we compute P(B)?• Consider the following two joint events:

– (A and B) P(A and B) = P(A).P(B/A) = (4/52)(3/51)

– (A‟ and B) P(A‟ and B) = (A‟).P(B/A‟) = (48/52)(4/51)

– These two are mutually exclusive events

– They include all possible outcomes favorable to event B.

– P(B) = P(A and B) + P(A‟ and B)

– P(B) = P(A).P(B/A) + P(A‟).P(B/A‟) = 0.0045+0.0724 = 0.0769

It is exactly the same as 4/52 = 0.0769. Why?

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Bayes‟ Theorem

• Revisit the Card drawing experiment again.

– Event A : King on first draw

– Event B : King on second draw

• P(A), P(B/A) are easily determined. P(A) = 4/52, P(B/A) = 3/51.

• Suppose we are interested in P(A/B)!!! How to find it?

• Bayes‟ theorem comes to our rescue.

• Recall the joint probability formula – P(A and B) = P(A).P(B/A)

• Reversing the role of events A and B, the following must also be true:

– P(A and B) = P(B).P(A/B)

• Combining the two, we get

– P(A).P(B/A) = P(B).P(A/B)

– Or P(A/B) = [P(A).P(B/A)]/P(B) Baye‟s Formula

• Let us substitute P(B) = P(A)P(B/A) + P(A‟)P(B/A‟)

• P(A/B) = [P(A).P(B/A)] / [P(A)P(B/A) + P(A‟)P(B/A‟)]

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Female Male Total

Admin (A) 20 30 50

Operations (O) 60 170 230

Sales (S) 100 70 170Totals 180 270 450

An employee is chosen at random:

What is the probability that the person is female?

that he is male working in administrative division?

works in operations, if the person is female?

is a female if the person works in operations?

Are S and M independent?

Are A and F independent?

Female Male Total

Admin (A) 0.0444 0.0666 0.111

Ops. (O) 0.1333 0.3777 0.511

Sales (S) 0.2222 0.1555 0.3777Totals 0.4 0.6 1.0

A bl l t i t f th li B B d B Fift

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• An assembly plant receives parts from three suppliers B1, B2 and B3. Fifty

percent of supply comes from B1, and 25 percent from B2 and B3 each. Parts

supplied by these three suppliers contain 5%, 10% and 12% defective parts

respectively.

– If a part is chosen at random from the stock room, what is the probability

that it is defective?

– If a randomly selected part was found to be good, what is the probability

that it came from B2?

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Random Variable• Experiment Outcome Numerical Value

• A random variable X maps the experimental outcome to a real number.

• Consider tossing a coin 3 times.• Possible outcomes:

– HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

– X – No. of heads appearing in three tosses, X = 0,1,2,3

– P(X = 0) = 1/8, P(X = 1) = 3/8, P(X=2) = 3/8, P(X=3) = 1/8

• Random Variables are of two types:

– Discrete: can take a finite number of fixed values

• No. of empolyees absent, No. of cars sold

– Continuous:

• Qty. of cement produced

• Diameter of the hole drilled.

• weight of fertilizer packed in a bag. – Sometimes it is o.k. to approximate a discrete variable as a continuous one.

• No. of customers visiting a bank branch.

• No. of cars passing through a toll booth.

• Bags of cement produced in a day by a cement plant.

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Probability Distribution

• Discrete Random variable

– Assigns a probability value to each possible value of the random variable.

– No. of head in 2 tosses 0 1 2

– Probability 0.25 0.5 0.25

– Sum of all probabilities must be 1.0.

• Continuous Random variable

– Can we assign a probability to each possible value? Why not?

– Suppose X has a specific range of values (0,10)? Is it meaningful to talk about

P(X=3.71)?

– We can assign a probability density.

– It is only meaningful to talk about probability of X lying in an interval.

– P(a X b) is given by the area under the density function curve between a and b.

a b

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Binomial Distribution

• The production process of a company produces 5% defective

items. If a random sample of 20 items is taken from the day‟s

production, what is the probability that

– There are no defective items in the sample.

– There is only one defective item in the sample.

– There are „k‟ defective items in the sample

• Define random variable X as the number of defective items in asample of n (=20 in this case). What is the probability distribution

of random variable X?

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• Bernoulli Trial

– An experiment that has only two possible outcomes (Success/Failure)

• e.g. tossing a coin – P(S) = p, P(F) = 1-p

– Probability p is unchanged from one trial to next.

– All trials are independent of each other.

• Suppose that n such trial are conducted.

– What is the number of successes in n trials? – It is a random variable. Range : 0 – n.

– What is the probability distribution of this random variable?

• Consider a specific sequence of Successes and Failures: FSSFF

– There are 2 Successes? X = 2

– What is the probability of this specific sequence?

– (1-p)pp(1-p)(1-p) = p2(1-p)3

– Is this the only sequence with 2 successes? SSFFF, SFSFF, SFFSF, SFFFS, ……

– How many? (52). The probability of occurrence for each one is p2(1-p)3

– What is the total probability of getting 2 Successes out of 5 trials?

– (52) p

2(1-p)3

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• f(x) = P(X=x) = p(x;n,p) = (nx) p

x(1-p)n-x

• n and p are the parameters of the distribution.• Expected Value of a Random Variable. (Mean value)

• E(X) = x.f(x)

• For Binomial distribution E(X) = x = n.p

• Variance V(X)• V(X) = (x – E(x))2.f(x)

• Variance of Binomial Distribution: V(X) = x2 = [n.p.(1-p)]

• x=[n.p.(1-p)]

• How is the variance affected by n and p?

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Binomial

Probabilities

N = 10, p = 0.05

x p(x)

0 0.5987

1 0.3152

2 0.0746

3 0.0105

4 0.0009

5 0.0001

6 0.0000

7 0.0000

8 0.0000

9 0.0000

10 0.0000

Binomial

Probabilities

N = 10, p = 0.50

x p(x)

0 0.0010

1 0.0097

2 0.0440

3 0.1172

4 0.2051

5 0.2460

6 0.2051

7 0.1172

8 0.0440

9 0.0097

10 0.0010

Binomial

Probabilities

N = 10, p = 0.95x p(x)

0 0.0000

1 0.0000

2 0.0000

3 0.0000

4 0.0000

5 0.0001

6 0.0009

7 0.0105

8 0.0746

9 0.3152

10 0.5987

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Applications of Binomial Distribution

• The production process of a particular product produces 5% defective units. Acustomer has ordered a batch of 20 units. If the batch contains 3 or more

defective units, the customer will reject the entire batch, and cancel the order. – What is probability that the order will be cancelled?

– What is the maximum permissible percentage of defectives in the production process so that the probability of rejection is reduced to less than 1%?

• P(Rejection) = 1- P(X 2) = 1- [P(X=0)+P(X=1)+P(X=2)]

• P(X 2) = (20

0

) p0(1-p)20 +(20

1

) p1(1-p)19 +(20

2

) p2(1-p)18

• = 0.3585+0.3774+0.1887 = 0.9246

• P(rejection) = 0.0754

• The management of a restaurant that operates on reservations

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only, knows from experience that 15% of persons making table

reservations will not show up. If the restaurant accepts 25

reservations, but has only 20 tables, what is the probability that all

who show up will be accomodated?


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only, knows from experience that 15% of persons making table

reservations will not show up. If the restaurant accepts 25

reservations, but has only 20 tables, what is the probability that all

who show up will be accomodated?

i i ib i

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Poisson Distribution

• Consider n and p 0

• But Lim n.p =

• There are 100000 vehicles on the streets of Ahmedabad

• On a given day, probability of a vehicle meeting with an accident is

0.00005.

• n.p = = 5 accidents per day (accident rate )

• The actual values of n and p are not important, as long as is known.

• A Bank has 10000 customers. The probability of customer arriving on a

given day is 0.005.

• = 50. Customer arrive at a rate of 50 per day (arr ival rate ).

• The actual number of customers arriving may range from 0 to (actually10000).

• A machine produces 50000 parts per day, with 0.001 probability of a part

being defective. What is the number of defective parts produced in a day?

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Random variable X : No. of customer arriving in one hour.

No. of accidents taking place in a day.

No. of defective parts produced in a day.

What is the probability distribution of X?Can we use Binomial here? What is the difficulty?

Poisson Distribution is a limiting case of Binomial

as n and p 0 Lim n.p =

P i Di ib i

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Poisson Distribution

P(x, ) =

e- x

----------x!

E(X) = , V(X) =

n = 10

p = 0.2

0.1074

0.2684

0.3020

0.2013

0.0881

0.02640.0055

0.0008

0.0001

0.0000

n=20

p = 0.1

0.1216

0.2702

0.2852

0.1901

0.0898

0.03190.0089

0.0020

0.0004

0.0001

n = 40

p = 0.05

0.1285

0.2706

0.2777

0.1851

0.0901

0.0342

0.0105

0.0027

0.0006

0.0001

n = 100

p = 0.02

0.1326

0.2707

0.2734

0.1823

0.0902

0.0353

0.0114

0.0031

0.0007

0.0002

X

0

1

2

3

4

56

7

8

9

Poisson

= 2

0.1353

0.2707

0.2707

0.1804

0.0902

0.0361

0.0120

0.0034

0.0009

0.0002

Poisson Tables

Poisson approximation of

Binomial,

n>20 and p<0.05

N i Bi i l Di ib i

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Negative Binomial Distribution

• Consider a sequence of Bernoulli trials.

• Binomial : X - No. of successes in n trials.• Negative Binomial

– Suppose you wish to continue conducting trials until a desired number of

successes are achieved.

– Random Variable X – How many failures take place before the desired

number, k , of successes occur?

– If k = 3, and the sequence FFSFSFFFS results, then X = 6.

• Last trial must be a success. Why?

• x+k-1 trials are conducted before the last one.

• What is the number of possible sequences with x failure and k-1

successes? x+k-1Cx.

• Each one has probability pk (1-p)x

• P(X=x) = P(x;k,p) = x+k-1Cx pk (1-p)x

N ti Bi i l Di t ib ti

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Negative Binomial Distribution

k(1-p)

-------

p2

V(X) =

k(1-p)

-------

p

E(X) =

A machine has a 60% chance that it will be able to produce a

part of acceptable quality. If five parts are to be produced, whatis the probability that no more than 7 attempts are required to

produce them? What is the average number of attempts

required?

G t i Di t ib ti

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Geometric Distribution

Geometric Distribution : A special case of Negative Binomial Dist.

Let k = 1

How many failures before the first success?

x+k-1Cx pk (1-p)x

P(x;p) = p(1-p)x

(1-p)

-------

pE(X) =

(1-p)

-------

p2 V(X) =

k(1-p)

-------

p2 V(X) =

k(1-p)

-------

pE(X) =

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• ISRO is testing a new design of satellite launching rocket. The

probability of a successful test launch is 0.7. What is the

probability that they will achieve a successful launch within 4

attempts?

• P(X=0) + P(X=1) + P(X=2) + P(X=3)

H t i Di t ib ti

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Hypergeometric Distribution

• Binomial

– Select sample of size n from a potentially infinite population.

– View selection of n items as a sequential process.

– Probability p is unaffected by previous selections.

• What if the population is finite?

– In a population of 10 items, 5 are defective.

– p = 0.5 for the first item.

– For second item, p depends on the outcome of first selection.

• First item defective p = 4/9

• Not defective p = 5/9

• Given a population of N items containing k=Np defectives, if n

items are selected at random, what is the probability of getting x

defective items? Hypergeometric Distribution.

– Range of x is from 0 to min(n,k)

H t i Di t ib ti

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Hypergeometric Distribution

• Red/Black Balls.

– N – total number of balls

– Np – No. of black balls (defective items)

– N-Np – No. of red balls (good items)

P(x; N,n,p) =

( Npx) (

N-Np n-x)

----------------( Nn)

As N becomes large, or n/N becomes small, these probabilities

tend to Binomial probabilities.

E l

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Example

• A box has 3 red and 4 black balls. If four balls are selected at

random, what is the probability of getting two red and two black

balls.

C ti P b bilit Di t ib ti

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Continuous Probability Distributions

• Uniform Distribution (Rectangular Distribution)

– Random Variable X lies in the interval (a,b), i.e. a≤X≤b

– All values of X are equally likely. (Uniform )

– Probability Density is the same everywhere.

a b

1/(b-a)

f(x;a,b) =

1/(b-a) a≤ x ≤b

0 elsewhere{

F(x;a,b) = {0 x < a

(x-a)/(b-a) a≤ x ≤b

1 x > b

t

P(x≤t)

E pectation and Variance

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Expectation and Variance

• E(X) = = a b x.f(x).dx

• V(X) = 2 = a

b (x-)2.f(x).dx

• For Uniform Distribution

– E(X) = (a+b)/2

– V(X) = (b-a)2/12, x = (b-a)/23

– A special case : a = 0, b = 1 Uniform(0,1)

– E(X) = 0.5, V(X) = 1/12, x = 1/23

Digression to “Moments” and “Kurtosis”

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Digression to “Moments” and “Kurtosis”

• E(X) = 1 = - x.f(x).dx

• V(X) = 2 = - (x-1)2 .f(x).dx

• 3 = - (x-1)3 .f(x).dx

• 4 = - (x-1)4 .f(x).dx

First Moment about zero (Mean)

Second central Moment (Variance)

Third Central Moment (Skewness)

Fourth Central Moment (Kurtosis)

• First moment defines the location

• Higher Moments define the shape

• Relative Values are more meaningful.

– Coefficient of Variation / • Relative Kurtosis 4 = 4/2

2

Relevance of Kurtosis

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Relevance of Kurtosis

Mesokurtic (Normal Dist.) 4 =3

Leptokurtic 4 >3

Platykurtic 4 < 3

Kurtosis indicates the “Peakedness” of the distribution.

Normal Distribution is viewed as a reference point, neither very

high, nor very low in terms of peakedness.

Leptokurtic – Higher Peakedness compared to Normal.

Platykurtic – Lower Peakedenss compared to Normal.

Uniform Distribution? Leptokurtic or Platykurtic?

4 =1.8

Normal Distribution

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Normal Distribution

• Also called Gaussian Disribution.

• Most important and most widely used.

• Symmetrical, bell shaped, extends infinitely in both directions.

• Many naturally occurring data follows Normal Distribution

– Temperature, rainfall, measurements of living organisms.

– Measurements of manufactured parts, errors and deviations from norms.

• Density Function

2

2

1exp2

1),;(

x x f

• Two parameters: - mean, - standard deviation

dx x

x X E

.

2

2

1

exp2

1

.)(

dx x

x X V .2

21exp

2

1.2)()(

=

= 2

2

2exp2

1

)1,0;(

x x f

Standard Normal Dist.

X

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= 5

= 15

= 0

= 10

Distribution shifts by , if is added to random variable X.

Distribution shrinks or expands if X is multiplied by . The mean stays at zero.

= 10

X

= 5

X-5

= 15

X+5

Computing Normal Probabilities

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F(b)

Computing Normal Probabilities

• How to find P(ax b)?

• Evaluate the integral of f(x) from a to b.

• Alternatively, P(ax b) = F(b) – F(a)

a bF(a)

• No closed form solution to integral of f(x).

• Numerical integration by computer is only way.

• Tables of F(x) available in text books are useful.

• What about the parameters and ?

• Do we need a separate table for each and combination?

• One table is enough for all. How?

Standard Normal Distribution

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• Suppose X ~ N(,)

• Consider Z = X- as a random variable.

• What is the mean and variance of Z? What is the distribution of Z?

• Z ~ N(0,)• P(X a) = ?

• Substitute X = Z+

• P(X a) = P(Z+ a) = P(Z a-)

aa-10 = 10

X

= 0

Z


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• X ~ N(0,)

• Consider Z = X/ as a random variable.

• What is the mean and variance of Z? What is the distribution of Z?

• Z ~ N(0,1)• P(X a) = ?

• Substitute X = Z

• P(X a) = P(Z a) = P(Z a/)

= 2

= 1

aa/2


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• Given X~N(,), define Z = (X- )/.

• Then Z ~ N(0,1)

• P(X a) = P(Z (a- )/)

• Table of F(z) = P(Z z) are available and can be used to findcumulative probability for Normal distribution with any mean andstandard deviation.

N(,) N(0,1)

aZa= (a- )/

z value corresponding to a

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• Daily cement production of a plant follows normal distribution

with mean of 50 tons and standard deviation of 3.5 tons. On how

many days in a year the production exceeds 58 tons.

• =50, =3.5

• X ~ N(50,3.5)

• Z58

= (55-50)/3.5 = 1.43

• P(X 55) = P(Z 1.43)

• = 1- P(Z 1.43)

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P(Z 1.43) = 0.9236

P(Z 1.43) = 0.0764 = P(X 55)

Ans = 365*0.0764 = 28 days

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Th f i bl i ll di ib d

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The amount of time to assemble a computer is normally distributed

with a mean of 50 minutes and standard deviation of 10 minutes.

Find the probability that the computer is assembled between 45

and 60 minutes.

Mensa is an organization whose members possess IQs that are in

the top 2% of the population. It is known that IQs are normally

distributed with a mean of 100 and standard deviation of 16. Findthe minimum IQ needed to be a Mensa member.

The annual rate of return on a mutual fund is normally distributed

with a mean of 14% and a standard deviation of 18%.

What is the probability that the fund returns more than 25%

next year?

What is the probability that the fund loses money next

year?

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A automatic pesticide packing machine is used to fill 500 g.

pesticide powder in a carton. The standard deviation of the amount

filled by the machine is 4 grams. The management wants that no

more than 2% of the cartons filled should have less than 500 grams

pesticide in it. What amount should the machine be programmed to

fill?

The lifetime of TVs produced by a company are normally

distributed with a mean of 75 months and a standard deviation of 8

months. If the manufacturer wants to have to replace only 1% of

the TVs, what should its warranty be?

Every day a bakery prepares its famous cup-cakes. The dailydemand for cakes is normally distributed with a mean of 850 and a

standard deviation of 90. How many cup-cakes should the bakery

make so that the chances of running short on any day is not more

than 20%.

Normal Approximation of Binomial

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Normal Approximation of Binomial

• Let X ~ Binomial(n,p)

• x

= np, x

= [np(1-p)]

• Define random variable Y = (X-np)/[n.p(1-p)]

• As n, distribution of Y approaches N(0,1).

• Though X is discrete, as n increases, it approaches continuity.

• Normal is a reasonable approximation of Binomial when – np > 5 if p < ½

– n(1-p) > 5 if p > ½

1 2 3 4 5 6 7

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• Example: What is the probability of 3 or fewer heads in 10 tosses?

– n = 10, p = 0.5, np = = 5, = 2.5 = 1.58

– Z3.5 = (3.5-5)/1.58 = -0.95

– FZ (-0.95) = 0.1711

– From Binomial Tables, P(X <= 3) = 0.1719

Use Excel to plot the F(x) for both.

Normal Approximation of Poisson

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Normal Approximation of Poisson

• If is large, Poisson distribution may also be approximated by

Normal.

• E(X) = V(X) = for Poisson. So = and = .

Example: If X follows Poisson distribution with =25, find the

probability of X=20 using normal approximation.

P(19.5<X<20.5) = P(-1.1<Z<-0.9) = F(-0.9) – F(-1.1)

= 0.1841 – 0.1357 = 0.0484

P(X=20) = 0.0519 using Poisson Formula

Exponential Distribution

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Exponential Distribution

• Also called Negative Exponential

• f(x;) = ()exp(-x) x 0

• E(X) = 1/, V(X) = 1/2, i.e. = 1/

• 3 = 2 (Skewed Right)

• 4 = 9 (Highly Leptokurtic)

F(x;) = 1-exp(-x)

a

is called Arrival

Rate or Failure Rate

Memory-less Property

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Memory less Property

• Exponential distribution is used to model customer arrivals in

queuing systems (e.g. a bank window)

• What is the probability that a customer will arrive in next 5

minutes? P(X 5).

• Suppose that no customer arrives in the 5 minutes.

• Given this, what is the probability that a customer will arrive in

next 5 minutes? P(X 10/ X 5)

• After 10 minutes? P(X 15/ X 10)

• For exponential distribution, it is the same as P(X 5), no matter

how long you have waited.

• P(X t+ / X t) = P(X ) for any value of t

• A : X t+

• B : X t

Memory-less Property

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Memory less Property

• P(A) = P(X t+) = 1-e -(t+)

• P(B) = P(X t) = e -t

• P(A/B) = P(AB)/P(B)

• P(AB) = P(tXt+) = F(t+)-F(t)

• (1-e-(t+)) – (1-e-t ) = e-t - e-(t+)

• P(A/B) = 1 - e-

= P(X )

• Conditional Distribution

Relationship between Poisson and Exponential

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e at o s p betwee o sso a d po e t a

• Consider Arrival of customers in bank.

• Time between two successive arrivals follows Exponential Dist.

• Consider an interval of time of duration t.

• How many customers will arrive in this interval?

• Discrete Random Variable.

• Follows Poisson distribution with parameter t.

P(x, ) =

e- x

----------

x!P(x, t ) =

e- t (t) x

----------

x!

Poisson Process

Probability Distributions

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Probability Distributions

Bernoulli Trial

Binomial(n,p)

Negative

Binomial(k,p)

Geometric(p)

k = 1

Poisson()n ∞, p 0

Normal(,)

np > 5

Exponential()

large

Uniform(a,b)

?

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