Discrete Probabilities4-Discrete Probabilities - Student

8/10/2019 Discrete Probabilities4-Discrete Probabilities - Student

1/75

DSC1007 Lecture 4

Discrete Probabilities


2/75

Random Variables

Discrete Probability Distributions

Summary Measures of Probability Distributions

Binomial Distribution

Poisson Distribution

Linear Functions of a Random Variable

Sums of Random Variables

Covariance and correlation

Joint probability distributions and independence

Discrete Probability


3/75

Random Variables

A random variable (r.v.) may be discreteor continuous.

A random variable is a rule that assigns a numerical value

to each possible outcome of a probabilistic experiment.

Discrete : can only assume values that are distinct and separate

Continuous: can take on any value within some interval of numbers

Example 1 : 0, 1, 2, 3, 4, 5, . . .

Examples : [ 0 , 100 ], [ 2 , 4 ]

Example 2 : 2.0, 2.5, 3.0, 3.5


4/75

Random Variables

A random variable (r.v.) may be discreteor continuous.

Examples

1.Total number of points in a throw of 2 dice

2.Number of cups of cappuccino Edward will sell today

3.Number of Bs you will score in this semester

4.Time between this and the next slide

5.Your present weight in kg




5/75

Random Variables

Example :

Lets say that 30 %of BBA students are internationalstudents. Suppose that each of the 3 BIZ1007 tutors randomly

selects one of his students to participate in a survey.

What are the possible outcomes of this experiment?

Outcome X

L L L 0

L L I 1

L I L 1

I L L 1

L I I 2

I L I 2

I I L 2

I I I 3

L : LocalI : International


6/75

Random Variables

Example :



Let X = number of international students chosen

Outcome X

L L L 0

L L I 1

L I L 1

I L L 1

L I I 2

I L I 2

I I L 2

I I I 3



7/75

Probability Distribution

A probability distribution for a random variable

describes how probabilities are distributed over the

values of the random variable



Recall


8/75

Discrete Probability Distribution

A probability distribution for a discrete random variable X

consists of

(i) possible values x1, x2, . . . , xn

(ii) corresponding probabilities p1,p2, . . . , pn

with the interpretation that

P(X= x1) =p1, P(X= x2) =p2, . . . , P(X= xn) =pn

Note:

Probabilities must sum to 1 : p1+p2+ . . . +pn= 1.0 ( pi 0 )


9/75


A probability distribution for a discrete random variable X

consists of

(i) possible values x1, x2, . . . , xn

(ii) corresponding probabilities p1,p2, ... , pn

with the interpretation that

P(X= x1) =p1, P(X= x2) =p2, . . . , P(X= xn) =pn


10/75

Example :




Outcome X

L L L 0

L L I 1

L I L 1

I L L 1

L I I 2

I L I 2

I I L 2

I I I 3


Q: What is its probability

distribution?

We know Xis a r.v.


Return to this example.


11/75

Example :




Outcome X

L L L 0

L L I 1

L I L 1

I L L 1

L I I 2

I L I 2

I I L 2

I I I 3

Probability distribution

X P(X)

0 P(X=0)

1 P(X=1)

2 P(X=2)

3 P(X=3)



12/75

Example :

Lets say that 30 %of BBA students are internationalstudents.Suppose that each of the 3 BIZ1007 tutors randomly selects one

of his students to participate in a survey.



Outcome X Probability of Outcome

L L L 0

L L I 1

L I L 1

I L L 1L I I 2

I L I 2

I I L 2

I I I 3


13/75


Example :




X P(X)

0 P(X=0) =

1 P(X=1) =

2 P(X=2) =

3 P(X=3) =


14/75


Example :




X P(X)

0

1

2

3

Probability Distribution of X


15/75

A histogramis a display of probabilities as a bar chart

Example 1:

X = number of international students chosen

0.00

0.10

0.20

0.30

0.40

0.50

0 1 2 3




16/75

Example 2:

Let Xbe the random variable that denotes the number of ordersfor Boeing 767 aircraft for next year.

Suppose that the number of orders for Boeing 767 aircraft fornext year is estimated to obey the following distribution:

Orders for Boeing 767Aircraft next year

xi

Probability

pi

42

43

44

45

46

47

48

0.05

0.10

0.15

0.20

0.25

0.15

0.10



17/75

Probability Distribution of the

Number of Orders of Boeing 767 Aircraft Next Year

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

43 44 45 46 47 48

Number of Orders

Probability

Page 9




18/75

EasternDivision

Xand Ydenote the sales next year in the easterndivision and

the westerndivision of a company, respectively.

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

3.0 4.0 5.0 6.0 7.0 8.0

Probability

Sales ($ million)

Probability Distribution Function of WesternDivision Sales

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

3.0 4.0 5.0 6.0 7.0 8.0

Probab

ility

Sales ($ million)

Probability Distribution Function of EasternDivision Sales

Sales($ million) Probability

3.0 0.15

4.0 0.20

5.0 0.25

6.0 0.15

7.0 0.15

8.0 0.10


3.0 0.05

4.0 0.20

5.0 0.35

6.0 0.30

7.0 0.10

8.0 0.00


WesternDivision


19/75


EasternDivision

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

3.0 4.0 5.0 6.0 7.0 8.0

Probability

Sales ($ million)


0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

3.0 4.0 5.0 6.0 7.0 8.0

Probability

Sales ($ million)



3.0 0.15

4.0 0.20

5.0 0.25

6.0 0.15

7.0 0.15

8.0 0.10


3.0 0.05

4.0 0.20

5.0 0.35

6.0 0.30

7.0 0.10

8.0 0.00

WesternDivision

Which division is going to perform better?


20/75

(Y = y)

0.00

0.10

0.20

0.30

0.40

0.50

0 1 2 3 4 5 6

x

(X =x)

0.00

0.10

0.20

0.30

0.40

0.50

0 1 2 3 4 5 6

Consider two r.v.s X, Y, with the following histograms:

Random variable X Random variable Y

Q: How to describe and compare Xand Y ?



21/75

Summary statistics are used to summarize a set ofobservations, in order to communicate the largest amount assimply as possible.

Statisticians commonly try to describe the observations in

a measure of location, or central tendency, such as the mean,median, modeetc

a measure of statistical dispersionlike the standard deviation,variance, rangeetc

a measure of the shapeof the distribution like skewnessorkurtosis



22/75

The mean (or expected value) of Xis

= =

=1


Suppose the discrete r.v.Xhas probability distribution

x1, x2, . . . , xn

p1, p2, . . . , pn


23/75


24/75

Example:

LetX

be the number that comes up on a roll of a die.Compute the mean,varianceand standard deviationofX.

Outcome P(X=x)

1 1/6

2 1/63 1/6

4 1/6

5 1/6

6 1/6



25/75

Example :

Let

s say that 30 %of BBA students are internationalstudents.Suppose that each of the 3 BIZ1007 tutors randomly selects oneof his students to participate in a survey.


Compute the mean,varianceand standard deviationof X.

X P(X)

0

1

2

3

= =

=1



26/75

Example :

Let

s say that 30 %of BBA students are internationalstudents.Suppose that each of the 3 BIZ1007 tutors randomly selects oneof his students to participate in a survey.



X P(X)

0

1

2

3

= =

= 2 = 2

=1

=



27/75

EasternDivision

Xand Ydenote the sales next year in the easterndivision and

the westerndivision of a company, respectively.

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

3.0 4.0 5.0 6.0 7.0 8.0

Probability

Sales ($ million)


0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

3.0 4.0 5.0 6.0 7.0 8.0

Probab

ility

Sales ($ million)



3.0 0.15

4.0 0.20

5.0 0.25

6.0 0.15

7.0 0.15

8.0 0.10


3.0 0.05

4.0 0.20

5.0 0.35

6.0 0.30

7.0 0.10

8.0 0.00

WesternDivision


Compute the mean,varianceand standard deviationof

X (eastern division sales) and Y(western division sales)


28/75

X (Eastern Division) Y (Western Division)

Sales ($m) Probability Sales ($m) Probability

3.0 0.05 3.0 0.15

4.0 0.20 4.0 0.20

5.0 0.35 5.0 0.25

6.0 0.30 6.0 0.15

7.0 0.10 7.0 0.15

8.0 0.00 8.0 0.10

Mean 5.20 5.25

Variance 1.06 2.3875

Std Dev 1.0296 1.5452


Compute the mean,varianceand standard deviationof

X (eastern division sales) and Y(western division sales)

Use Excel


29/75

Let X= number of pieces of chicken in an order.

Develop a probability distribution for X.

pieces in order orders2 170

3 200

4 260

8 165

12 120

16 5020 35

1,000


xi Probability

2 0.170

3 0.200

4 0.260

8 0.165

12 0.120

16 0.050

20 0.035

Where do probability distributions come from?

I. Empirically(from data)

Example: KFC sells chicken in bucketsof 2, 3, 4, 8, 12, 16 or 20 pieces.Over the last week, orders for fried chicken had the following data:

l b


30/75

Consider experiment consisting of nindependenttrials

Each trial has exactly two outcomes : successor failure


We say thatXis a binomialr.v. drawn from a sample sizen

and with probability of successp.

Where do probability distributions come from?

II.Theoretically

The Binomial Distribution

Each trial has same probability: successp, failure 1p

Let

X = number of successes in ntrials.

l b


31/75

We also say thatXobeys a binomial distribution with

parameters nandp: Binomial(n,p) or B(n,p)



= = !!! (1 ) for = 0, 1, . . . ,

Consider experiment consisting of nindependent trials

Each trial has exactly two outcomes : successor failureEach trial has sameprobability: successp, failure 1p

Let

X = number of successes in ntrials.

Bi i l Di ib i


32/75


IfXobeys a binomial distribution with parameters nand

p, then the mean, variance and standard deviation ofX

are:

Mean = =

Variance = 2

= (1 )Std deviation = (1 )

Expected Value and Variance

Bi i l Di ib i


33/75

Example :Lets say that 30 %of BBA students are internationalstudents.

Suppose that each of the 3 BIZ1007 tutors randomly selects oneof his students to participate in a survey.


Outcome X

L L L 0

L L I 1

L I L 1

I L L 1L I I 2

I L I 2

I I L 2

I I I 3

Note :Xis a binomial variable!

with :

n = 3 trials

Successselect international student

p= P(success) = 0.30

Recall Example


Bi i l Di ib i


34/75

Example :Lets say that 30 %of BBA students are internationalstudents.

Suppose that each of the 3 BIZ1007 tutors randomly selects oneof his students to participate in a survey.


Xobeys a binomial distribution with parameters n= 3

and p = 0.3: Binomial(3,0.3)

The probability distribution ofXis given by:

= = ! !! (.) (.) for = 0, 1, . . . , 3


Bi i l Di ib i


35/75

Xobeys binomial distribution with n = 3 and p = 0.3

The probability distribution of Xis given by:

= = ! !! (.) (.) for = 0, 1, . . . , 3

= 0 = !0!0! (.)0(.)0 = (.) = = 1 = !

1!1! (.)1(.)1 = (.)(.) =

Compare above with

probability distribution

we obtained earlier


Bi i l Di t ib ti


36/75

Example :Lets say that 30 %of BBA students are internationalstudents.Suppose that each of the 3 BIZ1007 tutors randomly selects oneof his students to participate in a survey.



Mean = = = 30.3 = .

Variance = 2 = 1 = 30.30.7 = .

Xobeys a binomial distribution with n= 3 and p = 0.3


Bi i l Di t ib ti


37/75

=

=

!!!

(1

) for x= 0, 1, . . . , n


EXCEL Function : BINOMDIST (x, n , p , cumulative)

cumulative = 0 ( or FALSE) P( X= x )

1 ( or TRUE) P ( X x )

Bi i l Di t ib ti


38/75

P (X= 15) = BINOMDIST (15, 15, 0.75, 0)

P (X14) = 1 P(X13)

= 1 BINOMDIST (13, 15, 0.75, 1)

Example Summary :

number of lasers (out of 15) that will pass the test

X Binomial (15, 0.75) P(X = 15) = 0.013363

P(X 14) = 0.0802


EXCEL Function : BINOMDIST (x, n , p , cumulative)


39/75

Case 1 - Application of Binomial

Distribution

An investment broker at the Michaels &

Dodson Company claims that he has

found a real winner out of 400 funds. He has tracked a mutual fund that has

beaten a standard market index in 37 out

of the past 52 weeks. Will you invest in the winnerfund?


40/75

Question 1

We say a fund beats the market purely by

chance if each week the fund has a fifty-

fifty chance of beating the market index,

independently of its performance in otherweeks.

What is the probability for such fund to

beat the market 37 out of 52 weeks?


41/75

Question 2

Suppose that all the 400 funds beat

market purely by chance. What is the

probability that the best of them beats the

market 37 out of 52 weeks?

Conclusion?



42/75

Useful for modelling the number of occurrences of an event over a

specified interval of time or space.

Examples :

number of customer orders received in one hour

number of failures in a large computer system per month

Properties

Probability of an occurrence is the same for any two intervals of

equal length.

Occurrences in nonoverlapping intervals are independent of one

another.




43/75


was derived by the

French Mathematician

Simon Poisson in1837.

His name is one of the

72 names inscribed onthe Eiffel Tower.


http://en.wikipedia.org/wiki/File:Simeon_Poisson.jpg


44/75

Examples and Applications:

The number of soldiers killed by horse-kicks

each year in each corps in the Prussian

cavalry.

The number of phone calls arriving at a callcentre per minute.

The number of goals in sports involving two

competing teams.

The number of infant death per year.




45/75

= = ! for = 0, 1, 2, . . .

= 0 =0

0! =

= 1 = 11!

=


Xis a discrete r.v. that takes on values 0, 1, 2, . . .

Note:

A random variable X is said to be a Poissonr.v. with parameter

(> 0) if it has the probability function



46/75


= = ! for = 0, 1, 2, . . .

It can be shown that

Mean E(X) =

Variance Var(X) =

Thus, parameter

can be interpreted as

the average number

of occurrences perunit time or space

A random variable X is said to be a Poissonr.v. with parameter

(> 0) if it has the probability function



47/75

= = ! for = 0, 1, 2, . . .


X is said to be a Poissonr.v. with parameter (> 0) if

ExamplePatients arrive at the A & E of a hospital at the average rate of 6 per hour

on weekend evenings. What is the probability of 4 arrivals in 30 minuteson a weekend evening?

Can expect patient arrivals to be approximately Poisson.

Average arrival rate is 6 / hour.

Let X be the number of patient arrivals in 30 minutes

X is Poissonwith parameter = 3



48/75


Excel Function : POISSON (x, , cumulative)

=

=

! for

= 0, 1, 2, . . .

cumulative = 0 ( or FALSE) P( X= x )

1 ( or TRUE) P ( X x )



49/75


P (X= 4) = POISSON (4, 3, 0)

Example Summary :

Let X be the number of patient arrivals in 30 minutes

X is Poisson with parameter = 3

P(X = 4) = 0.1680

Excel Function : POISSON (x, , cumulative)


50/75

Managing TV Inventory

Kriegland is a department store that sells

various brands of flat-screen TVs. One of

the managers biggest problems is to

decide on an appropriate inventory policyfor stocking TVs.


51/75

Discussion 1

Why is it important to decide a right

inventory level?

What are the factors to consider when

deciding the inventory level?


52/75

Discussion 2

The manager knows that the historical

average demand per month is

approximately 17.

If the manager attempts to find the

probability distribution of demand in a

typical month. How might he proceed?



53/75

Example

Suppose daily demand for croissants at a bakery shop is given by

Suppose it costs $135 per day to run the croissant operation, and

that the cost of producing one croissant is $0.75.

Daily cost of croissant operations = 0.75 X+ 135


Daily Demand Probability

60 0.05

64 0.15

68 0.20

72 0.25

75 0.15

77 0.10

80 0.10

Let X = daily demand for croissants

We can easily compute

E(X) = 71.15

Var(X) = 29.5275



54/75

Example

E(Y) = 0.75 E(X) + 135

Var(Y) = 0.752 Var(X)


E(X) = 71.15

Var(X) = 29.5275

X Probability Y= .75X+ 135

60 0.05 180.00

64 0.15 183.00

68 0.20 186.00

72 0.25 189.00

75 0.15 191.25

77 0.10 192.75

80 0.10 195.00

Y = 0.75 X + 135

E(Y) = ?

Var(Y) = ?

How are the means and variances related ?



55/75

If Y = aX + b

E(Y) = aE(X) + b

Var(Y) = a2 Var(X)


Note:

Formulas apply

to continuousr.v.s as well

Covariance and Correlation


56/75


How do we summarize the relationship between two variables?

Specifically : how do we summarize what we observe in a scatter

plot?

Examples:

Unemployment rate vs Crime rate

Stock market vs Property market

Time spent on DSC1007 vs DSC1007 Exam marks



57/75

Example: Chain of upscale cafs sells gourmet hotcoffees and coldbeverages.From past sales data, daily sales at one of their caf obey the followingprobability distribution for (X, Y),

X = # hotcoffees, Y = # coldbeverages sold per day

Q : How do we describe the relationship between two rvs?


Probability No. ofHotCoffees Sold No. of ColdDrinks Soldpi xi yi

0.10 360 360

0.10 790 110

0.15 840 30

0.05 260 90

0.15 190 450

0.10 300 230

0.10 490 60

0.10 150 290

0.10 550 140

0.05 510 290

Mean X = 457.00 Y = 210.00

Standard Deviation 244.28 145.64



58/75

Comment : It seems that smallersales of hotcoffees are

often accompanied by largersales of cold

beverages.

0

100

200

300

400

500

0 200 400 600 800 1000

ColdBeverage

Sales

Hot Coffee Sales

Scatter Plot of Daily Sales forHot Coffees and Cold Beverages




59/75

0

100

200

300

400

500

0 200 400 600 800 1000

ColdBeverage

Sales

Hot Coffee Sales

Scatter Plot of Daily Sales forHot Coffees and Cold Beverages

Comment : Hotcoffee sales greater than the average number

sold per day are typically accompanied by coldbeverage sales

that are smaller than the average sold per day.




60/75

Covariance, = = = , =


We now define the covarianceof two random variablesXand

Ywith means X

and Y:

Probability X Y

P(X=x1, Y=y1 ) x1 y1

P(X= x2, Y=y2 ) x2 y2

P(X= xN, Y=yN ) xN yN



61/75

Covariance

Observe from the above that:

(,) = 2 =


, = = = , =

, = ()=



62/75



Probability No. ofHotCoffees Sold No. of ColdDrinks Sold

pi xi yi

0.10 360 360

0.10 790 110

0.15 840 30

0.05 260 90

0.15 190 450

0.10 300 230

0.10 490 60

0.10 150 290

0.10 550 140

0.05 510 290

Mean X = 457.00 Y = 210.00




63/75



Cov(X,Y) = 0.10 (360457)(360210) + 0.10 (790457)(110210)

+ . . . + 0.05 (510

457)(290

210) = 27,260


pi xi yi

0.10 360 457 360 210

0.10 790 457 110 210

0.15 840 457 30 210

0.05 260 457 90 210

0.15 190 457 450 210

0.10 300 457 230 210

0.10 490 457 60 210

0.10 150 457 290 210

0.10 550 457 140 210

0.05 510 457 290 210

Mean X = 457.00 Y = 210.00




64/75

Correlation

Comments :

The measure of correlation is unit-free.

Corr (X, Y) is always between 1.0 and 1.0

, =(,)


We introduce a standardized measure of interdependencebetween two rvs :

Covariance

, = = = , =



65/75

Correlation , = (,)


Corr(X

,Y

) = 1.0= 0

= 1.0

perfect positive linear relationship

no linear relationship between Xand Y

perfect negative linear relationship



66/75

If higherthan average values ofXare apt to occur with higherthan average values of Y, then Cov(X, Y) > 0 and Corr(X, Y) > 0.

X and Yarepositivelycorrelated.


If higherthan average values ofXare apt to occur with lowerthan average values of Y, then Cov(X, Y) < 0 and Corr(X, Y) < 0.

X and Yare negatively correlated.



67/75

Common fallacy

Aoccurs in correlation with B

Therefore : Acauses B

Correlation is notthe same as Causality!

Joint Probability Distributions


68/75

J y

Consider two random variablesXand Y that assume values given by

Probability X Y

p1 P(X=x1, Y=y1 ) x1 y1

p2 P(X= x2, Y=y2 ) x2 y2

pN P(X= xN, Y=yN ) xN yN

Denote by f(xi,yi)

f is called the joint probability distribution function of (X,Y)

Joint Probability Distributions


69/75

J y

Two random variablesXand Y are said to be independentif

P(X=x,Y=y) = P(X=x) P(Y=y)

Thus, ifXand Y are independent, then

E(XY) = E(X)E(Y)

Roughly : Xand Yare independent if knowing the value of one

does not change the distribution of the other.

The concept of independenteventsleads quite naturally to a similar

definition for independentrandom variables.

It follows that ifXand Y are independent, then

Cov(X,Y) = 0 ( or Corr(X,Y)= 0 )



70/75

But : dependentrandom variables may also beuncorrelated!!

Example : Consider r.v.s Xand Ywith the following joint probability distribution

We know : independentrandom variables are always uncorrelated.

P(X ,Y) X Y

1/3 1 11/3 0 1

1/3 1 1

Check : are XandY independent?

Check : are XandY uncorrelated?

Sum of Random Variables


71/75

Mean

E( a X + b Y) = aE( X ) + bE( Y)

Variance

Var(aX+ bY) = a2Var(X) + b2Var(Y) + 2abCov(X,Y)

or:

Var(aX+ bY) = a2Var(X) + b2Var(Y) + 2abXY Corr(X,Y)

Note: Formulas apply to continuousr.v.s as well



72/75

If X and Yare independent

Cov(X,Y)= 0

Variance

Var(aX+ bY) = a2Var(X) + b2Var(Y) + 2abCov(X,Y)



73/75

Example: Chain of upscale cafs sells gourmet hotcoffees and coldbeverages.

From past sales data, daily sales at one of their caf obey the following

probability distribution for (X, Y),X = # hotcoffees, Y = # coldbeverages sold per day


pi xi yi

0.10 360 360

0.10 790 1100.15 840 30

0.05 260 90

0.15 190 450

0.10 300 230

0.10 490 60

0.10 150 290

0.10 550 140

0.05 510 290

Mean X = 457.00 Y = 210.00


Suppose : coldbeverages (Y) are $2.50/glass;

hotcoffees (X) $1.50/cup.



74/75

coldbeverages (Y) are $2.50/glass;


mean and standard deviation of total daily sales of allbeverages

mean and standard deviation of daily sales of coldbeverages

mean and standard deviation of daily sales of hotcoffees

E(2.5Y) = ? SD(2.5Y) = ?

E(1.5X) = ? SD(1.5X) = ?

E(1.5X+ 2.5Y) = ? SD(1.5X+ 2.5Y) = ?

Determine the following:

2.5 E(Y) 2.5 SD(Y) = 2.5Var (Y)

1.5 E(X) 1.5 SD(X) = 1.5Var(X)

E(X) = 457, Var(X) = 59,671

E(Y) = 210, Var (Y) = 21,210

Cov (X,Y) = 27,260



75/75

mean and standard deviation of total daily sales of allbeverages

mean and standard deviation of daily sales of coldbeverages

mean and standard deviation of daily sales of hotcoffees

E(2.5Y) = ? SD(2.5Y) = ?

E(1.5X) = ? SD(1.5X) = ?

E(1.5X+ 2.5Y) = ? SD(1.5X+ 2.5Y) = ?

Determine the following:

2.5 E(Y)

1.5 E(X)

Var (1.5X+ 2.5Y)

E(X) = 457, Var(X) = 59,671

E(Y) = 210, Var (Y) = 21,210

Cov (X,Y) = 27,260

2.5 SD(Y) = 2.5Var (Y)

1.5 SD(X) = 1.5Var(X)

coldbeverages (Y) are $2.50/glass;


Date post:	02-Jun-2018
Category:	Documents
Upload:	sherrysalvato130
View:	245 times
Download:	0 times

Discrete Probabilities4-Discrete Probabilities - Student

Documents