probability distribution part 2 - ca sri lanka...an experiment is any process that generates an...

Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.. Chap 5-1

Probability Distribution – PART 2

An experiment is any process that generates

an outcome.

For a random experiment, the outcomes

cannot be predicted (i.e. they’re uncertain)

A random variable is a variable whose

numerical value depends upon the outcome of

a random experiment.

Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc. Chap 5-2

Random Variable


Random Variable

Random Variable

Outcomes of an experiment expressed numerically

E.g., Toss a die twice; count the number of times the

number 4 appears (0, 1 or 2 times)

E.g., Toss a coin; assign $10 to head and -$30 to a

tail

= $10

T = -$30


Introduction to Probability Distributions

Random Variable

Represents a possible numerical value from

an uncertain event

Random

Variables

Discrete

Random Variable

Continuous

Random Variable


Discrete Random Variables

Can only assume a countable number of values

Examples:

Roll a die twice

Let X be the number of times 4 comes up

(then X could be 0, 1, or 2 times)

Toss a coin 5 times.

Let X be the number of heads

(then X = 0, 1, 2, 3, 4, or 5)


Discrete Probability Distribution

List of All Possible [Xj , P(Xj) ] Pairs

Xj = Value of random variable

P(Xj) = Probability associated with value

Mutually Exclusive (Nothing in Common)

Collective Exhaustive (Nothing Left Out)

0 1 1j jP X P X


Experiment: Toss 2 Coins. Let X = # heads.

T

T

Discrete Probability Distribution

4 possible outcomes

T

T

H

H

H H

Probability Distribution

0 1 2 X

X Value Probability

0 1/4 = 0.25

1 2/4 = 0.50

2 1/4 = 0.25

0.50

0.25

Pro

bab

ilit

y


Discrete Random Variable Summary Measures

Expected Value (or mean) of a discrete

distribution (Weighted Average)

Example: Toss 2 coins,

X = # of heads,

compute the mean of X:

μ = (0 x 0.25) + (1 x 0.50) + (2 x 0.25)

= 1.0

X P(X)

0 0.25

1 0.50

2 0.25

N

1i

ii )X(PX E(X)


Variance of a discrete random variable

Standard Deviation of a discrete random variable

where:

μ = The mean (Expected value) of the discrete random variable X

Xi = the ith outcome of X

P(Xi) = Probability of the ith occurrence of X


N

1i

i

2

i

2 )P(X][Xσ

(continued)

N

1i

i

2

i

2 )P(X][Xσσ


Example: Toss 2 coins, X = # heads, compute standard deviation (recall μ = 1)


)P(X][Xσ i

2 i

0.7070.50(0.25)1)(2(0.50)1)(1(0.25)1)(0σ 222

(continued)

Possible number of heads

= 0, 1, or 2


The Binomial Distribution

Binomial

Hypergeometric

Poisson

Probability

Distributions

Discrete

Probability

Distributions


Binomial Probability Distribution

A fixed number of observations, n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse

All outcomes may be classified as one of two mutually

exclusive categories

e.g., head or tail in each toss of a coin; defective or not defective light bulb

Generally called “success” and “failure”

Probability of success is p, probability of failure is 1 – p

Constant probability for each observation

e.g., Probability of getting a tail is the same each time we toss

the coin


Possible Binomial Distribution Settings

A manufacturing plant labels items as either defective

or acceptable

A firm bidding for contracts will either get a contract

or not

A student taking a multiple choice/true-false

A marketing research firm receives survey responses

of “yes I will buy” or “no I will not”

New job applicants either accept the offer or reject it

XCX)!(nX!

n!n


P(X) = probability of X successes in n trials,

with probability of success p on each trial

X = number of ‘successes’ in sample,

(X = 0, 1, 2, ..., n)

where

n = sample size (number of trials

or observations)

p = probability of “success”

P(X) n

X ! n X p (1-p)

X n X !

( ) !

Example: Flip a coin four

times, let x = # heads:

n = 4

p = 0.5

1 - p = (1 - 0.5) = 0.5

X = 0, 1, 2, 3, 4

Binomial Distribution Formula


Example: Calculating a Binomial Probability

What is the probability of one success in five

observations if the probability of success is .1?

X = 1, n = 5, and p = 0.1

0.32805

.9)(5)(0.1)(0

0.1)(1(0.1)1)!(51!

5!

p)(1pX)!(nX!

n!p)(1pC1)P(X

4

151

XnXXnX

X

n

Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.

The Binomial Distribution, An Example Worked by Equation

• A study by the International Coffee Association

found that 52% of the U.S. population aged 10 and

over drink coffee. For a randomly selected group of

4 individuals, what is the probability that 3 of them

are coffee drinkers?

So, p = 0.52, (1 – p) = 0.48, x = 3, (n – x) = 1 .


The Binomial Distribution, Working with the Equation

• To solve the problem, we substitute:

2700.0 269967. )48(.)140608(.4

1)48(.3)52(. )!3–4(!3

!4 )3(

–)–1()!–(!

! )(

P

xnpxpxnx

nxP


Binomial Distribution Characteristics

Mean

Variance and Standard Deviation

npE(x)μ

p)-np(1σ2

p)-np(1σ

Where n = sample size

p = probability of success

(1 – p) = probability of failure


Binomial Probability Distribution

A fixed number of observations, n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse

All outcomes may be classified as one of two mutually

exclusive categories

e.g., head or tail in each toss of a coin; defective or not defective light bulb

Generally called “success” and “failure”

Probability of success is p, probability of failure is 1 – p

Constant probability for each observation

e.g., Probability of getting a tail is the same each time we toss

the coin

XCX)!(nX!

n!n


P(X) = probability of X successes in n trials,

with probability of success p on each trial

X = number of ‘successes’ in sample,

(X = 0, 1, 2, ..., n)

where

n = sample size (number of trials

or observations)

p = probability of “success”

P(X) n

X ! n X p (1-p)

X n X !

( ) !

Example: Flip a coin four

times, let x = # heads:

n = 4

p = 0.5

1 - p = (1 - 0.5) = 0.5

X = 0, 1, 2, 3, 4

Binomial Distribution Formula


The Poisson Distribution

Binomial

Hypergeometric

Poisson

Probability

Distributions

Discrete

Probability

Distributions


Important Discrete Probability

Distributions

Discrete Probability

Distributions

Binomial Hypergeometric Poisson Siméon Poisson


Poisson Distribution Formula

where:

X = number of events in an area of opportunity

= expected number of events

e = base of the natural logarithm system (2.71828...)

!X

e)X(P

x


Poisson Distribution Characteristics

Mean

Variance and Standard Deviation

λμ

λσ2

λσ

where = expected number of events



• The Poisson distribution defines the probability that an event will occur exactly x times over a given span of time, space, or distance.

where = the mean number of occurrences over the span

e = 2.71828, a constant

Mean = Variance =

Example: During the 12 p.m. – 1 p.m. noon hour, arrivals at a curbside banking machine have been found to be Poisson distributed with a mean of 1.3 persons per minute. If x = number of arrivals per minute, determine the probability that less than two persons arrive at the banking machine during the next minute:

= 1.3 P(X < 2) = P(X 1)

!

– )(

x

xexP



Apply the Poisson Distribution when:

You wish to count the number of times an event occurs in a given area of opportunity

The probability that an event occurs in one area of opportunity is the same for all areas of opportunity

The number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunity

The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smaller

The average number of events per unit is (lambda)


The Poisson Distribution,

Working with the Equation • Example: During the 12 p.m. – 1 p.m. noon hour, arrivals at a

curbside banking machine have been found to be Poisson distributed with a mean of 1.3 persons per minute. If x = number of arrivals per minute, determine P(X < 2):

P(x < 2) = 0.2725 + 0.3543 = 0.6268

2725.0 1

2725.0

!0

0)3.1(3.1–)71828.2( )0( xP

3543.0 1

3543.0

!1

1)3.1(3.1–)71828.2( )1( xP


The Sum of Two Random Variables

Expected Value of the sum of two random variables:

Variance of the sum of two random variables:

Standard deviation of the sum of two random variables:

XY

2

Y

2

X

2

YX σ2σσσY)Var(X

)Y(E)X(EY)E(X

2

YXYX σσ


Portfolio Expected Return and Portfolio Risk

Portfolio expected return (weighted average

return):

Portfolio risk (weighted variability)

Where w = portion of portfolio value in asset X

(1 - w) = portion of portfolio value in asset Y

)Y(E)w1()X(EwE(P)

XY

2

Y

22

X

2

P w)σ-2w(1σ)w1(σwσ


Portfolio Example

Investment X: μX = 50 σX = 43.30

Investment Y: μY = 95 σY = 193.21

σXY = 8250

Suppose 40% of the portfolio is in Investment X and 60% is in Investment Y:

The portfolio return and portfolio variability are between the values for investments X and Y considered individually

77(95)(0.6)(50)0.4E(P)

133.30

)(8250)2(0.4)(0.6(193.71)(0.6)(43.30)(0.4)σ 2222

P

probability distribution part 2 - ca sri lanka...an experiment is any process that generates an...

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