Probability Distribution

Chapter 5 5-

Created by: Prabhat MittalE-mail Id: [email protected]

Statistics for Business Analysis

Day 4Session- I

PROBABILITY DISTRIBUTIONS

Learning Objectives

� The properties of a probability distribution

� To calculate the expected value and variance of a probability distribution

� To calculate the covariance and its use in finance

� To calculate probabilities from binomial and Poisson distributions

� How to use the binomial and Poisson distributions to solve business problems

Chapter 5 5-


Introduction to Probability Distributions

� Frequency Distribution

� Listing of the observed frequencies of all the outcomes of an experiment that actually occurred when the experiment was done.

Defining a Random Variable� A value is random if it takes different values as a

result of the outcomes of a random experiment � Or it represents a possible numerical value from an

uncertain event.

� Probability Distribution� Listing of the probabilities of all the possible

outcomes that could result if the experiment were done.

Introduction to Probability Distributions

� Table 1 illustrates the number of cars sold per day during the last 20 days. The table gives a Frequency Distribution. Number of cars sold daily during 20 days.

� We can use this historical record to assign a probability to each possible number of cars and find a Probability distribution (Table-2). This has been accomplished by normalizing the observed frequency distribution.

4105

3104

5103

7101

1100

Frequency of Occurance

Number of Cars sold per day (X)

1.00Total

4/20= .20105

3/20 =.15104

5/20 =.25103

7/20 =.35101

1/20 =.05100

Probability that the random variable will take on this value

Value of the random variable (X)

Chapter 5 5-


Introduction to Probability Distributions� Discrete Random Variable

If a random variable is allowed to take on only a limited number of values, which can be listed, it is a discrete random

variable.

Random

Variables

Discrete

Random Variable

Continuous

Random Variable

� Continuous Random Variable

If it allowed to assume any value within a given range, it is a continuous random variable.

Discrete Random Variables

� Can only assume a countable number of values

Examples:

� Roll a die twiceLet 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)

Chapter 5 5-


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) we multiply each value that the random variable can assume by the probability of occurrence of that value and sum these products.

� Example: Toss 2 coins, X = # of heads,

compute expected value of X:

E(X) = (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

∑=

==µN

1iii )X(PX E(X)

Chapter 5 5-


� Variance of a discrete random variable

� Standard Deviation of a discrete random variable

where:E(X) = Expected value of the discrete random variable X

Xi = the ith outcome of XP(Xi) = Probability of the ith occurrence of X


∑=

−=N

1ii

2i

2 )P(XE(X)][Xσ

(continued)

∑=

−==N

1ii

2i

2 )P(XE(X)][Xσσ

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


)P(XE(X)][Xσ i2

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

Chapter 5 5-


The Covariance

� The covariance measures the strength of the linear relationship between two variables

� The covariance:

)YX(P)]Y(EY)][(X(EX[σN

1iiiiiXY ∑

=

−−=

where: X = discrete variable XXi = the ith outcome of XY = discrete variable YYi = the ith outcome of YP(XiYi) = probability of occurrence of the

ith outcome of X and the ith outcome of Y

Numerical Problems

Ref. # 5-9 Page No.230: The only information available to you regarding the probability distribution of a set of outcomes is the following list of frequencies:

X 0 15 30 45 60 75

Frequency 25 125 75 175 75 25

Construct a probability distribution for the set of outcomes.

a. Find the expected value of an outcome.

b. Compute the variance and standard deviation for the distribution.

Observati

on (X) Frequency

Probability

P(x) X.P(X)

Deviation

(x-mean)

Deviation

Squared*P(X)

0 25 0.05 0.00 -36.75 67.53

15 125 0.25 3.75 -21.75 118.27

30 75 0.15 4.50 -6.75 6.83

45 175 0.35 15.75 8.25 23.82

60 75 0.15 9.00 23.25 81.08

75 25 0.05 3.75 38.25 73.15

500 1.00 36.75 370.69

Expected value Variance

Chapter 5 5-


Computing the Mean for Investment Returns

Return per $1,000 for two types of investments

P(XiYi) Economic condition Passive Fund X Aggressive Fund Y

0.2 Recession - $ 25 - $200

0.5 Stable Economy + 50 + 60

0.3 Expanding Economy + 100 + 350

Investment

E(X) = µX = (-25)(0.2) +(50)(0.5) + (100)(0.3) = 50

E(Y) = µY = (-200)(0.2) +(60)(0.5) + (350)(0.3) = 95

Computing the Standard Deviation for Investment Returns


0.2 Recession - $ 25 - $200



Investment

43.30

(0.3)50)(100(0.5)50)(50(0.2)50)(-25σ 222X

=

−+−+−=

193.71

(0.3)95)(350(0.5)95)(60(0.2)95)(-200σ 222Y

=

−+−+−=

Chapter 5 5-


Computing the Covariance for Investment Returns


0.2 Recession - $ 25 - $200



Investment

8250

95)(0.3)50)(350(100

95)(0.5)50)(60(5095)(0.2)200-50)((-25σXY

=

−−+

−−+−−=

Interpreting the Results for Investment Returns

� The aggressive fund has a higher expected return, but much more risk

µY = 95 > µX = 50

but

σY = 193.71 > σX = 43.30

� The Covariance of 8250 indicates that the two investments are positively related and will vary in the same direction

Chapter 5 5-


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:

XY2Y

2X

2YX σ2σσσY)Var(X ++==+ +

)Y(E)X(EY)E(X +=+

2YXYX 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) −+=

XY2Y

22X

2P w)σ-2w(1σ)w1(σwσ +−+=

Chapter 5 5-


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 2222P

=

++=

Statistics for Business Analysis

Day 4Session- II

PROBABILITY DISTRIBUTIONS

Chapter 5 5-


Probability Distributions

ContinuousProbability

Distributions

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions

Normal

Uniform

Exponential

The Binomial Distribution

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions

Chapter 5 5-


Binomial Probability Distribution

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

� Two mutually exclusive and collectively exhaustive

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

� Observations are independent� The outcome of one observation does not affect the outcome of

the other

Applications of Binomial Distribution

� A manufacturing plant labels items as either defective or acceptable

� A firm bidding for contracts will either get a contract or not

� 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

Chapter 5 5-


Rule of Combinations

� The number of combinations of selecting X objects out of n objects is

X)!(nX!

n!Cxn

−=

where:n! =(n)(n - 1)(n - 2) . . . (2)(1)

X! = (X)(X - 1)(X - 2) . . . (2)(1)

0! = 1 (by definition)

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)

n = sample size (number of trials or observations)

p = probability of “success”

P(X)n

X ! n Xp (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

Define a random variable

X ~ BIN (n,p)

Chapter 5 5-


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!1)P(X

4

151

XnX

=

=

−−

=

−−

==

−

−

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

Chapter 5 5-


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

Binomial Distribution

� The shape of the binomial distribution depends on the values of p and n

� Here, n = 5 and p = 0.1

� Here, n = 5 and p = 0.5

n = 5 p = 0.1

n = 5 p = 0.5

Mean

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

0.5(5)(0.1)npµ ===

0.6708

0.1)(5)(0.1)(1p)np(1-σ

=

−==

2.5(5)(0.5)npµ ===

1.118

0.5)(5)(0.5)(1p)np(1-σ

=

−==

Binomial Characteristics

Examples

Chapter 5 5-


Using Binomial Tables

xp=.50p=.55p=.60p=.65p=.70p=.75p=.80…

10

9

8

7

6

5

4

3

2

1

0

0.0010

0.0098

0.0439

0.1172

0.2051

0.2461

0.2051

0.1172

0.0439

0.0098

0.0010

0.0025

0.0207

0.0763

0.1665

0.2384

0.2340

0.1596

0.0746

0.0229

0.0042

0.0003

0.0060

0.0403

0.1209

0.2150

0.2508

0.2007

0.1115

0.0425

0.0106

0.0016

0.0001

0.0135

0.0725

0.1757

0.2522

0.2377

0.1536

0.0689

0.0212

0.0043

0.0005

0.0000

0.0282

0.1211

0.2335

0.2668

0.2001

0.1029

0.0368

0.0090

0.0014

0.0001

0.0000

0.0563

0.1877

0.2816

0.2503

0.1460

0.0584

0.0162

0.0031

0.0004

0.0000

0.0000

0.1074

0.2684

0.3020

0.2013

0.0881

0.0264

0.0055

0.0008

0.0001

0.0000

0.0000

…

…

…

…

…

…

…

…

…

…

…

0

1

2

3

4

5

6

7

8

9

10

p=.50p=.45p=.40p=.35p=.30p=.25p=.20…x

n = 10

Examples:

n = 10, p = 0.35, x = 3: P(x = 3|n =10, p = 0.35) = 0.2522

n = 10, p = 0.75, x = 2: P(x = 2|n =10, p = 0.75) = 0.0004

Some more facts about Binomial

When n is constant, notice the changes due to changes in p, probability of success.1. when p is small (0.1), the binomial distribution is skewed to the right.2. As p increases (to 0.3), the skewness is less noticeable.3. When p = 0.5, the binomial dist. Is symmetrical.4. When p is larger than 0.5, the distribution is skewed to the left.5. When p =0.7, the probabilities are the same as for p=0.3 except that the values are reversed.

Let us examine when p stays constant but n is increased. 1. As n increases, the vertical lines not only become more numerous but also tend to bunch up together to form a bell shape.

Note:

Look at the Excel worksheet

Chapter 5 5-


Numerical Problems

Ref. # 5-18 Page No.247: For a Binomial distribution with n = 7 and p = 0.2, find

a. P(X=5)

b. P(X>2)

c. P(X<8)

d. P(X≥4)

Ans.

• .0043

• .1480

• 1

• .0333

Numerical Problems

Ref. # 5-22 Page No.248: Harley Davidson, director of quality control for the Kyoto Motor company is conducting his monthly spot check of automatic transmissions. In that procedure, 10 transmissions areremoved from the pool of components and are checked for manufacturing defects. Historically, only 2% of the transmissions have such flaws. (Assume that flaws occur independently in different transmissions)

a. What is the probability that Harley’s sample contains more than two transmissions with manufacturing flaws?

b. What is the probability that none of the selected transmissions has manufacturing flaws.

Ans.

a. .0009

b. .8171

Chapter 5 5-


Numerical Problems

Ref. # 5-25 Page No.248: A recent study of how Indians spend their leisure time surveyed workers employed more than 5 years. They determined the probability an employee has 2 weeks of vacation time to be 0.45, 1 week of vacation time to be 0.10, and 3 or more weeks to be 0.20. Suppose 20 workers are selected at random. what is the probability that;

a. 8 have 2 weeks of vacation time?b. Only one worker has 1 week of vacation time?c. At most 2 of the workers have 3 or more weeks of vacation time?d. At least 2 workers have 1 week of vacation time?

Ans. a. 0.1623b. 0.2702c. 0.2061d. 0.6083

Numerical Problems

Ref. # 5-26 Page No.248: Harry Ohme is in charge of the electronics section of a large department store. He has noticed that the probability that a customer who is just browsing will buy something is 0.3. Suppose that 15 customers browse in the electronics sectioneach hour. What is the probability that;

a. At least one browsing customer will buy something during a specified hour?

b. At least four browsing customers will buy something during a specified hour?

c. No browsing customers will buy anything during a specified hour?d. No more than four browsing customers will buy something during a

specified hour? Ans.

a. 0.9953b. 0.7031c. 0.0047d. 0.5155

Chapter 5 5-


The Poisson Distribution

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions

The Poisson Distribution

� Apply the Poisson Distribution when:

� You wish to count the number of times an event occurs in a givenarea of opportunity

� Distribution of telephone calls going through a switchboard system.

� The demand (needs) of patients for service at a health institution.

� The arrivals of trucks and cars at a tollbooth.

The above examples all have common element that they can be described by a discrete random variable that takes on integer values 0,1,2,3,4…

� 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 average number of events per unit is λλλλ (lambda)

Chapter 5 5-


Poisson Distribution Formula

where:

X = number of events in an area of opportunity

λ = expected number of events

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

!X

e)X(P

xλ=

λ−

Poisson Distribution Characteristics

� Mean

� Variance and Standard Deviation

λµ =

λσ2 =

λσ =

where λ = expected number of events

Chapter 5 5-


Using Poisson Tables

λλλλ

0.90

0.4066

0.3659

0.1647

0.0494

0.0111

0.0020

0.0003

0.0000

0.4493

0.3595

0.1438

0.0383

0.0077

0.0012

0.0002

0.0000

0.4966

0.3476

0.1217

0.0284

0.0050

0.0007

0.0001

0.0000

0.5488

0.3293

0.0988

0.0198

0.0030

0.0004

0.0000

0.0000

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000

0.6703

0.2681

0.0536

0.0072

0.0007

0.0001

0.0000

0.0000

0.7408

0.2222

0.0333

0.0033

0.0003

0.0000

0.0000

0.0000

0.8187

0.1637

0.0164

0.0011

0.0001

0.0000

0.0000

0.0000

0.9048

0.0905

0.0045

0.0002

0.0000

0.0000

0.0000

0.0000

0

1

2

3

4

5

6

7

0.800.700.600.500.400.300.200.10X

Example: Find P(X = 2) if λ = 0.50

0.07582!

(0.50)e

X!

e2)P(X

20.50Xλ

====−−

λ

Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

0

1

2

3

4

5

6

7

X

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000

λλλλ =

0.50

P(X = 2) = 0.0758

Graphically:

λλλλ = 0.50

Chapter 5 5-


Poisson Distribution Shape

� The shape of the Poisson Distribution depends on the parameter λ :

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x

)0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

λ = 0.50 λ = 3.00

Numerical Problems

Ref. # 5-28 Page No.255: If the prices of new cars increase an average of four times every 3 years, find the probability of

a. No price hikes in a randomly selected period of 3 years. P(X=0)

b. Two price hikes P(X=2)c. Four price hikes P(X=4)d. Five or more P(X≥5)Ans. • .0183• .1465• .1954• .3711

Chapter 5 5-


Numerical Problems

Ref. # 5-32 Page No.256: Guy Ford, production supervisor for the Winstead company’s Charlottesville plant, is worried about an elderly employee’s ability to keep up the minimum work pace. In addition to the normal daily breaks, this employee stops for short rest periods an average of 4.1 times per hour. The rest period is a fairly consistent 3 minutes each time. Fordhas decided that if the probability of the employee resting for 12 minutes (not including normal breaks) or more per hour is greater than 0.5, he will move the employee to a different job. Should he do so?

Ans.

• Yes, the probability of resting at least 12 minutes is 0.5859

Numerical Problems

Ref. # 5-34 Page No.256: Southwestern Electronics has developed a new calculator that performs a series of functions not yet performed by any other calculator. The marketing department is planning todemonstrate this calculator to a group of potential customers, but it is worried about some initial problems, which have resulted in 4 percent of new calculators developing mathematical inconsistencies. The marketing VP is planning on randomly selecting a group of calculators for this demonstration and is worried about the chances of selecting a calculator that could start malfunctioning. He believes that whether or not a calculator malfunctions is Bernoulli process, and he is convinced that the probability of a malfunctions really about 0.04.

a. Assuming that the VP selects exactly 50 calculators to use in the demonstration, and using the Poisson distribution as an approximation of the binomial, what is the chance of getting at least three calculators that malfunction?

b. No calculators malfunctioning?Ans.

a. 0.3233b. 0.1353

Chapter 5 5-


Numerical Problems

Ref. # 5-36 Page No.256: The U.S. Bureau of printing and engraving is responsible for printing this country’s paper money. The BPE has an impressively small frequency of printing errors; only 0.5 percent of all bills are too flawed for circulation. What is the probability that out of a batch of 1000 bills;

a. None are too flawed for circulation?b. Ten are too flawed for circulation?c. Fifteen are too flawed for circulation?

Ans. a. 0.00674b. 0.01813c. 0.00016

Poisson Distribution as an Approximation of the Binomial Distribution

Poisson Distribution can be a reasonable approximation of the Binomial, but only under certain conditions. These conditions occur when n is large and p is small, that is when

The number of trials is large (n ≥ 20)and

The binomial probability of success is small (p < 0.05).

with above conditions the mean of Poisson can be substituted by mean of binomial i.e. np

!

)()(

X

npeXP

xnp−

=

Chapter 5 5-


The Hypergeometric Distribution

Binomial

Poisson


DiscreteProbability

Distributions

Hypergeometric

The Hypergeometric Distribution

� “n” trials in a sample taken from a finite

population of size N

� Sample taken without replacement

� Outcomes of trials are dependent

� Concerned with finding the probability of “X”

successes in the sample where there are “A”

successes in the population

Chapter 5 5-


Hypergeometric Distribution Formula

−

−

== −−

n

N

Xn

AN

X

A

C

]C][C[P(X)

nN

XnANXA

Where

N = population size

A = number of successes in the population

N – A = number of failures in the population

n = sample size

X = number of successes in the sample

n – X = number of failures in the sample

Properties of the Hypergeometric Distribution

� The mean of the hypergeometric distribution is

� The standard deviation is

Where is called the “Finite Population Correction Factor”from sampling without replacement from a finite population

N

nAE(x)µ ==

1- N

n-N

N

A)-nA(Nσ

2⋅=

1- N

n-N

Chapter 5 5-


Using the Hypergeometric Distribution

■ Example: 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded?

N = 10 n = 3A = 4 X = 2

0.3120

(6)(6)

3

10

1

6

2

4

n

N

Xn

AN

X

A

2)P(X ==

=

−

−

==

The probability that 2 of the 3 selected computers have illegal software loaded is 0.30, or 30%.

Probability Distribution

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