3+ +probability+distributions

8/8/2019 3+ +Probability+Distributions

1/32

Quantitative Methods 2010

3

Probability Distributions

1QM 2010 - Kingston


2/32

Frequency Distributions

When data is very voluminous, we may find it

useful to view it in compressed form. One

way to compress is to show data as a

frequency table or a frequency

distribution.

Divide the entire range of data into groups or

classes

Show how many data points fall into each class.

QM 2010 - Kingston 2


3/32


Store Stock, Days

1 2.02 3.8

3 4.1

4 4.7

5 5.5

6 3.4

7 4.08 4.2

9 4.8

10 5.5

11 3.4

12 4.1

13 4.3

14 4.9

15 5.5

16 3.8

17 4.1

18 4.7

19 4.9

20 5.5

0

1

2

3

4

5

6

7

8

9

2.0 to

2.5

2.6 to

3.1

3.2 to

3.7

3.8 to

4.3

4.4 to

4.9

5.0 to

5.5

More

Raw Data Frequency Distribution

As Table

Frequency Distribution

As Histogram

Stock, Days Frequency

2 0 2 5 1

2 6 3 1 0

3 2 3 7 2

3 8 4 3 84 4 4 9 5

5 0 5 5 4

0

If ins ad f c un s, y u sh w as

f ac i n p c n ag f al

numb f bs va i ns, y u g a

R la iv Frequency Dis ribu i n


4/32


They are related to Frequency Distributions.

Frequency Distributions show observedfrequencies of outcomes in an experiment.

Probability Distributions indicate expectedfrequencies of all possible outcomes in theexperiment.

Discrete Probability Distributions indicate expectedfrequencies directly

Continuous Probability Distributions indicate expectedfrequencies indirectly, by area under the curve.



5/32


Discrete Continuous


0 1 2 3 4 5 ------- 500

P[ X ]

X = No Of Parts Bad X = Weight Of Students, kg

f(x)

45 50 55 60 65 70 75

P[x] directly indicates probability

of particular x

f[x] does not directly indicate probability

of particular x.


6/32

Discrete Probability Distributions



7/32

Counts Of Events

Many situations, where we are interested in counts:how many times something happened. We receive 500 bolts every week.

How many bolts were defective each week?

Each possible no can be thought of as event, and it will have itsown probability.

Zero defective ------ P[0]

1 defective ------- P[1]

2 defective -------- P[2]

500 defective ------ P[500]

Counts are one example of discrete variables.

7QM 2010 - Kingston


8/32

Discrete Probability Distributions

A table or picture

(graph) showing

probabilities for all

possible values of a

discrete variable is

called a discrete

distribution.

8

Value P[ ]

0 0.02

1 0.15

2 0.18

3 0.25

500 0.0005

Total 1.0000

0 1 2 3 4 5 ------- 500

P[ ]

ValueQM 2010 - Kingston


9/32

Discrete Cumulative Probability Distributions

9

A table or picture

(graph) showing for all

possible values of a

discrete variable, thesum of probabilities is

called a cumulative

discrete distribution.

Value P[ ]

0 0.02

1 0.17

2 0.35

3 0.6

500 1.0000

0 1 2 3 4 5 ------- 500

P[ ]

1.00

QM 2010 - Kingston


10/32

Binomial Distribution

Describes discrete data resulting from aBernoulli Process

Bernoulli Process is a process that has only 2

outcomes (example: success, failure) withfixed probability p and q = (1 p), andindependent trials.

Gives the probability of r successes in n trials: P[r/n] =

Binomial tables are available to look up probabilities

Mean, = np, Variance 2 = npq

10

rnrqp

rnr

n

)!(!

!

QM 2010 - Kingston


11/32

Example Binomial Probabilities

There are 15 members of a certain political party in

the Bombay Municipal Corporation (BMC), who are

notorious for not being present when BMC is in

session. It has been found that the probability of amember of this party being absent is 0.35. What is

the probability that for an upcoming crucial budget

session, 10 of them will be absent?

Ans: P[10/15] = = 0.0096

11

510)65.0()35.0(

!5!10

!15

QM 2010 - Kingston


12/32

Extract From Binomial Tables

12

n r

1

2

21

22

2

2

2

2

2

2

2

1

2

1

1

2

1

2

11

11

al es

QM 2

1

- Kingston


13/32

Poisson Distribution

Occurs in situations where

Probability of an event in a time-interval is verysmall, and the same for all intervals

Probability of 2 or more events in that interval ~ 0 No of events in any interval is independent of where

the interval occurs

No of events in any interval is independent of no. in

any other interval P(x) = --- Tables available

Mean, = = Variance, 2

13

!/ xex PP

QM 2010 - Kingston


14/32

Example Poisson Probabilities

The Vikhroli intersection on Eastern ExpressHighway is notorious for accidents. Recordsshow a mean of 5 accidents per month, and

Poisson distribution. What is the probability ofmore than 3 accidents in a given month?

P[x > 3] = 1 P[x 3]

= 1 {P[0] + P[1} + P[2] + P[3]}

= 1 {0.00674 + 0.03370 + 0.08425 + 0.14042}

= 1 0.26511

= 0.73 or 73%

14QM 2010 - Kingston


15/32

Extract From Poisson Tables

15

x 1 5 5

1

5

1

1 55

QM

1 - Kingston


16/32

Poisson

As An Approximation To

Binomial In a Binomial situation, if n is large ( 20) and p

is small ( .05), you can use a Poisson with =np.

Example You are an investment banker, advising onIPO of 20 clients. Probability of any one of them beingunder-subscribed is .02. What is the probability of 3clients being under-subscribed?

Situation is really Binomial: n= 20, p = .02

P[3 out of 20] = .0065 ---- (from Binomial tables)

Using Poisson, with = np = .40 :

P[3] = .0072 ---- (from Poisson tables)



17/32

Continuous Probability Distributions



18/32

Normal Distribution N(,)

18

Roughly 1

QM 2010 - Kingston


19/32

Uniform Distribution U(a, b)


a b

= 95 = 100

X = mm of rainfall

f(x)


20/32

Exponential Distribution


X = Time Between Arrivals At Toll-booth, mins

f(x)


21/32

Expected Value Of A Random Variable



22/32

Random Variables

Random variable

Takes different values as the result of a random

experiment

Only limited number of values discrete random

variable

Any values within a range continuous random

variable



23/32

Expected Values For

Discrete Random Variables It is basically a weighted average of the value of each

possible outcome, weighted by its probability.

Sum of cross-products of variable value and itsprobability E(x) =

Example The no. of defective parts in boxes

of 1000 has the following

probability distribution. What is

the expected value of no. of

defective parts?

Ans: 4.13

23

!

n

i

ii xpx1

)(Variabl

cti

! art"

i # boxof $ % % %

Variabl

Val & "

' xi ( ) xi 0 1 ross- ! rod

0 0.01 0

1 0.07 0.07

2 0.22 0.44

3 0.18 0.54

4 0.14 0.56

5 0.11 0.55

6 0.09 0.54

7 0.08 0.56

8 0.05 0.4

9 0.03 0.27

10 0.02 0.2

E ) xi 0 4.13

CALCULATI 2

E3

(ECTE 4 VALUE 5 F

4 ISCRETE RANDOMVARIABLE

QM2010 - Ki 6 gston


24/32

Expected Values For

Continuous Random Variables

It is equivalent to that of discrete random

variables

Example - Rainfall in June is uniformly distributed between 95

and 100 mm. What is the expected rainfall in June?

f(x) = 1/(b a)


! dxxxfxE )()(

5.97

2

95100

2

)()(

2)(

1

2)(

1

)(

1

)(

1

)()(

222

!

!

!

!

!

!

!!

ab

abab

xab

xdxabdxabxdxxxfx

b

a

b

a

b

a


25/32

Use Of Normal Distribution

Example

A certain type of project activity takes an average

of 22 days to complete, and has a standard

deviation of 4 days. What is the probability that in

the current project it may take upto 30 days?



26/32

The Required Probability

26

= 22 30

Let x be no of days to complete the activity

We want to know P[x 30]

QM 2010 - Kingston


27/32

Standard Normal Distribution

Since there are an infinite number of Normal

distributions, we cannot have table for all of

them

Every Normal x with (, ) can be converted

into equivalent Standard Normal z with (0, 1)

by the conversion

We have table for z, Standard Normal.


W

Q)(

! xz


28/32

28

z F(z) z F(z) z F(z) z F(z)

-4 3.16712 -05 -2 0.02275 0.1 0.539828 2 0.97725

-3.9 4.80963 -05 -1.9 0.028717 0.2 0.57926 2.1 0.982136

-3.8 7.2348 -05 -1.8 0.03593 0.3 0.617911 2.2 0.986097

-3.7 0.0001078 -1.7 0.044565 0.4 0.655422 2.3 0.989276

-3.6 0.000159109 -1.6 0.054799 0.5 0.691462 2.4 0.991802

-3.5 0.000232629 -1.5 0.066807 0.6 0.725747 2.5 0.99379

-3.4 0.000336929 -1.4 0.080757 0.7 0.758036 2.6 0.995339

-3.3 0.000483424 -1.3 0.0968 0.8 0.788145 2.7 0.996533

-3.2 0.000687138 -1.2 0.11507 0.9 0.81594 2.8 0.997445-3.1 0.000967603 -1.1 0.135666 1 0.841345 2.9 0.998134

-3 0.001349898 -1 0.158655 1.1 0.864334 3 0.99865

-2.9 0.001865813 -0.9 0.18406 1.2 0.88493 3.1 0.999032

-2.8 0.00255513 -0.8 0.211855 1.3 0.9032 3.2 0.999313

-2.7 0.003466974 -0.7 0.241964 1.4 0.919243 3.3 0.999517

-2.6 0

.004661188 -0

.6 0

.274253 1

.5 0

.933193 3

.4 0

.999663

-2.5 0.006209665 -0.5 0.308538 1.6 0.945201 3.5 0.999767

-2.4 0.008197536 -0.4 0.344578 1.7 0.955435 3.6 0.999841

-2.3 0.01072411 -0.3 0.382089 1.8 0.96407 3.7 0.999892

-2.2 0.013903448 -0.2 0.42074 1.9 0.971283 3.8 0.999928

-2.1 0.017864421 -0.1 0.460172 2 0.97725 3.9 0.999952

-2 0.022750132 0 0.5 4 0.999968

AREA UNDER STANDARD NORMAL TO THE LEFT OF Z

QM 2010 - Kingston


29/32


30/32

The Standard Normal Distribution

30

= 22 X =30 = 0 Z = 2

= 1

Z = (x - ) /

From table or EXCEL =normsdist(2), P[z 2] = P[x 30] = .97725

QM 2010 - Kingston


31/32

P[x 25 days]

31

= 0

Z = 0.75

= 1

Z = (x - ) /

= 22X =25

= 4

From table or EXCEL =normsdist(2), P[z 0.75] = 0.7734. Then P[x 25] = 1

P[z < 0.75] = 0.2266QM 2010 - Kingston


32/32

End OfProbability Distributions


3+ +probability+distributions

Documents