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# Discrete Probability Distributions: Binomial and Poisson Distribution. Week 7 (2)

## 1. BBA182 Applied Statistics Week 7 (2) Discrete Probability Distributions: Binomial and Poisson Distribution

DR SUSANNE HANSEN SARALEMAIL: [email protected]

HT TPS://PIAZZA.COM/CLASS/IXRJ5MMOX1U2T8?CID=4#

WWW.KHANACADEMY.ORG

DR SUSANNE HANSEN SARAL

1

## 2. Mid-term exam

Bring:• Calculator

•Pen

•Eraser

23/03/2017

11:45 – 13:00 hours

## 3. Probability and cumulative probability distribution of a discrete random variable

In the last class we saw how to calculate the probabilityof a specific discrete random variable, such as P( x = 3)

and the cumulative probability such as P(x ≥ 3) in n trials

with the following formula:

We need to know n, x and P (probability of success)

## 4. The binomial distribution is used to find the probability of a specific or cumulative number of successes in n trials.

NUMBER OFHEADS (r)

PROBABILITY =

0

P(X = 0)

0.0778 =

1

P(X = 1)

0.2592 =

2

P(X = 2)

0.3456 =

3

P(X = 3)

0.2304 =

4

P(X = 4)

0.0768 =

5

P(X = 5)

0.0102 =

5!

(0.5)r(0.5)5 – r

r!(5 – r)!

5!

0!(5 – 0)!

5!

1!(5 – 1)!

5!

2!(5 – 2)!

5!

3!(5 – 3)!

5!

4!(5 – 4)!

5!

5!(5 – 5)!

DR SUSANNE HANSEN SARAL

(0.5)0(0.5)5 – 0

(0.5)1(0.5)5 – 1

(0.5)2(0.5)5 – 2

(0.5)3(0.5)5 – 3

(0.5)4(0.5)5 – 4

(0.5)5(0.5)5 – 5

4

## 5. Probability and cumulative probability distribution of a discrete random variable

Ali, real estate agent has 5 potential customers to buy a house or apartment with aprobability of 0.40. The probability and cumulative probability table of the sale of

houses here below. We are able to answer what is P( x=3), P(x ≤ 4), P(x >3)etc.

## 6. Shape of Binomial Distribution

MeanThe shape of the binomial distribution depends on the

values of P and n

Here, n = 5 and P = 0.1

.6

.4

.2

0

P(x)

x

0

Here, n = 5 and P = 0.5

.6

.4

.2

0

n = 5 P = 0.1

P(x)

2

3

4

5

n = 5 P = 0.5

x

0

DR SUSANNE HANSEN SARAL

1

1

2

3

4

5

Ch. 4-6

## 7. Binomial Distribution shapes

When P = .5 the shape of the distribution is perfectly symmetrical andresembles a bell-shaped (normal distribution)

When P = .2 the distribution is skewed right. This skewness increases as P

becomes smaller.

When P = .8, the distribution is skewed left. As P comes closer to 1, the

amount of skewness increases.

DR SUSANNE HANSEN SARAL

7

## 8. Using Binomial Tables instead of to calculating Binomial probabilities manually

Nx

…

p=.20

p=.25

p=.30

p=.35

p=.40

p=.45

p=.50

10

0

1

2

3

4

5

6

7

8

9

10

…

…

…

…

…

…

…

…

…

…

…

0.1074

0.2684

0.3020

0.2013

0.0881

0.0264

0.0055

0.0008

0.0001

0.0000

0.0000

0.0563

0.1877

0.2816

0.2503

0.1460

0.0584

0.0162

0.0031

0.0004

0.0000

0.0000

0.0282

0.1211

0.2335

0.2668

0.2001

0.1029

0.0368

0.0090

0.0014

0.0001

0.0000

0.0135

0.0725

0.1757

0.2522

0.2377

0.1536

0.0689

0.0212

0.0043

0.0005

0.0000

0.0060

0.0403

0.1209

0.2150

0.2508

0.2007

0.1115

0.0425

0.0106

0.0016

0.0001

0.0025

0.0207

0.0763

0.1665

0.2384

0.2340

0.1596

0.0746

0.0229

0.0042

0.0003

0.0010

0.0098

0.0439

0.1172

0.2051

0.2461

0.2051

0.1172

0.0439

0.0098

0.0010

Examples:

n = 10, x = 3, P = 0.35:

P(x = 3|n =10, p = 0.35) = .2522

n = 10, x = 8, P = 0.45:

P(x = 8|n =10, p = 0.45) = .0229

DR SUSANNE HANSEN SARAL

Ch. 4-8

## 9. Solving Problems with Binomial Tables

MSA Electronics is experimenting with the manufacture of a newUSB-stick and is looking into the number of defective USB-sticks

◦ Every hour a random sample of 5 USB-sticks is taken

◦ The probability of one USB-stick being defective is 0.15

◦ What is the probability of finding 3, 4, or 5 defective USB-sticks ?

P( x = 3), P(x = 4 ), P(x= 5)

n = 5, p = 0.15, and r = 3, 4, or 5

DR SUSANNE HANSEN SARAL

2–9

## 10. Solving Problems with Binomial Tables

TABLE 2.9 (partial) – Table for Binomial Distribution, n= 5,n

5

r

0

1

2

3

4

5

0.05

0.7738

0.2036

0.0214

0.0011

0.0000

0.0000

P

0.10

0.5905

0.3281

0.0729

0.0081

0.0005

0.0000

DR SUSANNE HANSEN SARAL

0.15

0.4437

0.3915

0.1382

0.0244

0.0022

0.0001

2 – 10

## 11. Solving Problems with Binomial Tables

MSA Electronics is experimenting with the manufacture of a newUSB-stick

◦ Every hour a random sample of 5 USB-sticks is taken

◦ The probability of one USB-stick being defective is 0.15

◦ What is the probability of finding more than 3 (3 inclusive) defective USBsticks? P( x ≥ 3)

n = 5, p = 0.15, and r = 3, 4, or 5

DR SUSANNE HANSEN SARAL

2 – 11

## 12. Solving Problems with Binomial Tables Cumulative probability

TABLE 2.9 (partial) – Table for Binomial Distributionn

5

r

0

1

2

3

4

5

0.05

0.7738

0.2036

0.0214

0.0011

0.0000

0.0000

P

0.10

0.5905

0.3281

0.0729

0.0081

0.0005

0.0000

0.15

0.4437

0.3915

0.1382

0.0244

0.0022

0.0001

We find the three probabilities in the table for

n = 5, p = 0.15, and r = 3, 4, and 5 and add

them together

DR SUSANNE HANSEN SARAL

2 – 12

## 13. Solving Problems with Binomial Tables Cumulative probabilities

TABLE 2.9 (partial) – Table for Binomial Distributionn

0.05

P(3

orr more defects)

=

5

0

1

2

3

4

5

0.7738

0.2036

0.0214

0.0011

0.0000

0.0000

=

=

P

P(3)0.10

+ P(4) + P(5) 0.15

0.5905

0.4437

0.0244

0.0001

0.3281+ 0.0022 +

0.3915

0.0729

0.1382

0.0267

0.0081

0.0244

0.0005

0.0022

0.0000

0.0001

We find the three probabilities in the table for n =

5, p = 0.15, and r = 3, 4, and 5 and add them

together

DR SUSANNE HANSEN SARAL

2 – 13

## 14. Suppose that Ali, a real estate agent, has 10 people interested in buying a house. He believes that for each of the 10 people the probability of selling a house is 0.40. What is the probability that he will sell 4 houses, P(x = 4)?

DR SUSANNE HANSEN SARAL14

## 15. Suppose that Ali, a real estate agent, has 10 people interested in buying a house. He believes that for each of the 10 people the probability of selling a house is 0.40. What is the probability that he will sell 4 houses?

DR SUSANNE HANSEN SARAL15

## 16. Suppose that Ali, a real estate agent, has 10 people interested in buying a house. He believes that for each of the 10 people the probability of selling a house is 0.20. What is the probability that he will sell 7 houses, P(x = 7)?

DR SUSANNE HANSEN SARAL16

## 17. Suppose that Ali, a real estate agent, has 10 people interested in buying a house. He believes that for each of the 10 people the probability of selling a house is 0.35. What is the probability that he will sell more than 7 houses, 7 houses included, P(x

Suppose that Ali, a real estate agent, has 10 people interested in buying ahouse. He believes that for each of the 10 people the probability of selling

a house is 0.35. What is the probability that he will sell more than 7

houses, 7 houses included, P(x ≥ 7)?

DR SUSANNE HANSEN SARAL

17

## 18.

Probability DistributionsProbability

Distributions

Discrete

Probability

Distributions

Continuous

Probability

Distributions

Binomial

Uniform

Poisson

Normal

Exponential

DR SUSANNE HANSEN SARAL

Ch. 4-18

## 19. Poisson random variable, first proposed by Frenchman Simeon Poisson (1781-1840)

A Poisson distributed random variable is oftenuseful in estimating the number of occurrences

over a specified interval of time or space.

It is a discrete random variable that may assume

an infinite sequence of values (x = 0, 1, 2, . . . ).

DR SUSANNE HANSEN SARAL

19

## 20. Poisson Random Variable - three requirements

Poisson Random Variable three requirements1. The number of expected outcomes in one interval of time or unit space

is unaffected (independent) by the number of expected outcomes in any

other non-overlapping time interval.

Example: What took place between 3:00 and 3:20 p.m. is not affected by

what took place between 9:00 and 9:20 a.m.

DR SUSANNE HANSEN SARAL

20

## 21. Poisson Random Variable - three requirements (continued)

Poisson Random Variable three requirements (continued)2.The expected (or mean) number of outcomes over any time period or unit space is

proportional to the size of this time interval.

Example:

We expect half as many outcomes between 3:00 and 3:30 P.M. as between 3:00 and 4:00

P.M.

3.This requirement also implies that the probability of an occurrence must be constant over any

intervals of the same length.

Example:

The expected outcome between 3:00 and 3:30P.M. is equal to the expected occurrence

between 4:00 and 4:30 P.M..

DR SUSANNE HANSEN SARAL

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## 22. Examples of a Poisson Random variable

The number of cars arriving at a toll booth in 1 hour (the time interval is 1 hour)The number of failures in a large computer system during a given day (the given day is

the interval)

The number of delivery trucks to arrive at a central warehouse in an hour.

The number of customers to arrive for flights at an airport during each 10-minute time

interval from 3:00 p.m. to 6:00 p.m. on weekdays

DR SUSANNE HANSEN SARAL

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## 23. Situations where the Poisson distribution is widely used: Capacity planning – time interval

Areas of capacity planning observed in a sample:A bank wants to know how many customers arrive at the bank in a given time period

during the day, so that they can anticipate the waiting lines and plan for the number of

employees to hire.

At peak hours they might want to open more guichets (employ more personnel) to

reduce waiting lines and during slower hours, have a few guichets open (need for less

personnel).

DR SUSANNE HANSEN SARAL

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