PROBABILITY DISTRIBUTIONS

Probability distributions → A listing of all the outcomes of an experiment and the probability assosiated with each outcome, the type :a. Discrete Probability Distributionsb. Continuos Probability DistributionsHow can we generate a probability distribution ?Suppose we are interested in the number of heads showing face up on the first tosses of a coin. This is the experiment.the possible results are: zero heads, one heads, two heads, and three heads. What is the probability distribution for the number of heads?

These results are listed below

Possible Coin Toss Number of

Result First Second Third Heads

1 T T T 0

2 T T H 1

3 T H T 1

4 T H H 2

5 H T T 1

6 H T H 2

7 H H T 2

8 H H H 3

Probability distribution for the events of zero,one, two, and three heads showing face up on three of a coin

Number of Heads

x

Probability of Outcome

P (x)

0

1

2

3

Total

1/8 = 0,125

3/8 = 0,375

3/8 = 0,375

1/8 = 0,125

8/8 = 1,000

Graphical presentation of the number of heads resulting from three tosses of a coin and the corresponding probability

0

0,05

0,1

0,15

0,2

0,25

0,3

0,35

0,4

0 1 2 3

The Mean, Variance, and Standard Deviation of a Probability Distributions

1. Mean : μ = Σ [xP(x)]

2. Variance : σ2 = Σ [(x-μ)2 P(x)]

3. Stand Dev : σ = √σ2

Example:

John sells new car for Pelican Ford. John usually sells the largest number of cars on Saturday. He has the following probability distributions for the number of cars he expects to sell on a particular Saturday.

Number of Cars Sold

x

Probability of Outcome

P (x)

0

1

2

3

4

Total

0,1

0,2

0,3

0,3

0,1

1,0

Questions:

1.What type of distributions is this?

This a discrete probability distribution

2.On a typical Saturday, how many cars does John expect to sell?

μ = Σ [xP(x)]

= 0(0,10)+1(0,20)+2(0,30)+3(0,30)+4(0,10)

= 2,1

Number of Cars Sold

X

Probability

P(x)

x . P(x)

0

1

2

3

4

Total

0,10

0,20

0,30

0,30

0,10

1,00

0,00

0,20

0,60

0,90

0,40

μ = 2,10

This value indicates that, over a large number on Saturday expects to sell a mean 2,1 cars a day. In a year he can expect to sell 50 x 2,1 = 105 cars.

Number of Cars

Sold

Probability

P(x)

(x-μ) (x-μ)2 (x-μ)2 P(x)

0

1

2

3

4

0,10

0,20

0,30

0,30

0,10

0-2,1

1-2,1

2-2,1

3-2,1

4-2,1

4,41

1,21

0,01

0,81

3,61

0,441

0,242

0,003

0,243

0,361

σ2 = 1,290

3. What is the variance of the distribution?

Recall that the standard deviation, σ, is the positive square root of the variance. In this example = 1,136 cars.

If other salesperson, Rita, also sold a mean of 2,1 cars on Saturday and have the standard deviation in her sales was 1,91 cars. We would conclude that there is more variability in Saturday sales of Rita than john .

A. Discrete Probability Distributions

→ a discrete can assume only a certain number of separated values. It there are 100 employees, then the count of the number absent on Monday can only be 0,1,2,3,…,100. A discrete is usually the result of counting something.

1. Binomial Probability DistributionCharacteristics :a. An outcome on each trial of an experiment is

classified into one of two mutually exclusive categories- success or failure

b. The random variable counts the number of successes in a fixed number of trials

c. The probability of success and failure stay the same for each trial

d. The trials are independent, meaning that the outcome of one trial does not affect the outcome of any other trial

Binomial PD : P(x) = nCx πx (1-π)n-x

Example:

There are five flights daily from pittsburgh via US Airways into the Bradford, Pennysylvania Regional Airport. Suppose the probability that any flight arrives late is 0.20. What is the probability that none of the flights are late today? What is the probability that exactly one of the flights is late today?

P(0) = nCx πx (1-π)n-x

= 5C0 (,20)0 (1-,20)5-0

= (1)(1)(,3277) = 0,3277

P(1) = nCx πx (1-π)n-x

=5C1 π1 (1-,20)5-1

= (5)(,20)(,4096) =n 0,4096Binomial Probability Distribution for n=5,n=,20

Number of Late Flights Probability

0

1

2

3

4

5

Total

0,3277

0,4096

0,2048

0,0512

0,0064

0,0003

1,0000

Mean of Binomial Distribution

μ = nπ

= (5)(,20) = 1,0

Variance of Binomial Distributions

σ2 = nπ (1- π)

= 5(,20)(1-,20) = 0,80

2. Hypergeometric Probability Distribution

Characteristics:

a. An outcome on each trial of an experiment is classified into one of two mutually exclusive categories- success or failure

b. The random variable is the number of successes in a fixed number of trials

c. The trials are not independent

d. We assume that we sample from a finite population without replacement. So the probability of a success changes for each trial.

Hypergeometric Distribution

P(x) = (SCx)(N-SCn-x)

NCn

Example:

Play Time Toys, Inc. employs 50 people in the assembly Departement. Forty of the employees belong to a union and ten do not. Five employees are selected at random to form a commite to meet with management regarding shift starting times. What is the probability that four of the five selected for the committee belong to a union?

The Answer:P(x) = (SCx)(N-SCn-x)

NCn

P(4) = (40C4)(50-40C5-4)

50C5

= (91.390)(10) = 0,431

2.118.760

3.Poisson Probability Distribution

The probability Poisson describes the number of times some event occurs during aspecified interval

Characteristics :

1. The random variable is the number of times some event occurs during a defined interval

2. The probability of the event is proportional to the size of the interval

3. The intervals which do not overlap and are independent

Poisson Distribution

P(x) = μxe-μ

x!

Where e = 2,71828

Mean of a Poisson Distribution

μ = nπ

B.Continuos Probability Distributions

→ A continuous probability distribution usually results from measuring something, such as the distance from the dormitory to the classroom, the weight of an individual, or the amount of bonus earned by Ceos. Suppose we select five student and find the distance, in miles, they travel to attend class as 12.2, 8.9, 6.7 and 14.6.

We consider two families of Continuous Distribution : a. Uniform Probability Distribution Uniform distribution : P(x) = 1

b – a Mean : μ = a+b

2 Standar Deviasi : σ = √(b-a)2

12

b. Normal Probability DistributionThe number of normal distributions is unlimited, each having a different mean, Standard deviation, or both, While it is possible to provide probability tables for discrete distributions such asa the binomial and the poisson, providing tables for infinite number of normal distributions is impossible. Fortunately, one member of the family can be used to determine the probabilities for all normal distributions. It is called the standard normal distribution, and it is unique because it has a mean of 0 and a standard deviation of 1.

Any normal distribution can be converted into a standard normal distribution by subtracting the mean from each observationand dividing this difference by the standarddeviation. The results are called z values.They are also referred to as z scores, the zstatistics, the standard normal deviates,thestandar normal values, or just the normaldeviate.

Z value → the signed distance between a

selected value, disigned x, and the

mean,divided by the standard deviation.

Formula :

Standard Normal Value : z = X –μ

σ

Where:

X is the value of any particular observation or measurement.

μ is the mean of the distribution

σ is the standard deviation of the distribution

Example:The weekly incomes of shift foremen in the glass industry are normally distributed with a mean of $1,000 and a standard deviation of $100. What is the z value foe the income X of a foreman who earns $1,100 per week? For a foreman who earns $900 per week?for X= $1,100 For X = $900z = X – μ z = X – μ

σ σ = $1,100 - $ 1,000 = $900 - $1,000 $100 $100

= 1,00 = - 1,00