# Probability Distributions.ppt

## Full text

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6th Edition

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### Probability Distributions

Continuous

Probability Distributions

Binomial

Hypergeometric

Probability Distributions

Discrete

Probability Distributions

Normal

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• A discrete random variable is a variable that can assume only a countable number of values

Many possible outcomes:

– number of complaints per day – number of TV’s in a household

– number of rings before the phone is answered Only two possible outcomes:

– gender: male or female – defective: yes or no

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### Continuous Probability Distributions

• A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values)

– thickness of an item

– time required to complete a task – temperature of a solution

– height, in inches

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### The Binomial Distribution

Binomial

Hypergeometric

Probability Distributions

Discrete

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### • Characteristics of the Binomial Distribution:

– A trial has only two possible outcomes – “success” or “failure”

– There is a fixed number, n, of identical trials

– The trials of the experiment are independent of each other

– The probability of a success, p, remains constant

from trial to trial

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### Binomial Distribution Settings

• A manufacturing plant labels items as either defective or acceptable

• A firm bidding for a contract will either get the 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

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### Counting Rule for Combinations

• A combination is an outcome of an experiment where r objects are selected from a group of n objects

### 

where:

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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)

p = probability of “success” per trial q = probability of “failure” = (1 – p)

n = number of trials (sample size)

x

### q

nx

Example: Flip a coin four times, let x = # heads:

n = 4

p = 0.5

q = (1 - .5) = .5

x = 0, 1, 2, 3, 4

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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 = .1

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2

### 

Where n = sample size

p = probability of success

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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)(.1) np

μ   

0.6708 .1) (5)(.1)(1 npq σ     2.5 (5)(.5) np

μ   

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### The Hypergeometric Distribution

Binomial

Probability Distributions

Discrete

Probability Distributions

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### The Hypergeometric Distribution

• “n” trials in a sample taken from a finite population

of size N

• Sample taken without replacement

• Trials are dependent

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### Formula

                    n N r n b r a x

P( )

Where

N = Population size (a + b)

a = number of successes in the population n = sample size

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### a

■ Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the

probability that 2 of the 3 selected are defective?

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### The Normal Distribution

Continuous

Probability Distributions

Probability Distributions

Normal

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## The Normal Distribution

‘Bell Shaped’

Symmetrical

Mean, Median and Mode

are Equal

Location is determined by the mean, μ

Spread is determined by the standard deviation, σ

The random variable has an infinite theoretical range:

+  to  

Mean = Median = Mode

x f(x)

μ

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By varying the parameters μ and σ, we obtain different normal distributions

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### Shape

x f(x)

μ σ

Changing μ shifts the distribution left or right.

Changing σ increases or decreases the

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### Finding Normal Probabilities

Probability is the area under the

curve!

a b x

f(x) P (a x b)

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f(x)

x

μ

### Area Under the Curve

0.5 0.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

0.5 )

x

P(μ    

0.5 μ)

x

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### Empirical Rules

μ ± 1σencloses about 68% of x’s

f(x)

x μ μσ

μσ

What can we say about the distribution of values around the mean? There are some general rules:

σ σ

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### The Empirical Rule

μ ± covers about 95% of x’s

μ ± 3σ covers about 99.7% of x’s

x μ

x μ

95.44% 99.72%

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z f(z)

0 1

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## x

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### Example

If x is distributed normally with mean of 100 and

standard deviation of 50, the z value for x = 250

is

• This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.

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### Comparing x and z units

z 100

3.0 0

250 x

Note that the distribution is the same, only the

scale has changed. We can express the problem in original units (x) or in standardized units (z)

μ = 100

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### The Standard Normal Table

The value within the table gives the

probability up to the desired z value

z 0.00 0.01 0.02 …

0.1

0.2

.9772

2.0

P(z < 2.00) = .9772

The row shows the value of z to the first

decimal point

The column gives the value of z to the second decimal point

2.0

. . .

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### The Standard Normal Table

• The Standard Normal table in the textbook

(Appendix B, p398)

gives the probability from the far left (-∞) up to a desired value for z

z 0 2.00

.4772

Example:

P(z < 2.00) = .9772

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### Finding Probabilities

Draw the normal curve for the problem in terms of x

• Translate x-values to z-values

• Use the Standard Normal Table

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### Z Table example

Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6) = P(x < 8.6) – P(x < 8)

= P(0 < z < 0.12) = P(z < 0.12) – 0.5

Z 0.12 0 x 8.6 8 0 5 8 8 σ μ x

z     

0.12 5 8 8.6 σ μ x

z     

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### Z Table example

Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z 0.12

0 x

8.6 8

P(8 < x < 8.6)

 = 8

 = 5

 = 0  = 1

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Z

0.12

z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

### Solution: Finding P(0 < z < 0.12)

.0478 .02

0.1 .5478

Standard Normal Probability Table (Portion)

0.00

= P(0 < z < 0.12)

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Z

8.6

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### • Now Find P(x > 8.6)…

(continued)

Z

0.12

0 Z

0.12

.0478

0

.5000 .50 - .0478 = .4522

P(x > 8.6) = P(z > 0.12) = 1 - P(z < 0.12)

= 1 - .5478 = .4522

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### Simple example of “Other”

Continuous

Probability Distributions

Probability Distributions

Normal

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### probabilities for all possible

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The Continuous Uniform Distribution:

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where

f(x) = value of the density function at any x value a = lower limit of the interval

b = upper limit of the interval

(continued)

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### Uniform Distribution

Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 66 - 21

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### Chapter Summary

• Reviewed key discrete distributions

– binomial, hypergeometric and general

• Reviewed key continuous distributions

– normal, and other

• Found probabilities using formulas and tables

• Recognized when to apply different distributions

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## References

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Related subjects : random variable