Probability Distributions.ppt

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Business Statistics:

A Decision-Making Approach

6th Edition

Unit 2

Discrete and Continuous

Probability Distributions

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Chapter Goals

After completing this unit, you should be able to:

• Apply the binomial distribution to applied problems

• Apply and recognize the hypergeometric

distribution

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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

– spreads peanut butter first vs. spreads jelly first

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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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The Binomial Distribution

• 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

)!

(

!

!

)

,

(

r

n

r

n

r

n

C

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)

P(x)

p

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

Binomial Distribution Formula

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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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Binomial Distribution

Example 1

: It is known that 10% of business

students are left-handed. There are five

students sitting in a row. What is the

probability that there is exactly one left

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Binomial Distribution

Characteristics

• Mean

• Variance and Standard Deviation

np

E(x)

μ

npq

σ

2

npq

σ

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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Hypergeometric Distribution

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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Hypergeometric Distribution

Formula

0.3

120

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3

10

1

6

2

4

2)

P(x

























n

N

r

n

b

r

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?

N = 10

n = 3

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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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The Normal Distribution

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

Probability as

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

1.0

)

x

P(



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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Importance of the Rule

• If a value is about 2 or more standard

deviations away from the mean in a normal

distribution, then it is far from the mean

• The chance that a value that far or farther

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The Standard Normal Distribution

• Also known as the “z” distribution

• Mean is defined to be 0

• Standard Deviation is 1

z f(z)

0 1

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The Standard Normal

• Any normal distribution (with any mean

and standard deviation combination) can

be transformed into the

standard normal

distribution (z)

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Translation to the Standard

Normal Distribution

• Translate from x to the standard normal (the

“z” distribution) by subtracting the mean of

x and dividing by its standard deviation:

σ

μ

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.

3.0

50

100

250

σ

μ

x

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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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General Procedure for

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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Upper Tail Probabilities

• Suppose x is normal with mean 8.0

and standard deviation 5.0.

• Now Find P(x > 8.6)

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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The Uniform Distribution

Simple example of “Other”

Continuous

Probability Distributions

Probability Distributions

Normal

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The Uniform Distribution

• The uniform distribution is a

probability distribution that has equal

probabilities for all possible

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The Continuous Uniform Distribution:

otherwise

0

b

x

a

if

a

b

1

where

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

b = upper limit of the interval

The Uniform Distribution

(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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The Exponential Distribution

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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

Figure

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References

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Related subjects : random variable