Chapter 3 Discrete Random Variables and Probability Distributions - PowerPoint Presentation

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Chapter 3 Discrete Random Variables and Probability Distributions - PPT Presentation

31 Random Variables 32 Probability Distributions for Discrete Random Variables 33 Expected Values 34 The Binomial Probability Distribution 35 Hypergeometric and Negative ID: 777551

type 461 probability 539 461 type 539 probability random blood binomial 007 population 1500 individuals 323 094 017 077

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Slide1

Chapter 3Discrete Random Variables and Probability Distributions

3.1 - Random Variables

3.2 - Probability Distributions for Discrete

Random Variables

3.3 - Expected Values

3.4 -

The Binomial Probability Distribution

3.5 - Hypergeometric and Negative

Binomial Distributions

3.6 - The Poisson Probability Distribution

Slide2

POPULATION

Discrete

random variable

Examples

: shoe size, dosage (mg), # cells,…

Pop values xProbabilitiesp(x)Cumul ProbsF (x)x1p(x1) p(x1) x2p(x2) p(x1) + p(x2) x3p(x3) px1) + p(x2) + p(x3)⋮⋮⋮1Total1

Mean

Variance

Total Area = 1

Recall…

Slide3

Classical Discrete Probability Distributions

Binomial

= # Successes in

n trials, P

(Success) = Poisson ~ As above, but n large,  small, i.e., Success RARENegative Binomial ~ X = # trials for k Successes, P(Success) = Geometric ~ As above, but specialized to k = 1Hypergeometric ~ As Binomial, but  changes between trialsMultinomial ~ As Binomial, but for multiple categories, with 1 + 2 + … + last = 1 and x1 + x2 + … + xlast = n

Slide4

~ The Binomial Distribution ~

Used

only

when dealing with

binary outcomes

(two categories: “Success” vs. “Failure”), with a

fixed probability of Success () in the population.Calculates the probability of obtaining any given number of Successes in a random sample of n independent “Bernoulli trials.”Has many applications and generalizations, e.g., multiple categories, variable probability of Success, etc.

Slide5

For any randomly selected individual, define a

binary

random variable:

POPULATION

RANDOMSAMPLE

size n Discrete random variableX = # Successes in sample(x = 0, 1, 2, 3, …, n)For x = 0, 1, 2, 3, …, nxp(x)0p(0)1p(1)……np(n)1F(x)F(0)F(1)…1

Slide6

For any randomly selected individual, define a

binary

random variable:

POPULATION

Discrete

random variable

X = # Heads in sample(x = 0, 1, 2, 3, …, n).… etc…..… etc….Reformulate this as n independent coin tosses.RANDOMSAMPLE size n Each such sequence has probability

There are such sequences of n

tosses, with x Heads.

Slide7

“Success” vs. “Failure”7

RANDOM SAMPLE

n “Bernoulli trials.”

independent, with constant probability () per trial Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability mass function” p(x) = , x = 0, 1, 2, …, n.POPULATIONFor any randomly selected individual, define a binary random variable:Discrete random variableX = # Successes in sample(x = 0, 1, 2, 3, …, n)

Slide8

Factor

Blood Type

–

.384.077.461A.323.065.388B.094.017.111AB.032.007.039.833.166.999Example: Blood Type probabilities, revisitedReasonably assume that outcomes “Type O” vs. “Not Type O” between two individuals are independent of each other. Suppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type O)Binomial model applies? Check:1. Independent outcomes?2. Constant probability  ?

From table,

 = P(Type O) = .461 throughout population. 

Slide9

Example:

Blood Type probabilities, revisited

~ Bin(10, .461)

p(x)F (x)0 (.461)0 (.539)10 = 0.002070.002071 (.461)1 (.539)9 = 0.017700.019772 (.461)2 (.539)8 = 0.068130.087903 (.461)3 (.539)7 = 0.155380.243284 (.461)4 (.539)6 = 0.232570.475855 (.461)5 (.539)5 = 0.238700.714556 (.461)6 (.539)4 = 0.170130.884687 (.461)7 (.539)3

= 0.08315

0.96783

(.461)8

(.539)

= 0.02667

0.99450

(.461)

9 (.539)

= 0.00507

0.99957

(.461)10 (.539)

= 0.00043

1.00000

Suppose

= 10 individuals are to be selected at random from the population.

Probability table for

= #(Type O)

Binomial model applies

dbinom

(0:10, 10, .461)

Factor

Blood Type

–

.384

.077

.461

.323

.065

.388

.094

.017

.111

.032

.007

.039

.833

.166

.999

Also, can show mean





(

) =

and variance



=  (

–



)

(

) =

n

(1 –



)

= (10)(.461)

4.61

(

) = (.461)

(.539)

– x

2.48

Slide10

(

)

F (x)0 (.461)0 (.539)10 = 0.002070.002071 (.461)1 (.539)9 = 0.017700.019772 (.461)2 (.539)8 = 0.068130.087903 (.461)3 (.539)7 = 0.155380.243284 (.461)4 (.539)6 = 0.232570.475855 (.461)5 (.539)5 = 0.238700.714556 (.461)6 (.539)4 = 0.170130.884687 (.461)7 (.539)3 = 0.083150.967838 (.461)8

(.539)

2 = 0.02667

0.99450

(.461)

(.539)

= 0.00507

0.99957

(.461)

10 (.539)

0 =

0.00043

1.00000

dbinom

(0:10, 10, .461)

Also, can show mean





(

) =

and variance



=  (

– 

) 2

x) =

Factor

Blood Type

–

.384

.077

.461

.323

.065

.388

.094

.017

.111

.032

.007

.039

.833

.166

.999

Example:

Blood Type probabilities, revisited

~ Bin(10, .461)

Suppose

= 10 individuals are to be selected at random from the population.

Probability table for

= #(Type O)

Binomial model applies

(

) = (.461)

(.539)

– x

n

(1 –



)

4.61

2.48

Slide11

n = 10p = .461

pmf

= function(x)(

dbinom

(x, n, p))

N = 100000x = 0:10bin.dat = rep(x, N*pmf

(x))hist(bin.dat, freq = F, breaks = c(-.5, x+.5), col = "green")axis(1, at = x)axis(2)

Slide12

X ~ Bin(10, .461)

~ Bin(1500, .007)

2.48

Factor

Blood Type+–O.384.077.461A.323.065.388B.094.017.111AB.032.007.039.833.166.999Example: Blood Type probabilities, revisitedSuppose n = 10 individuals are to be selected at random from the population. Probability table for X = #(Type AB–)n = 1500 individuals are toRh Factor

Blood Type

–

.384

.077

.461

.323

.065

.388

.094

.017

.111

.032

.007

.039

.833

.166

.999

Therefore,

(

) =

= 0, 1, 2, …, 1500.

RARE EVENT!

Binomial model applies

Also, can show mean





(

) =

and variance



=  (

–



)

(

) =

n

(1 –



)

10.5

10.43

Slide13

X ~ Bin(10, .461)

~ Bin(1500, .007)

2.48

Factor

Blood Type

–

.384

.077

.461

.323

.065

.388

.094

.017

.111

.032

.007

.039

.833

.166

.999

Therefore,

(

) =

= 0, 1, 2, …, 1500.

RARE EVENT!

Binomial model applies

Also, can show mean





(

) =

and variance



=  (

–



)

(

) =

n

(1 –



)

10.5

10.43

Long positive skew as

 1500

…but contribution

 0

Is there a better alternative?

RARE EVENT!

Slide14

X ~ Bin(1500, .007)

Also, can show mean





p(x) =and variance  2 =  (x – ) 2p(x) = Poisson distribution x = 0, 1, 2, …, where mean and variance are  = n and  2 = nIs there a better alternative?Rh FactorBlood Type+–O.384.077.461A.323.065.388B.094.017.111AB.032.007

.039

.833

.166

.999

Example:

Blood Type probabilities, revisited

Suppose

= 10 individuals are to

be selected at random from the population.

Probability table for

= #(Type AB

–

)

= 1500 individuals are to

Factor

Blood Type

–

.384

.077

.461

.323

.065

.388

.094

.017

.111

.032

.007

.039

.833

.166

.999

Therefore,

(

) =

= 0, 1, 2, …, 1500.

Binomial model applies

RARE EVENT!

10.5

~ Poisson(10.5)

10.5

10.43

n

(1 –



)



Notation

: Sometimes the symbol



(“lambda”) is used instead of



(“mu”).

Slide15

Suppose

= 1500 individuals are to

be selected at random from the population.

Probability table for

= #(Type AB–)Poisson distribution x = 0, 1, 2, …, where mean and variance are  = n and  2 = nRh FactorBlood Type+–O.384.077.461A.323.065.388B.094.017.111AB.032.007.039.833.166.999Example: Blood Type probabilities, revisited

Rh Factor

Blood Type

–

.384

.077

.461

.323

.065

.388

.094

.017

.111

.032

.007

.039

.833

.166

.999

RARE EVENT!

10.5

~ Poisson(10.5)

: Probability of

exactly

= 15 Type(AB–) individuals = ?

Binomial:

Poisson:

(both ≈ .0437)

Therefore,

(

) =

= 0, 1, 2, …, 1500.

Is there a better alternative?

Slide16

Example:

Deaths in Wisconsin

Slide17

Example: Deaths in Wisconsin

Assuming

deaths among young adults are relatively

rare

, we know the following:

Average

584 deaths per year λ = Mortality rate (α) seems constant.Therefore, the Poisson distribution can be used as a good model to make future predictions about the random variable X = “# deaths” per year, for this population (15-24 yrs)… assuming current values will still apply. Probability of exactly X = 600 deaths next year P(X = 600) =0.0131 Probability of exactly X = 1200 deaths in the next two years P(X = 1200) =0.00746R: dpois(600, 584) Mean of 584 deaths per yr  Mean of 1168 deaths per two yrs,so let λ = 1168: Probability of at least one death per day: λ == 1.6 deaths/dayP(X = 1) + P(X = 2) + P(X = 3) + … P(X ≥ 1) =True, but not practical.P(X ≥ 1) =

1 –

P(X = 0)

= 1 –

–1.6

0.798