Review Probability Basic definitions: Randomization experiment - - PowerPoint PPT Presentation

Review

• Probability
• Basic definitions:

Randomization experiment Sample spaces Elementary outcomes Event

• Basic operations—conditional probability
• Bayes Theorem

Objectives

• Random Variable

Discrete random variable Continuous random variable

• Two probability distributions

Binomial distribution Normal distribution

Random variables

• A random variable is a function that assigns numeric

values to different events in a sample space. Usually we denote a random variable using a capital letter X, Y or Z…

• NOTE: (1) Randomness; (2) Numeric values
• Example 1: Randomly select a student from a class.

X=student’s number of siblings. X could be 0, 1, 2 …

• Example 2: Randomly select a student from a class.

X=student’s height. X could be any value bigger than 0

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Two types of random variables

1.

Discrete random variable: their outcomes are set of discrete (isolated) values.

• Eg. X=number of siblings

2.

Continuous random variable: its possible values cannot be enumerated; infinite number of values, all

• utcomes have probability zero. p(x)=0 for every x.
• Eg. X=the student’ height

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• EG1. Tossing two coins

let X=number of heads

Notation: X: variable x: observed values

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Outcome TT HT TH HH x 1 2

Probability distribution function

• A probability distribution function (pdf) is a

mathematical relationship, or rule, that assigns to any possible value x of a discrete random variable X the probability Pr(X=x).

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Probability Distribution of the Random Variable

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X=number of heads.

Outcome TT WT TW WW x 1 2 P(X=x) 1/4 1/2 1/4

Probability histogram

• EG2. Tossing two dice

Y: the sum of the dots on the two Dice. What’s the possible values of Y?

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Probability Distribution of the Random Variable

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Y: the sum of the dots on the two Dice.

frequency of occurrences frequency of all possible occurrences Probability = 0 ≤ Probability ≤ 1

Relative frequency

In practice, the probability can be estimated by the relative frequency of an event “in a long run”. Relative frequency histogram should look very much like the probability histogram, if the experiment is repeated many times.

Data set vs. Probability distributions

 Sample properties—based on data set

Sample mean: Sample variance:

 Model or population properties—based on

probability distribution. Population mean: Population variance:

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1 2 2 1

/ 1 ( ) 1

n i i n i i

x x n s x x n

= =

= = − −

∑ ∑

1 2 2 1

Pr( ) ( ) Pr( )

R i i i R i i i

x X x x X x µ σ µ

= =

= = = − =

∑ ∑

Mean of Random Variable

 Mean or expected value of X, denoted as E(X)

• r µ, is defined as

 It is the sum of the possible values, each

weighted by its probability

 Expectation represents “average” value of the

random variable

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∑

=

= = =

R i i i

x X x X E

1

) Pr( ) ( µ

Mean of X

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X=number of heads.

Outcome TT WT TW WW x 1 2 P(X=x) 1/4 1/2 1/4 xP(x) 1/2 1/2

3 1

( ) Pr( ) 1

i i i

E X x X x µ

=

= = = =

∑

Variance of Random Variable

 The variance of X is the expected squared

distance from the population mean.

 The standard deviation σ is the square root of

variance

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∑

=

= − = =

R i i i

x X x X Var

1 2 2

) Pr( ) ( ) ( µ σ

) ( ) ( X Var X sd = = σ

Variance of X

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X=number of heads.

Thus, Summary, µ and σ are computed from probability

• distribution. They are population properties.

x P(x) (X-µ)2 P(x) 0.25 (0-1)2*0.25=0.25 1 0.5 (1-1)2*0.25=0 2 0.25 (2-1)2*0.25=0.25 Total 0.50

2

0.5 σ =

Two types of random variables

1.

Discrete random variable: their outcomes are set of discrete (isolated) values.

2.

Continuous random variable: its possible values cannot be enumerated; infinite number of values, all

• utcomes have probability zero. p(x)=0 for every x.

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Continuous random variables

• A balanced spinning pointer.

Can stop anywhere in the circle

• X—the proportion of the

total circumference it lands

• n.
• X can be any value between

0 and 1. Infinite values.

• p(0.25≤x ≤0.75)=0.5
• p(x=0.5)=0, for x can take on

an infinite number of values.

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Probability density function(pdf) of X

• The curve is

the probability density function (pdf) of the random variable X

• Pr(a≤X ≤b)= is the area

under the curve between the x value a and b.

• The total area under the

density function curve over the entire range of possible values for the random variable is 1

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( ) y f x =

( ) ( )

b a

P a X b f x dx ≤ ≤ = ∫

( ) y f x =

( ) ( ) 1 P X f x dx

∞ −∞

−∞ ≤ ≤ ∞ = =

∫

Probability density function(pdf) of X

• The pdf has large values in

regions of high probability and small values in regions of low probability

• Pr(X=x)=0 for any specific

value x

• Generally, a distinction is not

made between probabilities such as Pr(X<x) and Pr(X≤x), Pr(a≤X≤b) and Pr(a<X<b) when X is a continuous

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( ) y f x =

Expectation and variance of a continuous random variable

• Mean :

Center of the probability density

• Variance :

Spread of the probability density

• The standard deviation, or σ, is the square root of

the variance, that is,

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2

σ

) (X Var = σ

(X) ( ) E xf x dx µ

∞ −∞

= = ∫

2 2

(X) ( ) ( ) Var x f x dx σ µ

∞ −∞

= = −

∫

µ

Two distributions

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 Binomial --discrete  Normal -- continuous

Bernoulli trial

23 Examples:

 A heads-or-tails Coin toss  A win-or-lose football game  A pass-or-fail automotive smog inspection

Properties:

 Two outcomes: success or failure  Success probability(p) is the same in each

trial

 Trials are independent.

Binomial random variable

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• --X is the number of success in n repeated

Bernoulli trial with probability p of success.

Success probability(p) is the same in each

trial

Trials are independent.

Binomial random variable

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Probability Distribution: the probability of

• btaining k successes in n trial, with success

probability p: : counts all possible ways of getting k success and n-k failures : probability for getting k success and n-k failures

( ) (1 )

k n k

n P X k p p k

−

  = = −    

! !( )! n n k k n k   =   −  

(1 )

k n k

p p

−

−

! ( 1) ... 1 where n n n = × − × ×

Mean and Variance of the Binomial Distribution

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2

(1 ) np np p µ σ = = −

Exercise

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Newborns were screened for HIV in a Massachusetts

• hospital. The positive rate for inner-city baby is p=0.01.

If 500 newborns are screened,

• 1. what is the exact binomial probability of 5 HIV

positive test results?

Exercise

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Newborns were screened for HIV in a Massachusetts

• hospital. The positive rate for inner-city baby is p=0.01.

If 500 newborns are screened,

• 1. what is the exact binomial probability of 5 HIV

positive test results? Answer: EXCEL: BINOMDIST(5,500,0.01,FALSE)

5 495

500 ( 5) 0.01 (1 0.01) 5 0.176 P X   = = −     =

Exercise

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Newborns were screened for HIV in a Massachusetts

• hospital. The positive rate for inner-city baby is p=0.01.

If 500 newborns are screened,

• 2. What is the exact binomial probability of at least 5

HIV positive test results?

Exercise

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Newborns were screened for HIV in a Massachusetts

• hospital. The positive rate for inner-city baby is p=0.01.

If 500 newborns are screened,

• 2. What is the exact binomial probability of at least 5

HIV positive test results?

Answer: EXCEL: F(4)= BINOMDIST(4,500,0.01,TRUE)

( 5) 1 ( 4) 0.44 4) .5 1 6 ( 1 P X P X F ≥ = − ≤ = − = = −

Normal distribution

• Normal distribution is also called Gaussian

distribution, after the well-known mathematician Karl Gauss (1777-1855, “the Prince of Mathematicians“)

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Normal distribution

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• Normal distribution is very useful
• Many things closely follow a normal distribution
• Heights of people
• Errors in measurement
• Blood pressure
• Scores on a test
• Many other distributions can be made approximately

normal by transformation—Binomial et al.

• Most statistical methods considered in this text are

based on normal distribution

The pdf of normal distribution

• The normal distribution is defined by its pdf, which is

given as for some parameters µ and σ

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2 2

( ) 2

1 ( ) 2

x

f x e

µ σ

πσ

  − −     

=

Other properties of Normal pdf

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• Mean=median=mode
• Symmetry about the center
• 50% of values less than the mean

Location is measured by µ

• In the graph, µ2>µ1

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Spread is measured by σ2

• In the graph, σ2>σ1

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Standard normal distribution N(0, 1)

• A normal distribution with mean 0 and variance 1

is called a standard normal distribution. Denoted as N(0, 1)

• In the following, we will examine the standard

normal distribution N(0, 1) in details.

• We will see that any information concerning a

general normal distribution N(µ, σ2) can be

• btained from appropriate manipulations of an

N(0,1) distribution

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Density of standard normal N(0,1)

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1 µ σ = =

2

2

1 ( ) 2

x

f x e π

−

=

Properties of the standard normal N(0, 1)

• It can be shown that about 68% of the area under the standard

normal density lies between -1 and +1, about 95% of the area lies between -2 and +2, and about 99% lies between -2.5 and +2.5 NOTE: You will see that, more precisely, Pr(-1<x<1)=0.6827, Pr(-1.96<X<1.96)=0.95, Pr(-2.576<X<2.576)=0.99

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Cumulative probability

• The cumulative distribution function (cdf) for a

standard normal distribution is denoted by Φ(x)=Pr(X≤x), where Z~N(0,1)

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( ) ( ) F a P Z a = ≤

Excel: F(a): NORMSDIST(a);

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( ) ( ) ( ) P a Z b F b F a ≤ ≤ = −

Excel: F(1): NORMSDIST(1); F(-1): NORMSDIST(-1);

( 1 1) (1) ( 1) =0.8413-0.1587 =0.6826 P Z F F − ≤ ≤ = − −

• Eg.

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( ) 1 ( ) P Z a F a ≥ = −

( 1) 1 (1) =1-0.8413 =0.1587 P Z F ≥ = −

(1) NORMSDIST(1)

Excel: F(1): NORMSDIST(1);

How to standardize the normal distribution?

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How to standardize the normal distribution?

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X Z µ σ − =

Then Z has a standard normal distribution, Z ~ N(0, 1)

Standardization

• IF X~ N(µ, σ2) and

then Z~N(0,1) Then

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P( ) P( ) ( ) ( ) a b b a a X b Z F F µ µ µ µ σ σ σ σ − − − − < < = < < = −

X Z µ σ − =

Use standardization for many problems

• Example:If X~N(80, 12^2), what is Pr(90<X<100)?
• Solution:

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90 80 80 100 80 Pr(90 100) Pr( ) 12 12 12 Pr(0.83 1.67) =F(1.67)-F(0.83) =0.9522-0.7977 X X Z − − − < < = < < = < < =0.155

Always draw a graph…

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Exercise

• Suppose we know that among men aged 30-34 who have

ever smoked, the mean number of years they smoked is 12.8 with a standard deviation of 5.1 years. Assuming that the duration of smoking is normally distributed, what proportion of men in this age group have smoked for more than 20 years?

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Exercise

Suppose we know that among men aged 30-34 who have ever smoked, the mean number of years they smoked is 12.8 with a standard deviation of 5.1 years. Assuming that the duration of smoking is normally distributed, what proportion of men in this age group have smoked for more than 20 years? Answer: We have And we need to compute

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2

~ (12.8, 5.1 ) X N ( 20) P X >

( 20) 1 ( 20) 20 12.8 =1- (Z ) 5.1 1 (1.412) =1-0.9210=0.079 P X P X P F > = − ≤ − ≤ = −

EXCEL: NORMDIST(20,12.8,5.1,TRUE) Or NORMSDIST(1.412)