Random Variable & Discrete Distribution

Random Variable & Discrete Distribution

Random Variable - 1

1

Theoretical Probability Distribution

• Discrete Probability Distributions

• Continuous Probability Distributions

2

Discrete Distributions

• Bernoulli • Binomial • Poisson • Geometric • …

3

Bernoulli Distribution Model (Bernoulli Probability Distribution)

4

Bernoulli Trial

Definition: Bernoulli trial is a random experiment whose outcomes are classified as one of the two categories. (S , F) or (Success, Failure) or (1, 0)

Example:

Tossing a coin, observing Head or Tail

Observing patient’s status Died or Survived.

5

Example: (Tossing a balanced coin)

P(S) = P(X=1) = p = .5

P(F) = P(X=0) = 1 p = .5

Bernoulli Distribution

.5

0 1

Bernoulli Probability Distribution

6

Bernoulli Probability Distribution

Example: In a random experiment of casting a balanced die, we are only interested in observing 6 turns up or not. It is a Bernoulli trail.

P(6) = P(X=1) = p = 1/6

P(6’) = P(X=0) =1 1/6 = 5/6

Bernoulli Distribution

0 1


Random Variable - 2

7

Binomial Distribution Model (Binomial Probability Distribution)

8

Binomial Experiment

A random experiment involving a sequence of independent and identical Bernoulli trials.

Example:

Toss a coin ten times, and observing Head turns up.

Roll a die 3 times, and observing a 6 turns up or not.

In a random sample of 5 from a large population, and observing subjects’ disease status. (Almost binomial)

9

Binomial Probability Model

A model to find the probability of having x number successes in a sequence of n independent and identical Bernoulli trials.

10

Binomial Probability Model

In a binomial experiment involving n independent and identical Bernoulli trials each with probability of success p, the probability of having x successes can be calculated with the binomial probability mass function, and it

is, for x = 0, 1, …, n,

xnx

xnx

ppx

n

ppxnx

nxXP

)1(

)1()!(!

!)(

11

Factorial

n! = 1·2·3·... ·n

0! = 1

Example: 3! = 1·2·3 = 6

Example:

1012123

12345

!2)!25(

!5

2

5

12

Binomial Probability

Example: A balanced die is rolled three times (or three balanced dice are rolled), what is the probability to see two 6’s?

Identify n = 3, p = 1/6, x = 2

xnx ppxnx

nxXP

)1(

)!(!

!)(

(6, 6’, 6’)

(6’, 6’, 6)

(6’, 6, 6’)

069.

6/56/13

)6/5()6/1()!23(!2

!3)2(

12

232

XP


Random Variable - 3

13


Example: If 10% of the population in a community have a certain disease, what is the probability that 4 people in a random sample of 5 people from this community has the disease? (Assume binomial experiment.) Identify n = 5, p = .10, x = 4

xnx ppxnx

nxf

)1(

)!(!

!)(

0004.

9.1.5

)10.1()10(.)!45(!4

!5)(

14

454

xf

14


Example: In the previous problem, what is the probability that 4 or more people have the disease?

Identify n = 5, p = .10, x = 4

0004.

00046.00001.00045.

)9(.)1(.!0!5

!5)9(.)1(.

!1!4

!5

)5()4()5()4()4(

0514

ffXPXPXP

(What is this number telling us?)

15

Parameters of Binomial Distribution

Parameters of the distribution:

Mean of the distribution, = n·p

Variance of the distribution, 2 = n·p·(1 p)

Standard deviation, , is the square root of

variance.

16

0 1 2 3 4 5

P(0) = .5905

P(1) = .3281

P(2) = .0729

P(3) = .0081

P(4) = .0004

P(5) = .00001

Binomial Distribution

0.590

n = 5, p = .10

= 5 x .10 = .5

2 = 5 x .1 x (1.1) = .45

17

Continuous Distribution

• Normal Distribution

• Exponential Distribution

• …

18

Relative Frequency Histogram

Percent

10 20 30 40 50 60 70 80 90 100 110

Test scores


Random Variable - 4

19

Density Curve

Density

function, f (x)

A smooth curve that fit

the distribution

Percent

Use a mathematical model to describe the variable.

10 20 30 40 50 60 70 80 90 100 110

Test scores

20

Continuous Random Variable

Probability Is Area

Under Curve!

f(x)

X c d

P c X d f x dx c

d ( ) ( )

21

Uniform Skewed to right Symmetrical Skewed to left

22

Normal Distribution 1. ‘Bell-Shaped’ &

Symmetrical

X

f(X)

Mean

Median

Mode

2. Mean, Median, Mode Are Equal

3. Random Variable Has Infinite Range

< x <

23

Normal Probability Density Function

2

2

1

e2

1)(

x

xf

f (x ) = Density of Random Variable x = Mean of the Distribution = Standard Deviation of the Distribution = 3.14159…; e = 2.71828… x = Value of Random Variable ( < x < )

Notation: N , A normal distribution with

mean and standard deviation

24

Effect of Varying Parameters ( & )

X

f(X)

A C

B


Random Variable - 5

25

Normal Distribution Probability

dxxfdxcPd

c )()(

c dx

f(x)

Probability is

area under

curve!

26

Standard Normal Distribution

Standard Normal Distribution:

A normal distribution with mean = 0 and standard deviation = 1.

Notation:

Z ~ N ( = 0, = 1)

0 Z

= 1

Cap letter Z

27

Area under Standard Normal Curve

How to find the proportion of the are under the standard normal curve below z or say P ( Z < z ) = ?

Use Standard Normal Table!!!

0 z

Z

28

P (1 < Z < 3)

P (Z > 3)

1 0 3 Z

0 3 Z

29


P(Z > 0.32) = Area above .32 = .374

0 .32

Z .00 .01 .02

0.0 .500 .496 .492

0.1 .460 .456 .452

0.2 .421 .417 .413

0.3 .382 .378 .374

Areas in the upper tail of the standard normal distribution

30


Random Variable - 6

31


P(0 < Z < 0.32) = Area between 0 and .32 = .126

0

Area = .5 - .374 = .126

.32

Z .00 .01 .02

0.0 .500 .496 .492

0.1 .460 .456 .452

0.2 .421 .417 .413

0.3 .382 .378 .374


32


P(Z< 0.32) = Area below .32 = .626

0

Area = 1 - .374 = .626

.32

Z .00 .01 .02

0.0 .500 .496 .492

0.1 .460 .456 .452

0.2 .421 .417 .413

0.3 .382 .378 .374


33

-1.00 0 1.00

P ( -1.00 < Z < 1.00 ) = _____

.341 .341

.682

34

-2.00 0 2.00

P ( -2.00 < Z < 2.00 ) = _____

35

-3.00 0 3.00

P ( -3.00 < Z < 3.00 ) = _____

36

- 1.40 0 2.33

P ( -1.40 < Z < 2.33 ) = ____

.490 .419

.909


Random Variable - 7

37

Standardize the Normal Distribution

X

Normal

Distribution

ZX

One table!

= 0

= 1

Z

Standardized

Normal Distribution

38

N ( 0 , 1)

0

bZ

aPbXaP )(

Standardize the Normal Distribution

a b

N ( , )

Normal

Distribution

X

Standard

Normal

Distribution

Z

a

b

39

Normal Distribution Example

X= 5

= 10

6.2

Normal

Distribution

For a normal distribution that has

a mean = 5 and s.d. = 10, what

percentage of the distribution is

between 5 and 6.2?

40

Standardizing Example

X= 5

= 10

6.2

Normal

Distribution

12.10

52.6

XZ

Z= 0

= 1

.12

Standardized

Normal Distribution P(5 X 6.2

P(0 Z .12

41

Z= 0

= 1

.12

Z .00 .01 .02

0.0 .500 .496 .492

0.1 .460 .456 .452

0.2 .421 .417 .413

0.3 .382 .378 .3745

Obtaining the Probability

.452

Standardized Normal

Probability Table (Portion)

Area = .5 - .452 = .048 42

Example P(2.9 X 7.1)

5

= 10

2.9 7.1 X

Normal

Distribution

0

= 1

-.21 Z.21

Standardized

Normal Distribution

.166

Area = .083 + .083 = .166

P(2.9 X 7.1) =P.21 Z .21

21.10

51.7

21.10

59.2

XZ

XZ


Random Variable - 8

43

Example P(X > 8)

X = 5

= 10

8

Normal

Distribution

Standardized

Normal Distribution

30.10

58

XZ

Z = 0

= 1

.30

.382

Area = .382

P(X > 8) =P(Z > .30

44

Example P(X > 8)

X = 5

= 10

8

Normal

Distribution

62% 38%

Value 8 is the 62nd percentile

45

More on Normal Distribution

The work hours per week for residents in

Ohio has a normal distribution with =

42 hours & = 9 hours. Find the

percentage of Ohio residents whose work

hours are

A. between 42 & 60 hours.

P(42 X 60) =?

B. less than 20 hours.

P(X 20) = ?

46

P(42 X 60) = ?

Normal

Distribution

29

4260

XZ

Standardized

Normal Distribution

09

4242

XZ

X

= 200

2400 Z 0

= 1

2.0

9

42 60 2

P(42 Z 60

P0 Z 2

.477 47.7% .477

47

P(X 20) = ?

Normal

Distribution

Standardized

Normal Distribution

44.29

4220

XZ

X

= 200

2400 Z 0

= 1

2.0

9

20 42 -2.44

P(X 20) = P(Z 2.44

= 0.007 = 0.7%

0.007

48

Finding Z Values for Known Probabilities

Standardized Normal

Probability Table (Portion) What is z given

P(Z < z) = .80 ?

Z … .04 .05

…

.264 .261 .258

0.7 .233 .230 .227

0.8 .203 .200 .198

0.9 .176 .174 .171

0

.80 .20

z

Upper Tail Area = 1 - .80

= .20

z = .84

= .84


Random Variable - 9

49

Finding X Values for Known Probabilities

Example: The weight of new born

infants is normally distributed with a

mean 7 lb and standard deviation of

1.2 lb. Find the 80th percentile.

Area to the left of 80th percentile in 0.200.

In the table there is a area value 0.200

corresponding to a z-score of .84.

80th percentile = 7 + .84 x 1.2 = 8.008 lb

50

Finding X Values for Known Probabilities

Example: The Body Mass Index for a

particular population is normally

distributed with a mean 22 and

standard deviation of 4. Find the 80th

percentile.

Area to the left of 80th percentile in 0.200.

In the table there is a area value 0.200

corresponding to a z-score of .84.

80th percentile = 22 + .84 x 4 = 25.36

Random Variable & Discrete Distribution

Documents

Transcript of Random Variable & Discrete Distribution