jmb chapter 3 lecture 1 9th edegr 252 2013 slide 1 chapter 3: random variables and probability...

JMB Chapter 3 Lecture 1 9th ed EGR 252 2013 Slide 1

Chapter 3: Random Variables and Probability Distributions

Definition and nomenclature A random variable is a function that associates a real

number with each element in the sample space. We use a capital letter such as X to denote the

random variable. We use the small letter such as x for one of its

values. Example: Consider a random variable Y which takes

on all values y for which y > 5.


Defining Probabilities: Random Variables

Examples: Out of 100 heart catheterization procedures

performed at a local hospital each year, the probability that more than five of them will result in complications is

P(X > 5)

Drywall anchors are sold in packs of 50 at the local hardware store. The probability that no more than 3 will be defective is

P(Y < 3)


Discrete Random Variables

Pr. 2.51 P.59 (Modified) A box contains 500 envelopes (75 have $100, 150 have $25, 275 have $10) Assume someone spends $75 to buy 3 envelopes.

The sample space describing the presence of $10 bills (H) vs. bills that are not $10 (N) is:

S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH}

The random variable associated with this situation, X, reflects the outcome of the experiment X is the number of envelopes that contain $10 X = {0, 1, 2, 3} Why no more than 3? Why 0?


Discrete Probability Distributions 1

The probability that the envelope contains a $10 bill is 275/500 or .55

What is the probability that there are no $10 bills in the group?P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125

P(X = 1) = 3 * (0.55)*(1-0.55)* (1-0.55) = 0.334125

Why 3 for the X = 1 case? Three items in the sample space for X = 1 NNH NHN HNN


Discrete Probability Distributions 2

P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125

P(X = 1) = 3*(0.55)*(1-0.55)* (1-0.55) = 0.334125

P(X = 2) = 3*(0.55^2*(1-0.55)) = 0.408375

P(X = 3) = 0.55^3 = 0.166375

The probability distribution associated with the number of $10 bills is given by:

x 0 1 2 3

P(X = x)


Another View

The probability histogram

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0 1 2 3

f(x)

x


Another Discrete Probability Example Given:

A shipment consists of 8 computers 3 of the 8 are defective

Experiment: Randomly select 2 computers Definition: random variable X = # of defective computers selected What is the probability distribution for X?

Possible values for X: X = 0 X =1 X = 2 Let’s start with P(X=0) [0 defectives and 2 nondefectives are selected]

Recall that P = specified target / all possible

28

10

2

8

2

5

0

3

)0(

XP

(all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers)


Discrete Probability Example What is the probability distribution for X?

Possible values for X: X = 0 X =1 X = 2 Let’s calculate P(X=1) [1 defective and 1 nondefective are selected]

28

3

2

8

0

5

2

3

)2(

XP

(all ways to get 1 out of 3 defectives) ∩ (all ways to get 1 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers)

28

15

2

8

1

5

1

3

)1(

XP

x 0 1 2

P(X = x)


Discrete Probability Distributions

The discrete probability distribution function (pdf)

f(x) = P(X = x) ≥ 0

Σx f(x) = 1

The cumulative distribution, F(x) F(x) = P(X ≤ x) = Σt ≤ x f(t)

Note the importance of case: F not same as f


Probability Distributions

From our example, the probability that no more than 2 of the envelopes contain $10 bills is P(X ≤ 2) = F (2) = _________________ F(2) = f(0) + f(1) + f(2) = .833625 Another way to calculate F(2) (1 - f(3))

The probability that no fewer than 2 envelopes contain $10 bills is P(X ≥ 2) = 1 - P(X ≤ 1) = 1 – F (1) = ________ 1 – F(1) = 1 – (f(0) + f(1)) = 1 - .425 = .575 Another way to calculate P(X ≥ 2) is f(2) + f(3)


Your Turn … The output of the same type of circuit board from two assembly lines is

mixed into one storage tray. In a tray of 10 circuit boards, 6 are from line A and 4 from line B. If the inspector chooses 2 boards from the tray, show the probability distribution function for the number of selected boards coming from line A.

x P(x)

0

1

2333.0

45

1*15

2

10

0

4

2

6

)2()2(

335.045

4*6

2

10

1

4

1

6

)1()1(

331.045

6*1

2

10

2

4

0

6

)0()0(

XPf

XPf

XPf


Continuous Probability Distributions

The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is

The probability that a given part will fail before 1000 hours of use is

b

a

dxxfbXaP )()(In general,


Visualizing Continuous Distributions

The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is

The probability that a given part will fail before 1000 hours of use is

-5 -3 -1 1 3 5

0 5 10 15 20 25 30


Continuous Probability Calculations

The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈ R

The cumulative distribution, F(x)

1)( dxxf

b

a

dxxfbXaP )()(

x

dttfxXPxF )()()(


Example: Problem 3.7, pg. 92

The total number of hours, measured in units of 100 hoursx, 0 < x < 1

f(x) = 2-x, 1 ≤ x < 20, elsewhere

a) P(X < 120 hours) = P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) NOTE: You will need to integrate two different functions over two different ranges.

b) P(50 hours < X < 100 hours) = Which function(s) will be used?

{

jmb chapter 3 lecture 1 9th edegr 252 2013 slide 1 chapter 3: random variables and probability...

Documents