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Chapter 3: Random Variables and Probability DistributionsPowerPoint Presentation

Chapter 3: Random Variables and Probability Distributions

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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.

EGR 252.001 2011

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)

- 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

EGR 252.001 2011

Discrete Random Variables

- Problem 2.51 Page 59 (Modified)
- 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}

EGR 252.001 2011

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

EGR 252.001 2011

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:

EGR 252.001 2011

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

(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)

EGR 252.001 2011

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]

(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)

EGR 252.001 2011

Discrete Probability Distributions

- The discrete probability distribution function (pdf)
- f(x) = P(X = x) ≥ 0
- Σxf(x) = 1

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

- Note the importance of case: F not same as f

EGR 252.001 2011

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)

EGR 252.001 2011

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.

EGR 252.001 2011

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

In general,

EGR 252.001 2011

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

EGR 252.001 2011

Continuous Probability Calculations

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

- The cumulative distribution,F(x)

EGR 252.001 2011

Example: Problem 3.7, pg. 92

The total number of hours, measured in units of 100 hours

x, 0 < x < 1

f(x) = 2-x, 1 ≤ x < 2

0, elsewhere

- 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?

{

EGR 252.001 2011

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