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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.

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chapter 3 random variables and probability distributions
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
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)

EGR 252.001 2011

discrete random variables
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
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
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 view
Another View
  • The probability histogram

EGR 252.001 2011

another discrete probability example
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
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
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
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
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
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
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
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
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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