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.
P(X > 5)
P(Y < 3)
a random variable, X, is a function that associates a real number with each element in the sample space and x is one of the values that X can take.
JMB Chapter 3 Lecture 1 9th ed
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2. 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)
P(X > 5)
P(Y < 3)
a random variable, X, is a function that associates a real number with each element in the sample space and x is one of the values that X can take.
JMB Chapter 3 Lecture 1 9th ed
EGR

3. Discrete Random Variables
Pr 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?
Note: if the number of possible solutions is countable, the variable is discrete
S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH}
X = {0, 1, 2, 3}
the probability distribution function (x, f(x)) – see definition 3.4, pg. 66
JMB Chapter 3 Lecture 1 9th ed
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4. 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) =
P(X = 1) = 3 * (0.55)*(1-0.55)* (1-0.55) =
Why 3 for the X = 1 case?
Three items in the sample space for X = 1
NNH NHN HNN
P(X=0) = P(not in the 1st envelope ∩ not in the 2nd ∩ not in the 3rd) = (1-275/500)3 = (0.45)3 =
P(0) =(1-0.55)^3 =
P(1) =3*((0.55)*(1-0.55)^2) =
P(2) =3*(0.55^2*(1-0.55)) =
P(3) = 0.55^3 =
(students fill in the table)
JMB Chapter 3 Lecture 1 9th ed
EGR

5. Discrete Probability Distributions 2
P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) =
P(X = 1) = 3*(0.55)*(1-0.55)* (1-0.55) =
P(X = 2) = 3*(0.55^2*(1-0.55)) =
P(X = 3) = 0.55^3 =
The probability distribution associated with the number of $10 bills is given by:
P(X=0) = P(not in the 1st envelope ∩ not in the 2nd ∩ not in the 3rd) = (1-275/500)3 = (0.45)3 =
P(0) =(1-0.55)^3 =
P(1) =3*((0.55)*(1-0.55)^2) =
P(2) =3*(0.55^2*(1-0.55)) =
P(3) = 0.55^3 =
(students fill in the table) x 1 2 3
P(X = x)
JMB Chapter 3 Lecture 1 9th ed
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6. Another View The probability histogram JMB Chapter 3 Lecture 1 9th ed
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7. 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)
P(X = 0) = P(0 defectives and 2 nondefective) =
(all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefective)
(all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers)
JMB Chapter 3 Lecture 1 9th ed
EGR

8. 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) x 1 2
P(X = x)
JMB Chapter 3 Lecture 1 9th ed
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9. 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
JMB Chapter 3 Lecture 1 9th ed
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10. 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) =
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)) = = .575
Another way to calculate P(X ≥ 2) is f(2) + f(3)
F(2) = f(0) + f(1) + f(2) =  (OR 1 - f(3))
1 – F(1) = 1 – (f(0) + f(1)) = =  (OR f(2) + f(3))
JMB Chapter 3 Lecture 1 9th ed
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11. 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) 1 2
P(x = 0) =
JMB Chapter 3 Lecture 1 9th ed
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12. Continuous Probability Distributions
In general,
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
Probability density function f(x)
JMB Chapter 3 Lecture 1 9th ed
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13. 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
JMB Chapter 3 Lecture 1 9th ed
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14. Continuous Probability Calculations
The continuous probability density function (pdf)
f(x) ≥ 0, for all x ∈ R
The cumulative distribution, F(x)
Example: the uniform distribution (i.e., f(x) = 1, 1 < x < 2)
1. what is the area of the rectangle? (1) The total area under the curve is P(S) and so will always be 1.
JMB Chapter 3 Lecture 1 9th ed
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15. 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? {
P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) = ∫01xdx + ∫11.2 (2-x)dx = (x2/2)|01 + (2x- x2/2)|11.2 =0.68
P(.5 < X < 1) =
JMB Chapter 3 Lecture 1 9th ed
EGR