1. 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
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2. 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))
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3. 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)
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4. 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
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5. 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.
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6. 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) =
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7. 4.1 Mathematical Expectation
Example: Repair costs for a particular machine are represented by the following probability distribution:
What is the expected value of the repairs?
That is, over time what do we expect repairs to cost on average? x
$50
$200
$350
P(X = x)
0.3
0.2
0.5
The expected value or mean of a probability distribution is the long-run theoretical average.
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8. Expected Value – Repair Costs
μ = mean of the probability distribution
For discrete variables,
μ = E(X) = ∑ x f(x)
So, for our example,
E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230
E(x) <weighted average>
E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230
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9. Another Example – Investment
By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0.3 or take a loss of $1000 with a probability of 0.7. What is the investor’s expected gain on the stock?
X $ $1000
P(X)
E(X) = $4000 (0.3) -$1000(0.7) = $500 X
P(X)
E(X) = 4000 (0.3) -1000(0.7) = 500
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10. Expected Value - Continuous Variables
For continuous variables,
μ = E(X) = E(X) = ∫ x f(x) dx
Vacuum cleaner example: problem 7 pg. 92
x, 0 < x < 1
f(x) = 2-x, 1 ≤ x < 2
0, elsewhere
(in hundreds of hours.) {
= 1 * 100 = hours of operation annually, on average
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11. Functions of Random Variables
Ex 4.4. pg. 111: Probability of X, the number of cars passing through a car wash in one hour on a sunny Friday afternoon, is given by
Let g(X) = 2X -1 represent the amount of money paid to the attendant by the manager. What can the attendant expect to earn during this hour on any given sunny Friday afternoon?
E[g(X)] = Σ g(x) f(x) = Σ (2X-1) f(x)
= (2*4-1)(1/12) +(2*5-1)(1/12) …+(2*9-1)(1/6) = $12.67 x 4 5 6 7 8 9
P(X = x)
1/12
1/4
1/6
Σ (2x-1) f(x) = 7(1/12) + 9(1/12) … + 17(1/6) = $12.67
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12. 4.2 Variance of a Random Variable
Recall our example: Repair costs for a particular machine are represented by the following probability distribution:
What is the variance of the repair cost?
That is, how might we quantify the spread of costs? x
$50
200
350
P(X = x)
0.3
0.2
0.5
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13. Variance – Discrete Variables
For discrete variables,
σ2 = E [(X - μ)2] = ∑ (x - μ)2 f(x) = E (X2) - μ2
Recall, for our example, μ = E(X) = $230
Preferred method of calculation:
σ = [E(X2)] – μ2
= 502 (0.3) (0.2) (0.5) – = $17,100
Alternate method of calculation:
σ2 = E(X- μ)2 f(x)
= (50-230)2 (0.3) + ( )2 (0.2) + ( )2 (0.5) = $17,100
E(X2) – μ2 = 502 (0.3) (0.2) (0.5) – 2302 = $17,100
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14. Variance - Investment Example
By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0.3 or take a loss of $1000 with a probability of 0.7. What are the variance and standard deviation of the investor’s gain on the stock?
E(X) = $4000 (0.3) -$1000 (0.7) = $500
σ2 = [∑(x2 f(x))] – μ2
= (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000
σ = $ X
P(X)
E(X) = 4000 (0.3) -1000(0.7) = 500
σ2 = ∑(x2 f(x)) –μ2 = (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000
σ =$
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15. Variance of Continuous Variables
For continuous variables,
σ2 = E [(X - μ)2] =[∫ x2 f(x) dx] – μ2
Recall our vacuum cleaner example pr. 7 pg. 88
x, 0 < x < 1
f(x) = 2-x, 1 ≤ x < 2
0, elsewhere
(in hundreds of hours of operation.)
What is the variance of X? The variable is continuous, therefore we will need to evaluate the integral. {
E(X) = ∫x2 f(x)dx – μ2
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16. Variance Calculations for Continuous Variables
(Preferred calculation)
What is the standard deviation?
σ = hours
[∫01 x3 dx + ∫12x2 (2-x)dx] – μ2 = x4/4 |10 + (2x3/3 – x4/4)| =
σ =
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17. Covariance/ Correlation
A measure of the nature of the association between two variables
Describes a potential linear relationship
Positive relationship
Large values of X result in large values of Y
Negative relationship
Large values of X result in small values of Y
“Manual” calculations are based on the joint probability distributions
Statistical software is often used to calculate the sample correlation coefficient (r)
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