1. Continuous probability distributions
Examples:
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, __________
P(90 < T < 95)
P(X < 1000)
Probability density function f(x)
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2. Understanding 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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3. Continuous probability distributions
The continuous probability density function (pdf)
f(x) ≥ 0, for all x ∈ R
The cumulative distribution, F(x)
Expectation and variance
μ = E(X) = ______________ σ2 = ________________
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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4. { An example x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere
1st – what does the function look like?
P(X < 1.2) = ___________________
P(0.5 < X < 1) = ___________________
E(X) = -∞∫∞ x f(x) dx = ________________________ {
P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) =
= ∫01 x dx + ∫11.2(2-x)dx = (x2/2)|01 + (2x - x2/2)| 11.2 =0.5 + [ ] =0.68
P(.5 < X < 1) = ∫0.51 x dx = (x2/2)|.51 = 0.375
E(x) = ∫01 x2 dx + ∫12x(2-x)dx = x3/3 |10 + (x2 – x3/3)|12 = 1
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5. Continuous probability distributions
Many continuous probability distributions, including:
Uniform
Exponential
Gamma
Weibull
Normal
Lognormal
Bivariate normal
Ch. 6:
Uniform
Exponential
Gamma
Weibull -
Ch. 7:
Normal –
Lognormal
Bivariate normal
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6. Uniform distribution
Simplest – characterized by the interval endpoints, A and B.
A ≤ x ≤ B
= 0 elsewhere
Mean and variance:
and
draw distribution
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7. Example
A circuit board failure causes a shutdown of a computing system until a new board is delivered. The delivery time X is uniformly distributed between 1 and 5 days.
What is the probability that it will take 2 or more days for the circuit board to be delivered?
interval = [1,5]
f(x) = 1/(B-A) = 1/(5-1) = ¼, 1 < x < 5 (0 elsewhere)
First: show the distribution and demonstrate the “intuitive” answer
Then –
P(X>2) = ∫25 (1/4)dx = 0.75
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8. Gamma & Exponential distributions
Recall the Poisson Process
Number of occurrences in a given interval or region
“Memoryless” process
Sometimes we’re interested in the time or area until a certain number of events occur.
For example
An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process.
What is the probability that up to a minute will elapse before 2 calls arrive?
How long before the next call?
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9. Gamma distribution
The density function of the random variable X with gamma distribution having parameters α (number of occurrences) and β (time or region, 1/λ).
x > 0.
Describes the time until a specified # of Poisson events occurs.
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10. Exponential distribution
Special case of the gamma distribution with α = 1.
x > 0.
Describes the time until or time between Poisson events.
μ = β
σ2 = β2
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11. Example β = ________ α = ________
An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process.
What is the probability that up to a minute will elapse before 2 calls arrive?
β = ________ α = ________
P(X ≤ 1) = _________________________________
β = 1/λ = 1/(2.7) =
α = 2
P(X < 1) = 0∫1 (1/ β2) x e-x/ β dx = ∫1 x e -2.7x dx = [-2.7xe-2.7x – e-2.7x]01
= 1 – e-2.7 ( ) =
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12. Example (cont.) β = ________ α = ________
What is the expected time before the next call arrives?
β = ________ α = ________
μ = _________________________________
Exponential distribution
μ = β = 1/2.7 = min.
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13. Another example Example 6-6, page 136.
Note: in Excel, use =GAMMADIST(x,alpha,beta,cumulative) to find the probability associated with a specific value of x. Use =GAMMAINV(probability,alpha,beta) to find the value of x given a probability.
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14. Wiebull Distribution
Used for many of the same applications as the gamma and exponential distributions, but
does not require memoryless property of the exponential
Uses –
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15. Example
Designers of wind turbines for power generation are interested in accurately describing variations in wind speed, which in a certain location can be described using the Weibull distribution with α = and β = 2. A designer is interested in determining the probability that the wind speed in that location is between 3 and 7 mph.
P(3 < X < 7) = ___________________________
In Excel, =WEIBULL(x,alpha,beta,cumulative)
P(3 < X < 7) = F(7) – F(3)
= [1 – e – (0.02)7^2] – [1 – e – (0.02)3^2] = = 0.46
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16. Normal distribution The “bell-shaped curve”
Also called the Gaussian distribution
The most widely used distribution in statistical analysis
forms the basis for most of the parametric tests we’ll perform later in this course.
describes or approximates most phenomena in nature, industry, or research
Random variables (X) following this distribution are called normal random variables.
the parameters of the normal distribution are μ and σ (sometimes μ and σ2.)
note: nonparametric tests are distribution-free, assume no underlying distribution (see ch 16)
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17. Normal distribution
The density function of the normal random variable X, with mean μ and variance σ2, is
, all x.
In Excel, =NORMDIST(x,mean,standard_dev,cumulative) and =NORMINV(probability,mean,standard_dev)
(μ = 5, σ = 1.5)
properties of the curve:
peak is both the mean and the mode and occurs at x = μ
curve is symmetrical about a vertical axis through the mean
total area under the curve and above the horizontal axis = 1.
points of inflection are at x = μ + σ
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18. Standard normal RV …
Note: the probability of X taking on any value between x1 and x2 is given by:
To ease calculations, we define a normal random variable
where Z is normally distributed with μ = 0 and σ2 = 1 and,
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19. Standard normal distribution
Table II, pg : “Cumulative Standard Normal Distribution”
In Excel, =NORMSDIST(z) and =NORMSINV(probability)
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20. Examples P(Z ≤ 1) = Φ(1) = P(Z ≥ -1) = P(-0.45 ≤ Z ≤ 0.36) =
draw the area on the picture …
1. P(Z < 1) =
2. P(Z ≥ -1) = 1 – Φ(-1) =
3. P(-0.45 ≤ Z ≤ 0.36) = Φ(0.36) – Φ(-0.45) = – = 0.316
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21. Your turn … Use Excel to determine (draw the picture!) 1. P(Z ≤ 0.8) =
4. P(Z ≤ -2.0 or Z ≥ 2.0) =
1. P(Z ≤ 0.8) =
2. P(Z ≥ 1.96) = 1 – = (=P(Z < -1.96)) Note symmetry!!
3. P(-0.25 ≤ Z ≤ 0.15) = – =
4. P(Z ≤ -2.0 or Z ≥ 2.0) = 2 * =
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22. The normal distribution “in reverse”
Example:
Given a normal distribution with μ = 40 and σ = 6, find the value of X for which 45% of the area under the normal curve is to the left of X.
If Φ(k) = 0.45,
k = ___________
Z = _______
X = _________
Will X be greater than 40?
k =
Z = = (X – 40)/6
Z = (-0.125*6)+40 = 39.25
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23. Normal approximation to the binomial
If n is large and p is not close to 0 or 1, or
if n is smaller but p is close to 0.5, then
the binomial distribution can be approximated by the normal distribution using the transformation:
NOTE: add or subtract 0.5 from X to be sure the value of interest is included (draw a picture to know which)
Look at example 7-10, pg. 156
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24. Another example
The probability of a patient making a full recovery from a rare heart ailment is 0.4. If 100 people are diagnosed with this ailment, what is the probability that fewer than 30 of them recover? p = 0.4 n = 100 μ = ____________ σ = ______________ if x = 30, then z = _____________________ and, P(X < 30) = P (Z < _________) = _________
μ= np = 100*0.4 = 40
σ = sqrt(npq) = sqrt(100*0.4*0.6) = 4.899
draw the picture!
z = ((30-0.5) – 40)/4.899 = -2.14
P(Z < -2.14) =
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25. Your Turn Refer to the previous example, DRAW THE PICTURE!!
What is the probability that more than 50 survive?
What is the probability that exactly 45 survive?
x > 50, z = (( ) – 40)/4.899 = 2.14
P(Z > 2.14) = P(Z < -2.14) = (by symmetry)
2. x = 45, z1 = ( )/4.899 = 1.12
z2 = (44.5 – 40)/4.899 =
P(X = 45) = P(z < 1.12) – P(z < 0.98) = – =
NOTE: b(45;100,0.4) = (100 choose 45)* =
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26. Lognormal Distribution
When the random variable Y = ln(X) is normally distributed with mean μ and standard deviation σ, then X has a lognormal distribution with the density function,
Uses – reliability and maintainability; environmental engineers – concentration of pollutants, particle size in emissions; long term rate of return on stock investments 26
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27. Example
The average rate of water usage (in thousands of gallons per hour) by a certain community is know to involve the lognormal distribution with parameters μ = 5 and σ =2. What is the probability that, for any given hour, more than 50,000 gallons are used?
P(X > 50,000) = __________________________
P(X > 50,000) = 1 – P(X < 50,000) = 1 – P (Z < (ln(50,000) – 5)/ 2) = 1 – P (Z < 2.91) = = 27
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28. Bivariate Normal distribution
Random variables [X, Y] that follow a bivariate normal distribution have a joint density function,
and a joint probability,
This would be pretty nasty to calculate, except …
X ∼ N (μX, σ2X ) and Y ∼ N (μY , σ2Y )
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29. So… It can be shown (see pp. 161-162) that,
This allows us to use the standard normal approach to solve several important problems.
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30. Example
Example7-14, pg. 163
Given the information regarding shear strength and weld diameter in the example, what is the probability that the strength specification (1080 lbs.) can be met if the weld diameter is 0.18 in.?
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