Engineering 323 Beautiful Homework #7 Page 1 of 5 Tyler Oxford Problem 4-4

4.4 Let X denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The paper “Blade Fatigue Life Assessment with Application to VAWTS” (J. Solar Energy Engr., 1982, pp. 107-111) proposes the Rayleigh distribution, with pdf

>

=−

otherwisexex

xfx

00*)/(

);()2/(2 22 θθ

θ

as a model for the X distribution.

a) Verify that f(x;θ) is a legitimate pdf.

b) Suppose θ = 100 (a value suggested by a graph in the paper). What is the probability that X is at most 200? Less than 200? At Least 200?

c) What is the probability that X is between 100 and 200?

d) Give an expression for P(X ≤ x).

Definitions Rayleigh distribution is a standard in the industry of wind turbines for performance ratings. This is a special case of Weibull distribution where the shape parameter is two. We will be learning the Weibull Distribution later on in chapter four. It is also explained and can be seen at the site Describing Wind Variations: Weibull Distribution. There is a great graphing demonstration program within this site if you click HERE! Probability Distribution or Probability Density Function (pdf) of X, where X is a continuous random variable (rv), is a function f(x) such that for any two numbers a and b with a ≤ b,

∫=≤≤b

a

dxxfbXaP )()(

This equation means that the probability of X having any value in the interval [a to b] is the area under the graph of the density function. The graph is called the density curve (Devore page 140). For f(x) to work as a pdf the following two assumptions must be correct:

Engineering 323 Beautiful Homework #7 Page 2 of 5 Tyler Oxford Problem 4-4

xallforxf 0)()1 ≥

∫∞

∞−

== 1)()()2 xfofgraphentiretheunderareathedxxf

If X is a continuous rv, then for any number c, P(X = c) =0. So for any two number’s a and b with a < b, P(a ≤ X ≤ b) = P(a < X ≤ b) P(a ≤ X < b ) = P(a < X< b) This equation can be found in Devore page 141. It essentially tells us that with continuous rv’s, it does not matter if we use the “less than” or “less than or equal to” sign. This is because with a continuous rv, the probability of any given number is zero or P(X=200)=0, for any real number. Only the probability of an interval can be computed. Part A Verify that f(x: θ) is a legitimate pdf. Since X must be greater than zero, our limits of integration are zero to infinity. Here we must apply the formula associated with assumption two above to see if

∫∞

−

0

)22/2(2 *)/( dxex x θθ = ∫∞

∞−

=1)( dxxf

This integral looks pretty scary, so for those who do not own a TI-89, let’s jump back to Calc II and use “u” substitution Let 22 2/ θxu −= then dxxdu )/( 2θ−= so duxdx )/( 2θ−=

Now 11000

)2/(

0

22

=+−=−=−==

∞=

=

∞= −∞

−−∫ x

x

u

u xuu eedue θ

Engineering 323 Beautiful Homework #7 Page 3 of 5 Tyler Oxford Problem 4-4

Part B Suppose θ = 100 (a value suggested by a graph in the paper). What is the probability that X is at most 200? Less than 200? At least 200? (I’m sure Beth would appreciate it if everyone noticed that this last part screams complement.) Stated mathematically as P(X ≤ 200), P(X<200) = P(X ≤ 200), and P(X ≥ 200) = 1- P(X ≤ 200) and discussed in the previous section on continuous rv’s. We solve this part exactly like part b, except our limits of integration are from zero to 200 and theta is 100.

1*)100/()200( )2100*2/2200(200

0

)2100*2/2()2100*2/2(2

0100

+−=−==≤ −−−∫ ==

eedxexXPx

xxx

865.01 2 =−= −e So P(X ≤ 200) = .865 In the second part, P(X<200) = P(X ≤ 200) = .865 The last part is simply the complement, so P(X ≥ 200) = 1- P(X ≤ 200) = 1-.865 = .135 These results can be seen in the graph below.

pmf

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0 100 200 300 400

Blade Fatigue

f(x) f(x)

Engineering 323 Beautiful Homework #7 Page 4 of 5 Tyler Oxford Problem 4-4

Part C What is the probability that X is between 100 and 200? In symbols this is P(100 ≤ X ≤ 200). This problem is solved just like part b, except our limits of integration are from 100-200.

∫ =

=−− −==≤≤200

100

)100*2/()100*2/(2

100

2002222

*)100/()200100(x

xxx edxexXP

471.)2/1(2 =+−= −− ee These results can also be seen in the above pmf graph. Part D Give an expression for P(X ≤ x). We should all recognize this as the same idea as the cdf from chapter three. The trickiest part of applying the cdf to a continuous rv is deciding on the limits of integration. For most it seems counterintuitive, but we should go from -∞ to x, because we are looking for a general x, not x at 100. Since our lower limit is zero, we can sub it for infinity. One should also note that this graph is monotonically increasing, or always rising, as can be seen in figure II below. We now have

∫ ∫∞−

−− −===≤=x x

xx edxexdxxfxXPxF0

)100*2/()100*2/(2 22221*)100/()()()(

Therefore

>−=≤

−

otherwisexexXP

x

001)(

)100*2/( 2

This cdf can be seen on the graph on the next page.

Engineering 323 Beautiful Homework #7 Page 5 of 5 Tyler Oxford Problem 4-4

PDF

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100 150 200 250 300

Blade Fatigue

P(X

<x)

weibull