Let X denote the vibratory stress (psi) on a wind turbine blade at a particular

Question

Let X denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. Use the Rayleigh distribution, with pdf f (x; theta) = {x/theta^2 middot e^-x^2/(2 theta^2) x > 0 0 otherwise as a model for the X distribution. Verify that f (x; theta) is a legitimate pdf. integral^infinity _0 x/theta^2 middot e^-x^2/(2 theta^2) dx =]^theta _0 = 0 - () = Suppose theta = 129. What is the probability that X is at most 200? Less than 200? At least 200? (Round your answer to four decimal places.) at most 200 less than 200 at least 200 What is the probability that X is between 100 and 200 (again assuming theta = 129)? (Round your answer to four decimal places.) Give an expression for P (X lessthanorequalto x). P (X lessthanorequalto x) =

Explanation / Answer

Back-up Theory

If a continuous random variable, X, has pdf (probability density function) f(x), for a x b, and 0 otherwise, then, f(x) must satisfy 3 conditions:

1. f(x) 0      (2). f(x) 1      (3) integral from ‘a’ to ‘b’ of {f(x)} = 1.

Establishing conditions (1) and (3) necessarily implies condition (2)

Now, to work out solution,

Given function is f(x) = (1/2)x.e^-(x2/22)dx. Let y = x2/22. Then, dy = x(1/2)dx.

So, integral f(x)dx = integral e-ydy = - e-y …………………………………….(1)

Since at x = 0 and x = , y also = 0 and ,

Integral (0 to )f(x)dx = [- e-y](0, )…………………………………………………(2)

Part (a)

Given x positive and 2 being a square is non-negative, f(x) is non-negative for x > 0.

=> condition (1) is satisfied.

Now, by (2) above, Integral (0 to )f(x)dx = [- e-y](0, ) = - 0 – (- 1) = 1. So, condition (3) is also satisfied. => f(x) is a valid pdf

DONE

Part (b) Given = 129

P(X is at most 200) =P(X 200) = Integral (0 to 200)f(x)dx = [- e-y](0, t),

where t = (½)(200/129)2 = 1.2017[at x = 200, y = {(200)2/2(129)2]

= 1 – e-1,2017 = 1 – 0.3007 = 0.6993. ANSWER1

For a continuous variable probability for a specific value is not defined, pdf does not represent the probability at x, it represents the probability between (x - dx) and (x + dx)/

So, P(X = 200) cannot be given.

P(X is at least 200) = P(X 200) = 1 - P(X 200)

= 0.3007 [from ANSWER 1] ANSWER 2

Part (c)

P(X lies between 100 and 200) = P(100 < X < 200) = P(X < 200) - P(X < 100)

= (1 – e-t2) - (1 – e-t1) = e-t1 - e-t2, where t1 = (½)(100/129)2 = 0.3005 and t2 = (½)(200/129)2 = 1.2017.

So, required probability = e-0.3005 - e-1.2017 = 0.7404 - 0.3307 = 0.2449 ANSWER

Part (d)

       P(X x) = Integral (0 to x)f(x)dx = [- e-y](0, t) = 1 – e- t where t = (½)(x/)2

       So, P(X x) = 1 – e^(-x2/22) ANSWER

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