Let ?X ?denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article “Blade Fatigue Life Assessment with Application to VAWTS” (?J. of Solar Energy Engr., ?1982: 107–111) proposes the Rayleigh distribution, with pdf as a model for the ?X ?distribution. a.? ?Verify that? ? ?) is a legitimate pdf. b.? ?Suppose ? = 100 (a value suggested by a graph in the article). What is the probability that ?X ? s at most 200? Less than 200? At least 200? c.? ?What is the probability that ?X ?is between 100 and 200 (again assuming ? = 100)? d.?? ive an expression for P(X ? x).

Problem 4E Answer: Step1: We have Let X denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article “Blade Fatigue Life Assessment with Application to VAWTS” (J. of Solar Energy Engr., 1982: 107–111) proposes the Rayleigh distribution, with pdf as a model for the X distribution. We need to find, a. Verify that f(x; ) is a legitimate pdf. b. Suppose = 100 (a value suggested by a graph in the article). 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 (again assuming = 100) d. Give an expression for P(X x). Step2: a). To verify whether f(x:) is a legitimate pdf or not, we need to verify two conditions 1).If x < 0 then f(x; ) = 0 f(x; ) 0. 2 If x > 0 then f(x; ) > 0 since x > 0 and e 2 e > 0 2 Therefore,the exponential function is always positive. 2 x x2 2). (x;)dx = ( )e 2 2dx 0 0 x = e 22 | dx 0 0 = -0 + e = 1. Therefore, the given probability density function has unity hence f(x; ) is a legitimate pdf. b). 1).Suppose = 100 then the probability that X is at most 200 is given by 200 200 x2 P(X 200) = P(X < 200) = f(x;)dx = ( )e 2dx 2 0 0 x2 200 = e 22 0 dx 200 = e - e 2(100) 40000 = 1 - e 20000 = 1 - e 2 = 1 - 0.1353 = 0.8646. Therefore, the probability that X is at most 200 is 0.8646. 2).The probability that X is at least 200 is given by 200 200 2 x x2 P(X 200) = 1 - P(X < 200) =1- f(x;)dx = 1 ( 2 2 dx 0 0 x2 =1 (e 22 |200 )dx 0 200 0 2(100) =1- (e - e ) 40000 = 1 - (1 - e 20000) 2 = e = 0.1353 Therefore,The probability that X is at least 200 is 0.1353. Step2: c). The probability that X is between 100 and 200 is given by 200 200 x x2 P(100 < X 200) = f(x;)dx = ( )2 2 dx 100 100 2 x2 200 = e 2 |100 dx 2 2 1002 200 2 = e 2(100)- e 2(100) 10000 40000 = e 20000 - e 20000 0.5 2 = e - e = 0.6065 - 0.1353 = 0.4712. Therefore, The probability that X is between 100 and 200 is 0.4712. d). If x < 0 then f(x; ) = 0 f(x; ) 0. If x 0 then x P(Xx) = f(t;)dt x t t = ()e 2 22dt 0 2 t2 x = ( e 2 |0) x2 = 1 - e 22 Therefore, P(Xx) = {0 if x < 0 x2 1 e22 if x 0