Question

Suppose a professor never finishes his lecture before the end of the hour but always finishes within 4 minutes after the end of the hour. Let X - (elapsed time between end of hour and end of lecture) have the following probability density function: f(x) = kx for (0 SXS 4) with f(x) = 0 otherwise. a.) Determine the value of k. (Give decimal answer to three places past decimal.) Submit Answer Tries 0/2 b.) What is the probability the lecture ends within 2.2 minutes of the end of the hour? (Glve decimal answer to three places past decimal.) Submit Answer Tries 0/2 c.) What is the probability the lecture ends between 2 and 2.5 minutes after the end of the hour? (Give decimal answer to three places past decimal.) Submit Answer Tries 0/2 d.) What is the expected value of X? (Give decimal answer to three places past decimal.) Submit Answer Tries 0/2 e.) What is the second moment of X? (Give decimal answer to three places past decimal.) Submit Answer Tries 0/2 f.) What is the variance of X? (Give decimal answer to three places past decimal.) Submit Answer Tries 0/2 This discussion is closed. Send Feedback

Answer 1

som given that the distribution. fo) = SK.*?. 0<x<4 To ; done ☺ Ango. obtain the value of K = ? By the def of pudor. dx -1 k. 22 dx = 1 - $**•x?dx = 1 =K.*** K. (34_0) = 1 = K = 3/64 = .047 - Tk= 1047 D Any find PL XSP-2) = ? 52-2.647.x2 dx = .047.83) 2.2 -.047.52-7130} I P(x32.2) = 0.167] Lecture and within 2:2 having Prob is on 167.

o AS lie What is the prob" lecture end blo 2. 202-5. =? plesx<2.5= 5.647)22dx = (0044). 25 = (-647) (2.5}} } = (047) ( 5.208 - 2167) PC25842-5) = .119 ELX) = ? By def; E(X)= fx. tex. da. E(X) = .047 5* x3 dx @ ☺ =1047-254-0) TEX) = 3.0081 second moment of x = E(X2) live Elx?) = 19.047 * xt de = .041.689) 2.047 (204.8 -0) JE(X2) = 9.625 ) variance of x = ? V(X) = E(X2) - (ECX)= 9.625 – 9.048 V(X) = 3577