A random sample size of 5 is drawn from the pdf 

3.10.4: A random sample size of 5 is drawn from the pdf fy(y) = 2y, 0 0.6) U P(Y; < 0.6) These are disjoint events, so we sum their probabilities: P(Y{ > 0.6) + P(Y; < 0.6) P(Y{ > 0.6) = P (Y,,Y2, Y3, Y4, Ys > 0.6) = [P(Y > 0.6)]5 = 2ydy = (0.64)5 = 0.107 %3D r0.6 P(Y < 0.6) = P(Y,, Y2, Y3, Y4, Y5 < 0.6) = [P(Y < 0.6)]s = 2ydy) = (0.36)5 = 0.006 %3D %3D Thus, P(Y < 0.6 < Y;) = 1- [P(Y{ > 0.6) + P(Y; < 0.6)] = 1- (0.107 + 0.006) = 1- 0.113 = 0.887 x 88.7% Find the cdf for the density in problem 3.10.4. Use that to randomly generate a sample size 5, call it y. Sort the sample and determine if the interval Y to Y', contains 0.60. The interval is y[1] to y[5]. Repeat for 1,000 trials and compare with your theoretical result.