Let Y_n be the nth order statistic of a random sample of size n from a distribution that has pdf f(x) = (3x^2)/(t^3) with 0<x<t.
a) Show that P(c < (Y_n)/t < 1) = 1 - c^(3n) with 0<c<1.
b) If n is 4 and if the observed value of Y_4 is 2.3 what is a 95% confidence interval for t?
I'm not sue how to approach this problem. Any help would be appreciated.