Let Y be a continuous random variable whose PDF is as g(y)=y4125 for yϵ(0,a). Constant "a" makes g(y) a valid PDF. Find the value of a5.
(2) The area under the curve f(x) in the range (a,b)is 1, that is: ∫baf(x)dx=1.
Thus we have ∫a0g(y)dy=1
⇒∫a0y4125dy=1
⇒1125∫a0y4dy=1
⇒1125[y55]a0=1
⇒1625[y5]a0=1
⇒1625[a5−0]=1
⇒a5625=1
⇒a5=625