Answer and Step-by-step explanation:
The given function is:
F(y) =1 – 1/y2 , 1 ≤ y < ∞
Verify function is valid
Limy->-∞f(y) = Limy->-∞ 0 = 0
Limy->∞f(y) = Limy->∞ 1 – 1/y2
= 1, for y ≤ 1
F(y) = 0 is constant. For y > 1, f”(y) = 2 / y3 > 0
So, function is increasing. Therefore f(y) is cdf.
Probability density function
The probability density function is
F(y) = d/dy f(y) = {(2/y2 if y>1o if y≤1 )}
z = 5(y+1)
F (z) = p (z ≤z) = p (5(y+1) ≤z)
= p(y ≤ (z/5) – 1)
F (z) ={(0 if z≤0 and 1-1/(z/5-1)2 if z>0 )