Expected Value of a continuous random number

Question

Let X be a continuous random number with (cumulative) distribution function

F (x) =(0 x < 5)

(1-(25/x² x>5)

The expected value E(X) is:

a)6

b)8

c)4

d)10

e) 12

The answer is displayed as 10, however, I find that the integral of xf(x) (and f(x)) diverges, and furthermore the function doesn't satisfy the normalization condition. What am I missing? Is there a different formula for this type of problem? Thank you so much

Answer 1

F(x)=1-(25/x²), x ≥5 works OK as the distribution function.(0 at 5 and 1 at ∞)

f(x)=50/x³ is the density function

∫x*(50/x³⁾ dx between 5 and ∞ is 10...which is the expectation.