Continuous probability distribution

[thereddevils]

Homework Statement

A continuous random variable ,X represents the period, of a telephone call in the office. The cdf of x is given by

F(x)= x^2/8 for 0<=x<=2

=1-4/x^3 for x>2

Find pdf , mean and variance.

Homework Equations

The Attempt at a Solution

pdf:

f(x) = 1/4 x for 0<=x<=2

=12/x^4 for x>2

= 0 , otherwise

$$E(x)=\int^{2}_{0}x(\frac{1}{4})x dx +\int^{\infty}_{2}x(\frac{12}{x^4})dx$$

=20/3

$$E(x^2)=\int^{2}_{0}x(\frac{1}{4})x^2 dx +\int^{\infty}_{2}x^2(\frac{12}{x^4})dx$$

=7

Var(x)=7-(20/3)^2

Variance is negative here , how can that be? I have checked my working thoroughly but am not able to spot my mistakes.

 

[statdad]

Check your calculation of $$E(X)$$ once again - I get a different value.