I am not sure how to write it in latex.

Let u = mean / mu - the greek letter m. (this is because I don't know how to write it in latex)

The random variable X has a normal distribution. The mean is u (u > 0), and the variance is 1/4u^2.

(a) Find P(X > 1.5u)

So we have a distribution: X ~ N(u, 1/4u^2) .

Using standardization form:

Let Z = 1/0.25u (X - u)

=> P(Z > 1/0.25u(1.5u - u))

=> P(Z > 1/0.25u(0.5u))

=> P(Z > 0.5u/0.25u)

$\displaystyle P(Z > 2) = 1 - \phi (2.0)$

But it won't give me the right answer as I am obviously doing something wrong...

(b) Find the probability that X is negative.

Here, it's P(Z > 0) or P(Z < 0) .. right?