1. ## Normal distribution help..

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?

2. Originally Posted by struck
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?
(a) $\displaystyle Z = \frac{X - \mu}{\sigma} = \frac{\frac{3 \mu}{2} - \mu}{\frac{\mu}{2}} = 1$.

(b) $\displaystyle Z = \frac{X - \mu}{\sigma} = \frac{0 - \mu}{\frac{\mu}{2}} = -2$. Therefore $\displaystyle \Pr (X < 0) = \Pr(Z < -2)$.

3. So I think I missed the sqrt(1/4) part in (a)

Thanks