Doubt about Q function Hi have a doubt about the Q function; I have this problem: 
$Z\sim N(20; 500)$ and I have to find $P(Z>0)$, by Q function I have: $Q(\frac{0-20}{\sqrt{500}})=Q(-0.894)$. Now I have to find $Q(-0.894)$ or $1-Q(0.894)$ on the table?
Thank you
 A: Why not use the direct definition of cdf of Normal RV?
$$
P(X>0) = 1 - P(X<0) = 1 - \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{0}e^{-\frac{(x- \mu)^2}{2\sigma^2}}dx 
$$
Set $z = \frac{x- \mu}{\sigma}$,  so $dx =  \sigma dz$, so this expression is 
$$
1- \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{-\sqrt{\frac{2}{5}}}e^{-\frac{z^2}{2}}dz = 1- \Phi\bigg(-\sqrt{\frac{2}{5}} \bigg) \approx 1- 0.1855 = 0.8145
$$
A: Your notation is not standard. Usually, in such
problems $Z$ is reserved for a standard normal
random variable (normal with mean $0$ and standard
deviation $1.)$
So let $X \sim N(\mu=20, \sigma^2 = 500).$ Then suppose
you are to find $P(X < 0):$
$$P(X < 0) = P\left(\frac{X-\mu}{\sigma} < \frac{0-20}{\sqrt{500}}\right)\\
= P(Z <  -0.89443)\\ \approx P(Z < -.89) = 0.1867.$$
Then you need to look in the margins of a CDF table of the standard normal
distribution for $\pm 0.89$ (or some related number) and find the probability
0.1867. Printed normal tables come in a variety of styles, so I can't
tell you exactly how to use yours. [The notation $Q$ is not standard. Use an example in your book with a similar problem for orientation. Usually, there is a small diagram
on the table with a shaded area to let you know what probabilities
are in the body of your table.]
If you use statistical software, you can get a more exact answer.
Here is how to use R to get the exact answer $0.1855467:$
pnorm(0, 20, sqrt(500))
[1] 0.1855467
pnorm(-.8944272)
[1] 0.1855467

Here are two figures, In each figure, the probability computed here
is represented by the area under the curve to the left of the
vertical red dotted line.

