Does the negative distribute? interpreting notation so this is sort of a silly question but I want to be clear of the meaning of this for generalization purposes. This is concerning Z scores, where $\Phi(x)$ = the (c.d.f) of a standard normal distribution. 
For : 
$P(Z \ge 1.05 ) = 1- \Phi(1.05) = 1-.8531 = .1469 $
The above example seems clear to be but then the book shows : $$P(-1.5 \le Z \le 1.18) = \Phi(1.18)-\Phi(-1.50)= .8810 - .0668=.8142$$
In working this out myself : $$P(-1.5 \le Z) = 1 - \Phi(-1.5)$$
Now, $\Phi(-1.5) = .0668$, so $1-.0668 = .9332$, this cannot be correct according to the above book example. So then does $1-\Phi(-1.5)$ become $1- \Phi(1.5)$? When working it out as $1-\Phi(1.5)$, the correct answer is obtained. 
 A: The CDF $\Phi(x)$ is all the area to the left of $x$. $P(x \leq Z \leq y)$ is the area which is to the left of $y$ and to the right of $x$. This excludes the area which is to the left of $x$. So you subtract that off, getting $\Phi(y)-\Phi(x)$. Visually this is the area to the left $y$ minus the area to the left of $x$.
Alternately, going the other way, you want all the area to the right of $x$ which is to the left of $y$. This excludes the area to the right of $y$, so you get $(1-\Phi(x))-(1-\Phi(y))$. Visually this is the area to the right of $x$ minus the area to the right of $y$.
Sometimes because of the use of tables, we don't actually have normal distribution values which go into negative arguments. In these cases we have to use the symmetry of the normal distribution. For instance $\Phi(-1)$ is the area to the left of $-1$, which by symmetry is the same as the area to the right of $1$. Hence $\Phi(-1)=1-\Phi(1)$.
All of this is a lot easier to see if you draw a picture.
A: The general principle, for an interval, is $$P(a\le Z\le b)=\Phi(b)-\Phi(a)$$
Sometimes the interval is half-infinite, e.g. $b=\infty$.  Then $$P(a\le Z)=P(a\le Z\le \infty)=\Phi(\infty)-\Phi(a)=1-\Phi(a)$$
There is no inconsistency, and $\Phi(1.5)\neq \Phi(-1.5)$.
A: Just look to the symmatry in the area of the standard normal disturbution CDF, you can see that the right area of the CDF with respect to a point of abscissa say $x$ with $x$>0 is the same area of that on the left of the point of abscissa -$x$ and hence you can say that $\psi(-x)$= $P(Z \le -x)$ = $P$($Z \ge x) \ $= $\ 1 - P(Z \le x)$ =$ 1 - \psi(x).$
