normally distributed random variable computation I'm working on the following problem and I think that I have the entire problem correct, however I am slightly confused at reading the tables for normally distributed random variables. 
Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean value 15 kips and standard deviation 1.25 kips.
a. P(X < 15)
b. P(X >= 10)
c. P(14 < X < 18)

I am using the following table as a reference:
http://www.stat.tamu.edu/~twehrly/651/ztable.pdf
I've done the following, but have my doubts on the b and c when I run into negative numbers.
a. P(x < 15) = P(z < (15-15)/1.25) = P(z < 0) = .5
b. P(x >= 10) = P(z > (10-15)/1.25 ) = P(-4 < z) = 1
c. P(14 < x < 18) = P(z > (14-15)/1.25) – P(z < (18-15)/1.25) 
= P(-.8 < z < 2.4) = 0.7799
I was told there are calculators online that can compute this, however I am still confused at what to do when I run into a situation such as b where you have a -4 or in part c.
 A: We look at Problem c).  We have 
$$\Pr(14\lt X\lt 18)=\Pr(X\lt 18)-\Pr(X\le 14).$$
I think you have no difficulty with $\Pr(X\lt 18)$. But for completeness, 
$$\Pr(X\lt 18)=\Pr\left(Z\lt \frac{18-15}{1.25}\right)=\Pr(Z\lt 2.4).$$
My table gives $0.9918$. 
Now we want $\Pr(X\le 14)$. We have
$$\Pr(X\lt 14)=\Pr\left(Z\lt \frac{14-15}{1.25}\right)=\Pr(Z\lt -0.8).$$
However, we cannot find this probability in our table. But luckily, the standard normal density function is symmetric about $z=0$. So the probability of landing in the  left tail (to the left of $-0.8$) is the same as the probability of landing in the right tail (to the right of $0.8$) It may be helpful to look at a picture of the standard normal density function while reading this.
The probability that $Z$ is $\ge 0.8$ is $1$ minus the probability that $Z\lt 0.8$. And this is available from the table. So we have
$$\Pr(Z\le -0.8)=\Pr(Z\ge 0.8)=1-\Pr(Z\lt 0.8)\approx 1-0.7881\approx 0.2119.$$
Finally, put things together. The required probability is $\approx 0.9918-0.2119$.  
