If X~$\mathscr N(-5,1)$ what would be $\Bbb P(X \ge -4)$? Can anyone explain the following solution,
$=\Bbb P\left(\displaystyle\frac{X-\mu}{\sigma} \ge \frac{-4-(-5)}{1}\right)$
$=\Bbb P(Z \ge 1)$
$=0.5 - \Bbb P(0<Z<1)$
$=0.5 - 0.3413$
$=0.1587$ Ans.
 A: The idea behind that solution is that your original distribution is invariant to translation;  simplistically that one can add (or subtract) a quantity $a$ to the probability distribution of $X$ and the shape of that probability distribution of $X$ will remain the same. 
The solution you present effectively says  that the question $P(X \ge -4)$   for $X \sim N(-5,1)$ can be answered for $X_{new} \sim N(0,1)$ if you also shift your original question such that is asks: $P(X_{new} \ge 1)$; here $\alpha$ is simply $5$ so you have $X_{new} = X + 5$. The answer emphasises the fact that the new distribution is $N(0,1)$ by using the symbol $Z$ to show that the new variable is standardised.
In any case you need to use the cumulative distribution function of the normal distribution to get your answer (function pnorm in R).
In R you could do something like: 
1- pnorm( -4, mean = -5, sd = 1)
# 0.1586553

to get the exact answer directly without resorting to translating the probability distribution $X \sim N(-5,1)$. The answer you describe prompts you to use: 
1- pnorm( 1, mean = 0, sd = 1)
# 0.1586553

so you query the distribution $X_{new} \sim  N(0,1)$. Both methods will return the same results: 
all.equal( 1- pnorm( 1, mean = 0, sd = 1), 1- pnorm( -4, mean = -5, sd = 1))
# TRUE

My suggestion would be to forget about tables, tabulations and related stuff and focus to the idea that each probability is an area under a curve. If you appreciate that, then you just look around for a function that gives the size of the slice you are examining. 
