I tackled a question that asked
Given the mean and std dev for rainfall in a city,
Mean = 1126mm of rain
std Dev. = 355mm of rain
What is the probability that a year would have more than 2191mm of rainfall.
The way I went about this was
Using the Value - Mean/ Std Dev formula
I retrieved a Z score from the table in my textbook as 0.0013 as the answer to the calculation was 3.00. So I interpreted this as, to get a rainfall value of larger than 2191mm there is a 0.13% chance as the Z score is 0.0013.
So my Question is, how would I calculate this for LESS THAN not MORE THAN 2191, do I just calculate like so?
1 - P where P is 0.13%?
EDIT: This is confusing for me then, if that is how you calculate it what if you retrieve a negative Z value
If I take 685mm for a graph distributed like so
And 685 is clearly towards the left side of the 1126mm mean value.
If i retrieve a negative Z score of -1.24 using the same formula above to calculate it, do I just use the Z table normally? If so then I get a value of .1075 and to find out the P of less than 685mm would be 1-.1075 but that would mean the probability of it being less than 685mm is 0.8925
It is obvious that it's not 0.8925.
So I assume that for values less than the mean of the given data and you're calculating less than you don't have to do 1-P but if you were calculating the probability of MORE THAN 685mm of rain it would be 1-P. Is this correct thinking or am I going mad?
Regards
I dealt with your first question in a comment. The added part about $Z$-scores takes a bit more space.
Suppose you want the probability that the rainfall is $\le 685$. That corresponds to a $Z$-score of $-1.24$. Information for negative values is ordinarily not tabulated. That's because it is easy to find using information about positives.
The probability that $Z\le -1.24$ is equal to the probability that $Z\ge 1.24$. This is because the density function of the standard normal $Z$ is symmetric about the horizontal axis. The probability of being in the left tail of points $\le -1.24$ is the same as the probability of being in the right tail of points $\ge 1.24$.
Similarly, $\Pr(Z\ge -1.24)=\Pr(Z\le 1.24)$.
It is best, I think, not to try to memorize formulas. Just be aware of the geometry, and "figure out" what to do each time. Remember, we can think of it as evaluating areas.