Answer:
a) There is a 9% probability that a drought lasts exactly 3 intervals.
There is an 85.5% probability that a drought lasts at most 3 intervals.
b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation
Step-by-step explanation:
The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.
It has the following probability density formula:

In which p is the probability of a success.
The mean of the geometric distribution is given by the following formula:

The standard deviation of the geometric distribution is given by the following formula:

In this problem, we have that:

So


(a) What is the probability that a drought lasts exactly 3 intervals?
This is 



There is a 9% probability that a drought lasts exactly 3 intervals.
At most 3 intervals?
This is 




Previously in this exercise, we found that 
So

There is an 85.5% probability that a drought lasts at most 3 intervals.
(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?
This is
.
We are working with discrete data, so 3.66 is rounded up to 4.
Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:



There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation