Answer:
a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?
We want this probability"

At most 3 hours?

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

What is the probability that it is less than the mean value by more than one standard deviation?

Step-by-step explanation:
Previous concepts
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
The cumulative distribution for this function is given by:

We know the value for the mean on this case we have that :


Solution to the problem
Part a
What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?
We want this probability"

At most 3 hours?

Part b
What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?
The variance for the esponential distribution is given by: 
And the deviation would be:

And the mean is given by 
Two deviations correspond to 5.540, so we want this probability:

What is the probability that it is less than the mean value by more than one standard deviation?
For this case we want this probablity:
