Using the Poisson distribution, we have that:
- There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
- There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
- There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.
<h3>What is the Poisson distribution?</h3>
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
is the mean in the given interval.
For this problem, the mean is given as follows:

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:


There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
The probability of having less than three days of precipitation in the month of April is:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
In which:




Then:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022
There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
For more than 15 days, the probability is:
P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)
Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:
There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.
More can be learned about the Poisson distribution at brainly.com/question/13971530
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