Respuesta :
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:
[tex]f(x) = (1-p)^{x}p[/tex]
In which p is the probability of a success.
The mean of the geometric distribution is given by the following formula:
[tex]\mu = \frac{1-p}{p}[/tex]
The standard deviation of the geometric distribution is given by the following formula:
[tex]\sigma = \sqrt{\frac{1-p}{p^{2}}[/tex]
In this problem, we have that:
[tex]p = 0.383[/tex]
So
[tex]\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61[/tex]
[tex]\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05[/tex]
(a) What is the probability that a drought lasts exactly 3 intervals?
This is [tex]f(3)[/tex]
[tex]f(x) = (1-p)^{x}p[/tex]
[tex]f(3) = (1-0.383)^{3}*(0.383)[/tex]
[tex]f(3) = 0.09[/tex]
There is a 9% probability that a drought lasts exactly 3 intervals.
At most 3 intervals?
This is [tex]P = f(0) + f(1) + f(2) + f(3)[/tex]
[tex]f(x) = (1-p)^{x}p[/tex]
[tex]f(0) = (1-0.383)^{0}*(0.383) = 0.383[/tex]
[tex]f(1) = (1-0.383)^{1}*(0.383) = 0.236[/tex]
[tex]f(2) = (1-0.383)^{2}*(0.383) = 0.146[/tex]
Previously in this exercise, we found that [tex]f(3) = 0.09[/tex]
So
[tex]P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855[/tex]
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 [tex]P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)[/tex].
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:
[tex]P(X \leq 3) + P(X \geq 4) = 1[/tex]
[tex]0.855 + P(X \geq 4) = 1[/tex]
[tex]P(X \geq 4) = 0.145[/tex]
There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation
