Respuesta :
Answer:
(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780
(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.
Step-by-step explanation:
The random variable Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.
The random variable Y follows a Geometric distribution with parameter p = 0.409.
The probability mass function of a Geometric distribution is:
[tex]P(Y=y)=(1-p)^{y}p;\ y=0,12...[/tex]
(a)
Compute the probability that a drought lasts exactly 3 intervals as follows:
[tex]P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844[/tex]
Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.
Compute the probability that a drought lasts at most 3 intervals as follows:
P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)
[tex]=(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780[/tex]
Thus, the probability that a drought lasts at most 3 intervals is 0.8780.
(b)
Compute the mean of the random variable Y as follows:
[tex]\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445[/tex]
Compute the standard deviation of the random variable Y as follows:
[tex]\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88[/tex]
The probability that the length of a drought exceeds its mean value by at least one standard deviation is:
P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)
= P (Y ≥ 3.325)
= P (Y ≥ 3)
= 1 - P (Y < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
[tex]=1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064[/tex]
Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.
