Respuesta :
Answer:
P(μ-σ ≤ X ≤ μ+σ) = [tex]0.61 + 0.1275 = 0.7375[/tex]
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 = 0.61[/tex].
So
[tex]\mu = \frac{1-p}{p} = 0.6393[/tex]
[tex]\sigma = \sqrt{\frac{1-p}{p^{2}} = 1.0234[/tex]
P(μ-σ ≤ X ≤ μ+σ) = [tex]P(-0.3841 \leq X \leq 1.6627) = P(0 \leq X \leq 1.6627) = f(0) + f(1.6627)[/tex]
[tex]f(x) = (1-p)^{x}p[/tex]
[tex]f(0) = (0.39)^{0}(0.61) = 0.61[/tex]
[tex]f(1.6627) = (0.39)^{1.6627}(0.61) = 0.1275[/tex]
P(μ-σ ≤ X ≤ μ+σ) = [tex]0.61 + 0.1275 = 0.7375[/tex]
