Respuesta :
Answer:
a) 0.0224
b) 0.4735
c) 0.1844
Step-by-step explanation:
We use the Poisson probability distribution to solve this problem.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this problem, we have that:
[tex]\mu = 3.8[/tex]
(a) no alarms
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.8}*(3.8)^{0}}{(0)!} = 0.0224[/tex]
(b) fewer than four alarms
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = 0) = \frac{e^{-3.8}*(3.8)^{0}}{(0)!} = 0.0224[/tex]
[tex]P(X = 1) = \frac{e^{-3.8}*(3.8)^{1}}{(1)!} = 0.0850[/tex]
[tex]P(X = 2) = \frac{e^{-3.8}*(3.8)^{2}}{(2)!} = 0.1615[/tex]
[tex]P(X = 3) = \frac{e^{-3.8}*(3.8)^{3}}{(3)!} = 0.2046[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0224 + 0.0850 + 0.1615 + 0.2046 = 0.4735[/tex]
(c) more than five alarms
Either there are five or fewer alarms, or there are more than five. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 5) + P(X > 5) = 1[/tex]
We want to find [tex]P(X > 5)[/tex]. So
[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]
In which
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
[tex]P(X = 0) = \frac{e^{-3.8}*(3.8)^{0}}{(0)!} = 0.0224[/tex]
[tex]P(X = 1) = \frac{e^{-3.8}*(3.8)^{1}}{(1)!} = 0.0850[/tex]
[tex]P(X = 2) = \frac{e^{-3.8}*(3.8)^{2}}{(2)!} = 0.1615[/tex]
[tex]P(X = 3) = \frac{e^{-3.8}*(3.8)^{3}}{(3)!} = 0.2046[/tex]
[tex]P(X = 4) = \frac{e^{-3.8}*(3.8)^{4}}{(4)!} = 0.1944[/tex]
[tex]P(X = 5) = \frac{e^{-3.8}*(3.8)^{5}}{(5)!} = 0.1477[/tex]
So
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0224 + 0.0850 + 0.1615 + 0.2046 + 0.1944 + 0.1477 = 0.8156[/tex]
Then
[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8156 = 0.1844[/tex]
