Respuesta :
Answer:
a. 2.30 minutes
b. 0.761
c. 0.239
Step-by-step explanation:
Since this is a uniform distribution, the probability distribution function for any given time between 0 and 4.6 minutes is:
[tex]P(X \leq x)=\frac{x-0}{4.6-0}=\frac{x}{4.6}[/tex]
a. The expected arrival time is the average waiting time, at an uniform distribution, the average is:
[tex]E(X) = \frac{4.6+0}{2}=2.30\ minutes[/tex]
b. The probability for X ≤ 3.5 is:
[tex]P(X \leq 3.5)=\frac{3.5}{4.6} \\P(X \leq 3.5)=0.761[/tex]
c. The probability for X ≥ 3.5 is:
[tex]P(X \geq 3.5)=1-P(X \leq 3.5)=1-0.761\\P(X \geq 3.5)=0.239[/tex]
Using the uniform distribution, it is found that:
a) The expected arrival time is of 2.3 minutes.
b) There is a 0.7609 = 76.09% probability that an elevator arrives in less than 3.5 minutes.
c) There is a 0.2391 = 23.91% probability that the wait for an elevator is more than 3.5 minutes.
What is the uniform probability distribution?
It is a distribution with two bounds, a and b, in which each outcome is equally as likely.
The expected value of the uniform distribution is given by:
[tex]E(X) = \frac{a + b}{2}[/tex]
The probability of finding a value of at lower than x is:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
The probability of finding a value above x is:
[tex]P(X > x) = \frac{b - x}{b - a}[/tex]
In this problem, the arrival time is equally likely at any time range during the next 4.6 minutes, hence a = 0, b = 4.6.
Item a:
[tex]E(X) = \frac{0 + 4.6}{2} = 2.3[/tex]
The expected arrival time is of 2.3 minutes.
Item b:
[tex]P(X < 3.5) = \frac{3.5 - 0}{4.6 - 0} = 0.7609[/tex]
There is a 0.7609 = 76.09% probability that an elevator arrives in less than 3.5 minutes.
Item c:
[tex]P(X > 3.5) = \frac{4.6 - 3.5}{4.6 - 0} = 0.2391[/tex]
There is a 0.2391 = 23.91% probability that the wait for an elevator is more than 3.5 minutes.
More can be learned about the uniform distribution at https://brainly.com/question/13889040
