Respuesta :
Answer:
a) 0.045 = 4.5% probability that exactly two arrivals occur during a particular hour
b) 0.983 = 98.3% probability that at least two people arrive during a particular hour
c) 4.5 arrivals during a 45-min period.
Step-by-step explanation:
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.
Poisson process with a rate parameter of six per hour.
This means that [tex]\mu = 6[/tex]. So
(a) What is the probability that exactly two arrivals occur during a particular hour?
This is P(X = 2).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-6}*(6)^{2}}{(2)!} = 0.045[/tex]
0.045 = 4.5% probability that exactly two arrivals occur during a particular hour
(b) What is the probability that at least two people arrive during a particular hour?
Either less than two people arrive during a particular hour, or at least two do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-6}*(6)^{0}}{(0)!} = 0.002[/tex]
[tex]P(X = 1) = \frac{e^{-6}*(6)^{1}}{(1)!} = 0.015[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.002 + 0.015 = 0.017[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.017 = 0.983[/tex]
0.983 = 98.3% probability that at least two people arrive during a particular hour
(c) How many people do you expect to arrive during a 45-min period? people
An hour has 60 mins.
45/60 = 3/4.
So
[tex]\mu = \frac{3*6}{4} = 4.5[/tex]
4.5 arrivals during a 45-min period.
The people arriving for treatment is an illustration of a Poisson distribution
The probability that exactly two arrivals occur during a particular hour
We have:
[tex]\mu = 6[/tex] --- the mean
A Poisson probability is represented as:
[tex]P(x) = \frac{e^{-\mu} * \mu^x}{x!}[/tex]
When x  = 2, the equation becomes
[tex]P(2) = \frac{e^{-6} * 6^2}{2!}[/tex]
Evaluate the expression
[tex]P(2) = 0.045[/tex]
Hence, the probability that exactly two arrivals occur during a particular hour
The probability that at least two people arrive during a particular hour
This is represented as:
P(x >= 2)
To calculate this, wemake use of the following complement rule:
[tex]P(x \ge 2) = 1 - P(x < 2)[/tex]
This gives:
[tex]P(x \ge 2) = 1 - P(0) - P(1)[/tex]
Calculate P(0) and P(1)
[tex]P(0) = \frac{e^{-6} * 6^0}{0!}= 0.002[/tex]
[tex]P(1) = \frac{e^{-6} * 6^1}{1!}= 0.015[/tex]
So, we have:
[tex]P(x \ge 2) = 1 - 0.002 - 0.015[/tex]
[tex]P(x \ge 2) = 0.983[/tex]
Hence, the probability that at least two arrivals occur during a particular hour is 0.983
How many people do you expect to arrive during a 45-min period?
This is the expected value in a 45-min period
It is calculated using:
[tex]E(x) = \frac{45\ mins}{1\ hour} * \mu[/tex]
Express hour as minute, and substitute 6 for [tex]\mu[/tex]
[tex]E(x) = \frac{45\ mins}{60\ mins} * 6[/tex]
Evaluate
[tex]E(x) = 4.5[/tex]
Approximate
[tex]E(x) = 5[/tex]
Hence, 5 people are expected to arrive during a 45-min period
Read more about Poisson distribution at:
https://brainly.com/question/7879375
