Respuesta :
Answer:
a) 0.5 per minutes
b) 5 arrivals expected in 10 minutes
c) P ( x = 0 ) = 0.00673 , P ( x = 1 ) = 0.03368 , P ( x = 2 ) = 0.08422 ,P ( x = 3 ) = 0.14037
d) P ( X >= 4 ) = 0.735
Step-by-step explanation:
Given:
- The number of customer arriving at window is modeled by Poisson distribution. The distribution is given by:
P(x) = ( λ^x ) (e^-λ) / x! x = 0 , 1 , 2 , 3 , ......
- Average rate λ = 30 / hr
Find:
a. How many customers arrive per minute, on average?
b. How many customers would you expect to arrive in a 10-minute interval?c. Use equation 13.1 to determine the probability of exactly 0, 1, 2, and 3 arrivals in a 10-minute interval.
d. What is the probability of more than three arrivals occurring in a 10-minute interval?
Solution:
- The average rate λ in number of customers that arrive in a minute is given by:
λ1 = 30 / 60 = 0.5 arrival per minutes
- The average number of customer that are expected to arrive in 10-minutes window is:
λ2 = 10*λ1 = 10*0.5 = 5 arrivals expected in 10 minutes
- The probability of exactly 0,1 , 2 , and 3 arrivals in 10 minute windows:
P ( x = 0 ) = ( 5^0 ) (e^-5) / 0! = 0.00673
P ( x = 1 ) = ( 5^1 ) (e^-5) / 1! = 0.03368
P ( x = 2 ) = ( 5^2 ) (e^-5) / 2! = 0.08422
P ( x = 3 ) = ( 5^3 ) (e^-5) / 3! = 0.14037
- The probability of more than three arrivals occuring in 10-minute interval is:
P ( X >= 4 ) = 1 - P ( X =< 3 )
P ( X >= 4 ) = 1 - [ P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) ]
P ( X >= 4 ) = 1 - [ 0.00673 + 0.03368 + 0.08422 + 0.14037 ]
P ( X >= 4 ) = 1 - [ 0.265 ]
P ( X >= 4 ) = 0.735
Using the Poisson distribution, it is found that:
a) 0.5 customers per minute.
b) 5 customers are expected to arrive.
c)
0.0068 = 0.68% probability of exactly 0 arrivals in a 10-minute interval.
0.0337 = 3.37% probability of exactly 1 arrivals in a 10-minute interval.
0.0842 = 8.42% probability of exactly 2 arrivals in a 10-minute interval.
0.1404 = 14.04% probability of exactly 3 arrivals in a 10-minute interval.
d) 0.7349 = 73.49% probability of more than three arrivals occurring in a 10-minute interval.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
Item a:
30 in one-hour(60 minutes), hence 0.5 customers per minute.
Item b:
0.5 customers per minute, hence, in a 10 minute interval, 5 customers are expected to arrive.
Item c:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}(5)^{0}}{(0)!} = 0.0068[/tex]
[tex]P(X = 1) = \frac{e^{-5}(5)^{1}}{(1)!} = 0.0337[/tex]
[tex]P(X = 2) = \frac{e^{-5}(5)^{2}}{(2)!} = 0.0842[/tex]
[tex]P(X = 3) = \frac{e^{-5}(5)^{3}}{(3)!} = 0.1404[/tex]
0.0068 = 0.68% probability of exactly 0 arrivals in a 10-minute interval.
0.0337 = 3.37% probability of exactly 1 arrivals in a 10-minute interval.
0.0842 = 8.42% probability of exactly 2 arrivals in a 10-minute interval.
0.1404 = 14.04% probability of exactly 3 arrivals in a 10-minute interval.
Item d:
This probability is:
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which:
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
From item c:
[tex]P(X \leq 3) = 0.0068 + 0.0337 + 0.0842 + 0.1404 = 0.2651[/tex]
Then:
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.2651 = 0.7349[/tex]
0.7349 = 73.49% probability of more than three arrivals occurring in a 10-minute interval.
A similar problem is given at https://brainly.com/question/16912674
