Respuesta :
Answer:
(a) λ = 2.
(b) P (X = 0) = 0.1353; P (X = 1) = 0.2706;
P (X = 2) = 0.2706; P (X = 3) = 0.1804
(c) P (Delay Problems) = 0.1431.
Step-by-step explanation:
Let X = number of arrivals at the drive-up teller window.
The average number of arrivals at the drive-up teller window per minute is,
p = 0.4 customers/ minute.
(1)
Compute the expected number of customers at the drive-up teller window in n = 5 minutes as follows:
[tex]E(X)=\lambda\\=np\\=5\times 0.4\\=2[/tex]
Thus, the mean number of customers that will arrive in a five-minute period is λ = 2.
(2)
The random variable X follows a Poisson distribution with parameter λ = 2.
The probability mass function of X is:
[tex]P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...[/tex]
Compute the probability of exactly 0 arrivals in 5 minutes as follows:
[tex]P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.1353\times 1}{1}=0.1353[/tex]
Compute the probability of exactly 1 arrivals in 5 minutes as follows:
[tex]P(X=1)=\frac{e^{-2}2^{1}}{1!}=\frac{0.1353\times 2}{1}=0.2706[/tex]
Compute the probability of exactly 2 arrivals in 5 minutes as follows:
[tex]P(X=2)=\frac{e^{-2}2^{2}}{2!}=\frac{0.1353\times 4}{2}=0.2706[/tex]
Compute the probability of exactly 3 arrivals in 5 minutes as follows:
[tex]P(X=3)=\frac{e^{-2}2^{3}}{3!}=\frac{0.1353\times 8}{6}=0.1804[/tex]
Thus, the values are:
P (X = 0) = 0.1353
P (X = 1) = 0.2706
P (X = 2) = 0.2706
P (X = 3) = 0.1804
(3)
Delays occur in the service time if there are more than three customers arrive during any five-minute period.
Compute the probability that there are more than 3 customers as follows:
P (X > 3) = 1 - P (X ≤ 3)
[tex]=1-\sum\limits^{3}_{x=0}{\frac{e^{-2}2^{x}}{x!}}\\=1-(0.1353+0.2706+0.2706+0.1804)\\=1-0.8569\\=0.1431[/tex]
Thus, the probability that delays will occur is 0.1431.
