Answer:
(a) <em>λ</em> = 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 <em>X</em> = number of arrivals at the drive-up teller window.
The average number of arrivals at the drive-up teller window per minute is,
<em>p</em> = 0.4 customers/ minute.
(1)
Compute the expected number of customers at the drive-up teller window in <em>n</em> = 5 minutes as follows:
![E(X)=\lambda\\=np\\=5\times 0.4\\=2](https://tex.z-dn.net/?f=E%28X%29%3D%5Clambda%5C%5C%3Dnp%5C%5C%3D5%5Ctimes%200.4%5C%5C%3D2)
Thus, the mean number of customers that will arrive in a five-minute period is <em>λ</em> = 2.
(2)
The random variable <em>X</em> follows a Poisson distribution with parameter λ = 2.
The probability mass function of <em>X</em> is:
![P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%5Cfrac%7Be%5E%7B-2%7D2%5E%7Bx%7D%7D%7Bx%21%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
Compute the probability of exactly 0 arrivals in 5 minutes as follows:
![P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.1353\times 1}{1}=0.1353](https://tex.z-dn.net/?f=P%28X%3D0%29%3D%5Cfrac%7Be%5E%7B-2%7D2%5E%7B0%7D%7D%7B0%21%7D%3D%5Cfrac%7B0.1353%5Ctimes%201%7D%7B1%7D%3D0.1353)
Compute the probability of exactly 1 arrivals in 5 minutes as follows:
![P(X=1)=\frac{e^{-2}2^{1}}{1!}=\frac{0.1353\times 2}{1}=0.2706](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%5Cfrac%7Be%5E%7B-2%7D2%5E%7B1%7D%7D%7B1%21%7D%3D%5Cfrac%7B0.1353%5Ctimes%202%7D%7B1%7D%3D0.2706)
Compute the probability of exactly 2 arrivals in 5 minutes as follows:
Compute the probability of exactly 3 arrivals in 5 minutes as follows:
![P(X=3)=\frac{e^{-2}2^{3}}{3!}=\frac{0.1353\times 8}{6}=0.1804](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5Cfrac%7Be%5E%7B-2%7D2%5E%7B3%7D%7D%7B3%21%7D%3D%5Cfrac%7B0.1353%5Ctimes%208%7D%7B6%7D%3D0.1804)
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)
![=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](https://tex.z-dn.net/?f=%3D1-%5Csum%5Climits%5E%7B3%7D_%7Bx%3D0%7D%7B%5Cfrac%7Be%5E%7B-2%7D2%5E%7Bx%7D%7D%7Bx%21%7D%7D%5C%5C%3D1-%280.1353%2B0.2706%2B0.2706%2B0.1804%29%5C%5C%3D1-0.8569%5C%5C%3D0.1431)
Thus, the probability that delays will occur is 0.1431.