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:
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:
Compute the probability of exactly 0 arrivals in 5 minutes as follows:
Compute the probability of exactly 1 arrivals in 5 minutes as follows:
Compute the probability of exactly 2 arrivals in 5 minutes as follows:
Compute the probability of exactly 3 arrivals in 5 minutes as follows:
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)
Thus, the probability that delays will occur is 0.1431.
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