Respuesta :
Answer:
(a) The probability of no customers are waiting in a line is 0.01832.
(b) The probability of 4 customers are waiting in a line is 0.19537.
(c) The probability of 4 or fewer customers are waiting in a line is 0.62885.
(d) The probability of 4 or more customers are waiting in a line during the visit is 0.56652.
Step-by-step explanation:
The number of customers waiting in a line between 4 PM and 7 PM (X) follows a Poisson distribution with parameter λ = 4.
The probability mass function of a Poisson distribution is:
[tex]P(X=x)=\frac{e^{-4}(4)^{x}}{x!} ;\ x=0, 1, 2,...[/tex]
(a)
Compute the probability that no customers are waiting in a line during the visit as follows:
[tex]P(X=0)=\frac{e^{-4}(4)^{0}}{0!}=\frac{0.01832\times1}{1}=0.01832[/tex]
Thus, the probability of no customers are waiting in a line is 0.01832.
(b)
Compute the probability that 4 customers are waiting in a line during the visit as follows:
[tex]P(X=4)=\frac{e^{-4}(4)^{4}}{4!}=\frac{0.01832\times256}{24}=0.19537[/tex]
Thus, the probability of 4 customers are waiting in a line is 0.19537.
(c)
Compute the probability that 4 or fewer customers are waiting in a line during the visit as follows:
P (X ≤ 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
[tex]=\frac{e^{-4}(4)^{0}}{0!}+\frac{e^{-4}(4)^{1}}{1!}+\frac{e^{-4}(4)^{2}}{2!}+\frac{e^{-4}(4)^{3}}{3!}+\frac{e^{-4}(4)^{3}}{3!}+\frac{e^{-4}(4)^{4}}{4!}\\=0.01832+0.07326+0.14653+0.19537+0.19537\\=0.62885[/tex]
Thus, the probability of 4 or fewer customers are waiting in a line is 0.62885.
(d)
Compute the probability of 4 or more customers are waiting in a line during the visit as follows:
P (X ≥ 4) = 1 - P (X < 4)
= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)
[tex]=1-\frac{e^{-4}(4)^{0}}{0!}+\frac{e^{-4}(4)^{1}}{1!}+\frac{e^{-4}(4)^{2}}{2!}+\frac{e^{-4}(4)^{3}}{3!}+\frac{e^{-4}(4)^{3}}{3!}\\=1-0.01832-0.07326-0.14653-0.19537\\=0.56652[/tex]
Thus, the probability of 4 or more customers are waiting in a line during the visit is 0.56652.
