Answer:
The probability that 2 or 3 customers will arrive in a 15-minute period is 0.4703
Explanation:
Firstly, we have to determine the segment unit, since the mean is 10 per hour, the segment unit is 1 hour.
The mean(m) = 10
since the period is 15 minutes = 0.25 hour, t= 0.25 hour / 1 hour. Therefore mt= 2*10 = 2.5
The poisson distribution formula P(x) = [tex]\frac{(mt)^{x}e^{-mt} }{x!}[/tex]
Therefore the probability that 2 or 3 customers will arrive in a 15-minute period
P(x=2) or P(x=3) = P(x=2) + P(x=3) = [tex]\frac{(2.5)^{2}e^{-2.5} }{2!}+\frac{(2.5)^{3}e^{-2.5} }{3!}[/tex] = 0.2565 + 0.2138 = 0.4703
Therefore P(x=2) or P(x=3) = 0.4703
The probability that 2 or 3 customers will arrive in a 15-minute period is 0.4703
Answer: 0.4704
Explanation:
Poisson formula for the distribution could be written as:
P(x) = [((nT) ^x) * (e^-nT)] ÷ x!
mean(n) = average number of customers per hour = 10
Length of time(T) = 15 minutes = 15÷60 = 0.25 hour
x = number of customers
e = exponential constant
Therefore chances of 2 or 3 customers arriving in a 0.25 hour period can be defined as ;
P(2) or P(3) = P(2) + P(3)
P(2)=[((10*0.25)^2) *(e^-10*0.25)] ÷ 2!
P(3)=[((10*0.25)^3) *(e^-10*0.25)] ÷ 3!
P(2) = 0.2566
P(3) = 0.2138
P(2)+P(3)= 0.2566 + 0.2138 =0.4704
