Question

The number of customers who enter a bank is thought to be Poisson distributed with a mean equal to 10 per hour. What are the chances that 2 or 3 customers will arrive in a 15-minute period

Answer 1

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

Answer 2

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) = $\frac{(mt)^{x}e^{-mt} }{x!}$

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) = $\frac{(2.5)^{2}e^{-2.5} }{2!}+\frac{(2.5)^{3}e^{-2.5} }{3!}$ = 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