Answer:
a) 0.012 or 1.2% probability
b) 0.2817 or 28.17%
Step-by-step explanation:
a) In this case, we will apply Poisson distribution because of the given period of time as distribution.
We use the Poisson distribution formula:
[tex] P(X) = u^x - e^-^u / X! [/tex]
Where;
X= Random variable (number of customers arriving in the interval)
u= mean
Applying the formula we have;
[tex] P(6) = (2^e e^-^2) / (6!) [/tex]
[tex]= 64 * 0.135 / 720 = 0.012 [/tex]
Therefore, there is a 0.012 probability of 6 customers arriving in a randomly selected 1 minute interval.
b) Here, we use exponential problem formula
[tex] P(x<x0) = 1 - e^-^(^x^0^/^u) [/tex]
[tex] P(>_ 2) = 1 - e^-^(^2^*^1) [/tex]
= 1 - 0.71828 = 0.28172
Therefore, there is a probability of 0.28172 of 2 minutes gap
