Question

Consider a Poisson distribution with an average of three customers per minute at a local grocery store. IF X = the number of arrivals per minute, find the probability of more than 7 customers arriving within a minute.

Answer 1

Answer: 0.0216

Step-by-step explanation:

Given : Average arrivals of customers at a local grocery store = 3 per minute

The Poisson distribution formula :-

$\dfrac{e^{-\lambda}\lambda^x}{x!}$, where $\lambda$ is the mean of the distribution.

If X = the number of arrivals per minute, then the probability of more than 7 customers arriving within a minute will be :-

$\dfrac{e^{-3}(3)^7}{7!}=0.0216040314525\approx0.0216$

Hence, the probability of more than 7 customers arriving within a minute = 0.0216

Answer 2

Answer:

0.012

Step-by-step explanation:

The probability of more than 7 customers arriving within a minute is obtained by taking the probability at X equal to 0, 1, 2, 3, 4, 5, 6, and 7 then subtracting from the total probability. It can be expected about 1.2% of times that more than 7 customers arriving within a minute.