Question

A. X is a random variable denotes number of customers visiting a local coffee shop, which follows a Poisson distribution. The mean number of customers per 10 minutes is 6.
a. What is the probability that there are 8 or less customers in the next 20 minutes?
b. What is the probability that there are more than 4 customers in the next 10 minutes?
B. X is a random variable denotes number of customers visiting a local coffee shop, which follows a Poisson distribution. The mean number of customers per 6 minutes is 6?
a. What is the probability the associate have to wait less than 1 minute to have the next customer showing up?
C. X is a random variable denotes number of customers visiting a local coffee shop, which follows a Poisson distribution. The mean number of customers per 6 minutes is 12?
a. What is the probability the associate have to wait more than 1 minutes to have the next customer showing up?

Answer 1

A

(a) You're looking for

$P(X\le 8) = \displaystyle \sum_{x=0}^8 P(X=x)$

where

$P(X=x) = \begin{cases}\dfrac{\lambda^x e^{-\lambda}}{x!}&\text{if }x\in\{0,1,2,\ldots\}\\0&\text{otherwise}\end{cases}$

Customers arrive at a mean rate of 6 customers per 10 minutes, or equivalently 12 customers per 20 minutes, so

$\lambda = \dfrac{12\,\rm customers}{20\,\rm min}\times(20\,\mathrm{min}) = 12\,\mathrm{customers}$

Then

$\displaystyle P(X\le 8) = \sum_{x=0}^8 \frac{12^x e^{-12}}{x!} \approx \boxed{0.155}$

(b) Now you want

$P(X\ge4) = 1 - P(X$

This time, we have

$\lambda = \dfrac{6\,\rm customers}{10\,\rm min}\times(10\,\mathrm{min}) = 6\,\mathrm{customers}$

so that

$P(X\ge4) = 1 - \displaystyle \sum_{x=0}^3 \frac{6^x e^{-6}}{x!} \approx \boxed{0.849}$

B

(a) In other words, you're asked to find the probability that more than 1 customer shows up in the same minute, or

$P(X > 1) = 1 - P(X \le 1) = 1 - P(X=0) - P(X=1)$

with

$\lambda = \dfrac{6\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 1\,\mathrm{customer}$

So we have

$P(X > 1) = 1 - \dfrac{1^0 e^{-1}}{0!} - \dfrac{1^1 e^{-1}}{1!} \approx \boxed{0.264}$

C

(a) Similar to B, you're looking for

$P(X \le 1) = P(X=0) + P(X=1)$

with

$\lambda = \dfrac{12\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 2\,\mathrm{customers}$

so that

$P(X\le1) = \dfrac{2^0e^{-2}}{0!} + \dfrac{2^1e^{-2}}{1!} \approx \boxed{0.406}$