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}(\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 = (12\,\rm customers)/(20\,\rm min)*(20\,\mathrm{min}) = 12\,\mathrm{customers}`

Then

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

(b) Now you want

`P(X\ge4) = 1 - P(X<4) = 1 - \displaystyle\sum_(x=0)^3 P(X=x)`

This time, we have

`\lambda = (6\,\rm customers)/(10\,\rm min)*(10\,\mathrm{min}) = 6\,\mathrm{customers}`

so that

`P(X\ge4) = 1 - \displaystyle \sum_(x=0)^3 (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 = (6\,\rm customers)/(6\,\rm min)*(1\,\mathrm{min}) = 1\,\mathrm{customer}`

So we have

`P(X > 1) = 1 - (1^0 e^(-1))/(0!) - (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 = (12\,\rm customers)/(6\,\rm min)*(1\,\mathrm{min}) = 2\,\mathrm{customers}`

so that

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