Homework Help is Here – Start Your Trial Now!
Math
Probability
The number of people arriving at a coffee shop is a Poisson process with a rate of 2 per 10 minutes. (a) Find the probability that: i. Between 9 and 10 am no customers arrive. il. more than 5 customers arrive between 9 and 9:30 am. i. two or less customers arrive between 10 and 10:30 am.

The number of people arriving at a coffee shop is a Poisson process with a rate of 2 per 10 minutes. (a) Find the probability that: i. Between 9 and 10 am no customers arrive. il. more than 5 customers arrive between 9 and 9:30 am. i. two or less customers arrive between 10 and 10:30 am.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question

Write this answer so i can just copy and paste it.

question with answer is given below 

The number of people arriving at a coffee shop is a Poisson process with a rate of 2 per 10 minutes. (a) Find the probability that: i. Between 9 and 10 am no customers arrive. il. more than 5 customers arrive between 9 and 9:30 am. i. two or less customers arrive between 10 and 10:30 am.
Transcribed Image Text:The number of people arriving at a coffee shop is a Poisson process with a rate of 2 per 10 minutes. (a) Find the probability that: i. Between 9 and 10 am no customers arrive. il. more than 5 customers arrive between 9 and 9:30 am. i. two or less customers arrive between 10 and 10:30 am.
The number of people arriving at a coffee shop is a Poisson process with a rate of 2 per 10 minutes X ~ P(2) per 10 ,minutes i.e X~ P(12) per hour e-12 12" P(X) = x! i) Between 9 and 10 am no customers arrive. P(X = 0) = e-12 P(X = 0) = 6.3 * 10–6 ii) more than 5 customers arrive between 9 and 9:30 am P(X > 5) = 1 – P(X < 5) P(X > 5) = 1- [P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+ P(X = 4) + P(X = 5)| 6°e-6 + 2 6²e-6 64e-6 65e-6 P(X > 5) = 1 – [e -6 + 6e -6 3! 4! 5! P(X > 5) = 1 – [e-6(1+6+ 18+36+ 34 + 64 +8)] P(X >5) = 1 – 179.8e-6 P(X > 5) = 1 – 0.4579 | P(X > 5) = 0.5460 iii) two or less customers arrive between 10 and 10:30 am. P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) %3D 62. P(X < 2) = e¬"[1+6+¬P(x < 2) = 2.5 * e-6 P(X < 2) = 0.06311
Transcribed Image Text:The number of people arriving at a coffee shop is a Poisson process with a rate of 2 per 10 minutes X ~ P(2) per 10 ,minutes i.e X~ P(12) per hour e-12 12" P(X) = x! i) Between 9 and 10 am no customers arrive. P(X = 0) = e-12 P(X = 0) = 6.3 * 10–6 ii) more than 5 customers arrive between 9 and 9:30 am P(X > 5) = 1 – P(X < 5) P(X > 5) = 1- [P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+ P(X = 4) + P(X = 5)| 6°e-6 + 2 6²e-6 64e-6 65e-6 P(X > 5) = 1 – [e -6 + 6e -6 3! 4! 5! P(X > 5) = 1 – [e-6(1+6+ 18+36+ 34 + 64 +8)] P(X >5) = 1 – 179.8e-6 P(X > 5) = 1 – 0.4579 | P(X > 5) = 0.5460 iii) two or less customers arrive between 10 and 10:30 am. P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) %3D 62. P(X < 2) = e¬"[1+6+¬P(x < 2) = 2.5 * e-6 P(X < 2) = 0.06311
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer