Let's say you check a website at the top of the hour every hour (both day and night). The posts arriveindependently at an average rate of 0.1 posts per hour, which means that the number of new posts each time you look is a Poisson random variable with intensity λ = 0.1. Answer each of the following questions with an expression of form AeB and then do evaluate the expression as a number.Fed up with this unsatisfying routine, you decide to sign up for daily digest emails. What is the probability that on a givenday you do not receive a daily digest email at 1pm? In other words, what is the probability that at 1pm there were no newposts in the preceding 24 hours?

Answered: Let's say you check a website at the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Let's say you check a website at the top of the hour every hour (both day and night). The posts arrive

independently at an average rate of 0.1 posts per hour, which means that the number of new posts each time you look is a Poisson random variable with intensity λ = 0.1. Answer each of the following questions with an expression of form AeB and then do evaluate the expression as a number.
1. Fed up with this unsatisfying routine, you decide to sign up for daily digest emails. What is the probability that on a given
  
  day you do not receive a daily digest email at 1pm? In other words, what is the probability that at 1pm there were no new
  
  posts in the preceding 24 hours?

Expert Solution

Step 1: Overview

From the previous part, probability that there are no new posts when you look the website: $P left parenthesis X equals 0 right parenthesis equals fraction numerator e to the power of negative 0.1 end exponent asterisk times 0.1 to the power of 0 over denominator 0 factorial end fraction equals e to the power of negative 0.1 end exponent equals 0.9048$

Note that for part 1, the expert did a little mistake in the calculation, the correct value of probability P(X=0) = 0.9048 = 0.905. Part 2 and 3 are correct.