Question

3. Bing is a diligent graduate student. She is a proud member of the American Statistical Association (ASA). ASA frequently e-mails their members with updates and information of interest. Each e-mail is sent independently of one another. Bing realizes ASA sends her an e-mail on average every 12 hours.
a) Let W be the number of e-mails Bing receives in the next twenty-four hours. What are the distribution, parameter(s), and support of W?
b) What is the probability Bing receives 3 e-mails on Tuesday AND no e-mails on Wednesday?
c) It’s 7:30 AM and Bing receives an update from ASA. What is the probability that Bing will receive her next e-mail from ASA by 1:30 PM? What distribution and parameter(s) are you using?
d) Knowing that the next e-mail from the ASA takes more than 10 hours, what is the probability that it will take more than 18.5 hours?
e) Bing receives 15 e-mails in one week (168 hours). What is the probability that 3 e-mails were sent from ASA in the last 24 hours of that one week period?

Answer 1

a)

disrtibution of W is poisson with parameter $\lambda$ =expected number of emails in 12 hours

suport of W is from 0 to infinity

b)

xpected number of emails in one day =24*1/12 =2= $\lambda$

therefore probability Bing receives 3 e-mails on Tuesday AND no e-mails on Wednesday

=P(X1=3)*P(X2=0) =(e^-2*2³/3!)*(e^-2*2⁰/0!) =0.0244

c)

here tis is exponential distribution with parameter =waiting time = $\beta$ =12 Hours

probability that Bing will receive her next e-mail from ASA by 1:30 PM(6 hours) =P(X<6) =1-e^{-x/ $\beta$} =1-e^-6/12

=0.3935

d)

P(X>18.5|X>10) =P(X>8.5) =e^-8.5/12 =0.5075

e)for 24 Hours expected number of emails =15*24/168=2.143

probability that 3 e-mails were sent from ASA in the last 24 hours of that one week period =P(X=3)

=e^-2.143*2.143³/3! =0.1924