A local store opens at 6:00am every day. Suppose that there is one customer comi
ID: 3220577 • Letter: A
Question
A local store opens at 6:00am every day. Suppose that there is one customer coming to the store at each minute with probability 0.02 Let X be the time until the first customer comes to the store once it is opened. Find the PMF of X and E[X]? Given that there is not any customer for the first 1 hour since it is opened and let Y be the time until the first customer comes after the first 1 hour since it is opened. Find E[Y]? Suppose that there is one customer coming to the store every 10 minutes on the average. 1. Find the probability that there are no customer coming to the store between 6:00am and 7:00am? 2. Find the probability that there are at least 2 customers coming to the store between 6:00am and 7:00am?Explanation / Answer
a)
1)Given probability P=0.02
Let X be time in minutes and X= 1,2,3....... min (until customer come to the store)
Probability of mass function P(X) =e-m.mx /X!
= e-1.1x /X! (since m=1, mean number of customer )
= e-1/X!
E(X) = e-1/1! + e-1/2! + e-1/3! + e-1/4! ...............
2) given that no customer for 1 hour(60 minuts)
n=60 min
m=n.P =60*0.02
=1.2
P(Y) = e-m.my /Y!
P(0)= e-1.2(1.2)0 /0!
= e-1.2
=0.3012
E(Y)=0.3012
b. given that there are 1 customer for every 10 minutes, so 6customers for 1 hour (time taken hours here) m=6
1.Probability of no customer coming between 6:00 to 7:00
X=0, m=6,
P(X=0)= e-m.mx /X!
= e-660 /0! =e-6
=0.00247
P(X=0) =0.00247
2)Probability of at least 2 customer between 6:00 to 7:00
P(X2) = 1-P(X<2)
= 1- [ P(X=0)+ P(X=1)]
=1-[ e-660 /0! + e-661 /1! ]
= 1-[0.00247+0.01487]
=0.9827
=98.27%
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.