suppose that customers arrive at a store according to a pois
suppose that customers arrive at a store according to a poisson process with rate one customer per minute. suppose that time t=0 is 8am t in minutes.
a) find the probability that exactly three customers arrive from 8 a.m to 8.05am.
b) find the probability of the single event that excatly 3 customers arrive between 8 a.m and 8.02 am and excatly 2 customes arrive between 8.01 am and 8.04 am.
c) find the probability that the fourth customer arreives by 8.05 am.
d) its known that excatly 6 customers have arrived by 8.10am. find the variance of the arrival time point of the fourth customer.
Solution
Average customer per minute = 1
a) Prob (3 customers in 5 minutes ) = 0.1404
b) P(3 customers in 2 minutes) x P(2 in 4 minutes)
=0.1804(0.1465)(since independent)
= 0.0264
c) P( 4th customer at 8.05)
= P(3 customers in 4 minutes)(p 1 customer at last 1 minute)
= 0.1954(0.3679)
= 0.0719
d) 6 customers arrive in 10 minutes
Gap between two cusomers = 1 minute
Hence fourth customer can arrive in 1, 2, 3,4,5,6, 7,8,9 or 10th minute
d)
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
| 0.0153 | 0.0902 | 0.168 | 0.1954 | 0.1755 | 0.1339 | 0.0912 | 0.0572 | 0.0337 | 0.0189 | 0.9793 | |
| 12.02382=variance |
