See attached file for clarity.
#3. Assume that the number of uninspected cars caught at a state police checkpoint is Poisson distributed with average 2.1 per hour. (a) What is the average number of cars caught in t hours? (b) What are P(no cars caught
in 14 hours? (c) P(at least 3 in 1.5 hours); (d) P(at least 1 car caught within 10 minutes of setting up the checkpoint)?
#7 a. Assume that batteries last for a time that is exponentially distributed with average 2 Months. If a transmitter needs one battery at a time and 4 (including the original) are taken on a expedition, what are (a)
P(Transmitter last at least 11 Months)?
#9. Louise sells encyclopedias door to door. She stops for the day after 4 sales. Assume that the time between sales is exponentially distributed. (a) If she can expect to work for 7 hours to make the 4 sales, what is the
parameter / ? What is the probability that she will stop (b) before 6 hours; (c) after 9 hours?
#10. Suppose that customers enter a store according to a Poisson stream with average 40.7 for a whole day. Suppose that 2 out of5 customers result in a sale. What is the distribution for the number of sales in half a day.
See the attached file for full solution.
#3. Given the number of cars caught is a Poisson process at an average rate 2.1 per hour. Then if X denotes the number of cars caught in an interval of duration t hours, X follows Poisson distribution with parameter (i.e., mean )
a) Average number of cars caught in t hours is =
b) Number of cars caught in ¼ hours follows Poisson distribution with mean
2.1 * 0.25(= 0.525).
Probability that no cars caught in ¼ hours = = 0.5916
Note: can be obtained by using the MS excel function POISSON (0, 0.525, 0)
c) Number of cars caught in 1.5 hours follows Poisson distribution with mean
2.1*1.5 = 3.15.
Probability that at least 3 cars caught within 1.5 hours
= = 1-0.3904 = 0.6096
Note: can be obtained by using the MS excel ...
The solution discusses four Poisson distribution problems.