would mean in 5 minutes we should expect to count about 6,000 events B . 12,000 events

1. Example: a measured rate of 1200 Hz = 1200/sec • would mean in 5 minutes we • should expect to count about • 6,000 eventsB. 12,000 events • C. 72,000 events D. 360,000 events • E. 480,000 eventsF. 720,000 events

2. Example: a measured rate of 1200 Hz = 1200/sec • would mean in 3 millisec we • should expect to count about • 0 eventsB. 1 or 2 events • C. 3 or 4 events D. about 10 events • E. 100s of eventsF. 1,000s of events 1 millisec = 10-3 second

3. Example: a measured rate of 1200 Hz = 1200/sec • would mean in 100 nanosec we • should expect to count about • 0 eventsB. 1 or 2 events • C. 3 or 4 events D. about 10 events • E. 100s of eventsF. 1,000s of events 1 nanosec = 10-9 second

4. The probability of a single COSMIC RAYpassing through a small area of a detector within a small interval of time Dt is p << 1 the probability that none pass in that period is ( 1 -p )1 While waiting N successive intervals (where the total time is t = NDt ) what is the probability that we observe exactlyn events? pn n “hits” × ( 1 -p )??? ??? “misses” × ( 1 -p )N-n N-n“misses”

5. While waiting N successive intervals (where the total time is t = NDt ) what is the probability that we observe exactlyn events? P(n) = nCNpn( 1 -p )N-n From the properties of logarithms you just reviewed ln (1-p)N-n = ln (1-p) ln (1-p)N-n = (N-n)ln (1-p) ??? ln x  loge x e=2.718281828

6. ln (1-p)N-n = (N-n)ln (1-p) and since p << 1 ln (1-p) - p ln (1-p)N-n = (N-n) (-p) from the basic definition of a logarithm this means e???? = ???? e-p(N-n)= (1-p)N-n

7. P(n) = pn( 1 -p )N-n P(n) = pn e-p(N-n) If we have to wait a large number of intervals, N, for a relatively small number of counts,n n<<N P(n) = pn e-pN

8. P(n) = pn e-pN And since N - (n-1)  N (N) (N) … (N) = Nn for n<<N

9. P(n) = pn e-pN P(n) = pn e-pN P(n) = e-Np

10. P(n) = e-Np If the average rate of some random event is p = 24/min = 24/60 sec = 0.4/sec what is the probability of recordingn events in 10 seconds? P(0) = P(4) = P(1) = P(5) = P(2) = P(6) = P(3) = P(7) =

11. P(n) = e- 4 e-4 = 0.018315639 If the average rate of some random event is p = 24/min = 24/60 sec = 0.4/sec what is the probability of recordingn events in 10 seconds? P(0) = 0.018315639 P(4) = 0.195366816 P(1) = 0.073262556 P(5) = 0.156293453 P(2) = 0.146525112 P(6) = 0.104195635 P(3) = 0.195366816 P(7) = 0.059540363

12. P(n) = e-Np Hey! What does Np represent?

13. Another useful series we can exploit m, mean = n=0 term n / n! = 1/(???)

14. ¥ m å ( N p ) - e N p = (N p ) ( m )! = m 0 m, mean let m = n-1 i.e., n = what’s this?

15. m, mean m = (Np) e-Np eNp m = Np

16. m = Np P(n) = e-m Poisson distribution probability of finding exactlyn events within time t when the events occur randomly, but at an average rate of m(events per unit time)

17. Recall: The standard deviation s is a measure of the mean (or average) spread of data away from its own mean. It should provide an estimate of the error on such counts.

18. The standard deviation s should provide an estimate of the error in such counts

19. What is n2 for a Poisson distribution? first term in the series is zero factor out e-m which is independent of n

20. What is n2 for a Poisson distribution? Factor out a m like before Let j = n-1  n = j+1

21. What is n2 for a Poisson distribution?

22. What is n2 for a Poisson distribution? This is just em again!

23. The standard deviation s should provide an estimate of the error in such counts In other words s 2 = m s = m