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Example: a measured rate of. 1200 Hz = 1200 /sec. would mean in 5 minutes we should expect to count about 6,000 events B . 12,000 events C . 72,000 events D . 360,000 events E . 480,000 events F . 720,000 events. Example: a measured rate of. 1200 Hz = 1200 /sec.
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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
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
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
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”
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
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
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
P(n) = pn e-pN And since N - (n-1) N (N) (N) … (N) = Nn for n<<N
P(n) = pn e-pN P(n) = pn e-pN P(n) = e-Np
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) =
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
P(n) = e-Np Hey! What does Np represent?
Another useful series we can exploit m, mean = n=0 term n / n! = 1/(???)
¥ m å ( N p ) - e N p = (N p ) ( m )! = m 0 m, mean let m = n-1 i.e., n = what’s this?
m, mean m = (Np) e-Np eNp m = Np
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)
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.
The standard deviation s should provide an estimate of the error in such counts
What is n2 for a Poisson distribution? first term in the series is zero factor out e-m which is independent of n
What is n2 for a Poisson distribution? Factor out a m like before Let j = n-1 n = j+1
What is n2 for a Poisson distribution? This is just em again!
The standard deviation s should provide an estimate of the error in such counts In other words s 2 = m s = m