* This calculates how many sigma needed

from scipy.stats import norm

# def findSigma(errval):
#   tar = (1.0 - errval/2.0)
#   return(norm.ppf(tar))

def findSigma(errval):
  tar = errval/2.0
  return(-norm.ppf(tar))


for i in range(1, 16):

  gsiz = 10**i
  erra = 1/gsiz
  nsig = findSigma(erra)

  print("1e-%02d   %0.4f" % (i, nsig))


greg@DBK8S5D3:~$ python3 junk.py
1e-01   1.6449
1e-02   2.5758
1e-03   3.2905
1e-04   3.8906
1e-05   4.4172
1e-06   4.8916
1e-07   5.3267
1e-08   5.7307
1e-09   6.1094
1e-10   6.4670
1e-11   6.8065
1e-12   7.1305
1e-13   7.4409
1e-14   7.7393
1e-15   8.0269
greg@DBK8S5D3:~$ 




* **********************************************************************************
Poisson Distribution
* **********************************************************************************



In the World Cup, an average of 2.5 goals are scored each game. Modeling this situation with a Poisson distribution, what is the probability that kk goals are scored in a game?

Conditions for Poisson Distribution:

  * An event can occur any number of times during a time period.

  * Events occur independently. In other words, if an event occurs, it does not affect the probability of another event occurring in the same time period.

  * The rate of occurrence is constant; that is, the rate does not change based on time.

  * The probability of an event occurring is proportional to the length of the time period. For example, it should be twice as likely for an event to occur in a 2 hour time period than it is for an event to occur in a 1 hour period.


Let X be how many events happen of a fixed time interval.
Let lamda be the average number of events that are known to happen over that interval.
(X is an integer, lamda can be a real number)

Example average number of interruptions per hour = 3.217
Then lamda = 3.217
X could be 0, 1, 2, 3, 4, 5, 6, 7, 8, ..... 20
X could be bigger than 20, but practically speaking it coundn't
We thus must have
P(X=0) + P(X=1) + ... + P(X=20) = 1.0

P(X=k) = lamda^k * exp(-lamda) / k!

Example : In the World Cup, an average of 2.5 goals are scored each game. Modeling this situation with a Poisson distribution, what is the probability that k goals are scored in a game?

P(X=0) = 2.5^0 * exp(-2.5) / 0! = 0.082
P(X-1) = 2.5^1 * exp(-2.5) / 1! = 0.205
P(X-1) = 2.5^2 * exp(-2.5) / 2! = 0.257
.
.
.

* **********************************************************************************