September 4, 2007 - Cosmic Telescope

Plateau Zeroing
Time Start	Time Stop	Time elapsed (min.)	Thresholds on both Scintillators	Singles Count (on top)	Coinc. Counts	Coinc. per minute	Angle measure
1:35 pm on 8/31	1:10 pm on 9/04	5725 min	225	626816	1654	.2889	75 degrees

Binomial distribtuion

best example is a coin toss, its either heads or tails

mean ([math]\mu[/math]) = number of tries [math]n[/math](coin flips) * probability of success[math]p[/math] (head, 1/2)

standard deviation([math]\sigma[/math]) = [math]np(1-p)[/math]

Poisson Distribution

standard deviation ([math]\sigma[/math]) = root of the mean ([math]\sqrt{\mu}[/math])

use in counting experiments

the distribtuion approximates the Binomial Distribution for the special case when the mean ([math]\mu[/math]) is a lot less than the number of attempts to measure ([math]n[/math]) because the probability of the event occurrring is small.

In the cosmic ray telescope experiment the mean number of detected cosmic rays is much smaller than the number of cosmic rays passing by.

Gaussian/Normal Distribution

Full WIdth at Half Max (FWHM) = width of the distribution at half the value of the maximum probabilty (distibution peak) = [math]\Gamma[/math]

standard deviation ([math]\sigma[/math]) = [math]\frac{\Gamma} {2.354}[/math]

error = [math]2\ln(2) \sigma[/math]

Time Start	Time Stop	Time elapsed (hour)	Singles Count (on top)	Coinc. Counts	Coinc. per hour	[math]\sqrt{N}[/math]	[math](x_i-\bar{x})^2[/math]
1:20 pm	2:20 pm	1	5694	127	127	11.3	121
2:20 pm	3:20 pm	1	4896	136	136	11.6	4
3:20 pm	4:20 pm	1	4655	151	151	12.2	169

Mean [math]\mu[/math] = 138 counts per hour

Instrumental Uncertainty =[math]\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{N-1}} = 12.12[/math] counts per hour

Your instrumental uncertainty is approximately equal to the Poisson sigma ([math]\sqrt{138}= 11.7[/math]) counts per hour.

The cosmic ray telescope sounting experiment appears to be following Poisson statistics.