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June 20th, 2007, 03:57 PM  #1 
Senior Member Joined: Dec 2006 From: New Jersey Posts: 378 Thanks: 1  Standard Deviation
A Police department reports the probabilities that 0,1,2,and 3 burglaries will be reported in any given day are 0.46, 0.44, 0.08, and 0.02, respectively. Find the Standard Deviation for the Probability Distribution. Round to nearest hundredths. 
September 5th, 2007, 08:47 AM  #2 
Member Joined: Aug 2007 Posts: 93 Thanks: 0 
Perhaps number of burglaries per day is best modelled with a Poisson distribution where the variance is equal to the mean, but using your numbers: x = 0,1,2 3 E(x) = Sum (x * p(x)) = 0.66 the Variance is Sum ( (E(x)  x)^2*p(x)) = 0.50 So the standard deviation is 0.50^.5 = 0.71 
September 5th, 2007, 01:12 PM  #3  
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716  Re: Standard Deviation Quote:
The variance can be calculated as the second moment minus square of first moment. First moment (mean) = .46*0+.44*1+.08*2+.02*3 Second moment=.46*0+.44*1+.08*4+.02*9 I'll let you do the arithmetic.  
September 6th, 2007, 07:23 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
So are we supposed to assume that the police department is reporting the population probabilities rather than a sample probability?

September 6th, 2007, 08:00 AM  #5  
Member Joined: Aug 2007 Posts: 93 Thanks: 0  Quote:
 

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