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June 20th, 2007, 03:57 PM   #1
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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.
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September 5th, 2007, 08:47 AM   #2
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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
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September 5th, 2007, 01:12 PM   #3
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Re: Standard Deviation

Quote:
Originally Posted by symmetry
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.
Standard deviation is defined as the square root of the variance.

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.
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September 6th, 2007, 07:23 AM   #4
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So are we supposed to assume that the police department is reporting the population probabilities rather than a sample probability?
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September 6th, 2007, 08:00 AM   #5
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Quote:
Originally Posted by CRGreathouse
So are we supposed to assume that the police department is reporting the population probabilities rather than a sample probability?
Yes, we can assume from the probability distribution that there are only 3 burglarable properties in the town
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