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Mean/Variance of Uniform Probability Distribution 
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#1
Sep2808, 11:07 AM

P: 26

Find the mean and variance of the uniform probability distribution:
f(x) = 1/x for x = 1,2,3,...,n Hint: The sum of the first positive n integers is n(n + 1)/2, and the sum of their squares is n(n + 1)(2n + 1)/6 I know mu/mean will be the sum of products of x and its probability of occurring over all x (through n in this case). I just don't know how to incorporate the formulas given in the hint into the general formulas for mean and variance: [tex]\Sigma[/tex]xf(x) is the mean [tex]\Sigma[/tex] (x  [tex]\mu[/tex])^{2} f(x) is the variance. Thank you for your help, (I don't know why mu is elevated.) Jeff 


#2
Sep2808, 03:38 PM

Sci Advisor
P: 6,031

Since you have the first 2 moments, the variance formula can be simplified.
variance=second moment  square of first moment. (This comes directly from the definition). Notes: f(x)=1/n, not 1/x. To get the moments, your sums must be divided by n. 


#3
Sep2808, 10:20 PM

P: 26

Mathman  can you show me how to calculate the first and second moments of the probability distribution?



#4
Sep2908, 04:02 PM

Sci Advisor
P: 6,031

Mean/Variance of Uniform Probability Distribution
The moments in your case are simply the sums divided by n. In general if the probabilites are unequal, then you sum the numbers (first moment) or squares (second moment) multiplied by the probabilities.



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