# Mean/Variance of Uniform Probability Distribution

by JeffNYC
Tags: probability, statistics
 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: $$\Sigma$$xf(x) is the mean $$\Sigma$$ (x - $$\mu$$)2 f(x) is the variance. Thank you for your help, (I don't know why mu is elevated.) Jeff
 Sci Advisor P: 5,935 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.
 P: 26 Mathman - can you show me how to calculate the first and second moments of the probability distribution?