(New page: When calculating the expected value of X in part B, the summation is solved by using the sum of an arithmetic series. In an arithmetic series, each successive term has a constant differen...) |
m (2b Gregory Pajot moved to 4.2b Gregory Pajot: Improperly named originally) |
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Latest revision as of 06:47, 15 October 2008
When calculating the expected value of X in part B, the summation is solved by using the sum of an arithmetic series. In an arithmetic series, each successive term has a constant difference, which in this case is just 1. Here we're summing from 1 to n, which equals n(n+1)/2.
In response to Arie, part B's random variable doesn't really have a typical classification (i.e., not binomial or geometric or anything) since the PMF is just 1/n.