|
|
#11
03-14-2007, 12:06 AM
|
|||
|
|||
Re: some probability 101 help (will give $)
[ QUOTE ]
1 I guess they mean exponential distribution with parameter 5 (gamma has 2 parameters). Then mean =1/5 mode = 0 median = 0.693/5=0.1386 exponential [/ QUOTE ] mean is 5 median is you solve by integrating the density function of a gamma distribution. in this case e^(-y/5)/5 which will give us a probability. you integrate this from 0 to y and make it equal to .5 because it is the median. Median is (half of the probability distribution). I believe that's how it works. Mode is 0 that is correct. 
|
|
#12
03-14-2007, 12:09 AM
|
|||
|
|||
|
Re: some probability 101 help (will give $)
[ QUOTE ]
Question 3: Let x = the fraction of a crate that we choose to stock. Then, our profits = $2.25*50 * integral from 0 to x of 3y^2 dy - $.90*50*x, = 112.5x^3 - 45x. We want to maximize this function, where x is from 0 to 1. Looking at the graph, the maximum occurs at x = 1, so we should buy 1 crate = 50 bins of candy. [/ QUOTE ] This is not correct but you are very close. Are you a math grad student?
|
 |
|
|
Linear Mode
