Hi!
As I was going about my day today I stumbled across a problem I became curious about. The context is in a video game, where an item exists that skips time forward forward for you between 5 and 500 hours. The description of the item states the following:
"Gives 24 hrs on average. Odds are: 54% for 5-10 hrs, 25% for 10-25 hrs, 18% for 25-100 hrs, 2.1% for 100-499 hrs, 0.4% for 500 hrs."
Notice it adds up to 0.995, there is some rounding.
What I want to find out is, what probability density function produces these probabilities?
The integrals I thought were then integral (eh?) were:
[math]\int_5^{10} f(x)\,dx = 0.54, \int_{10}^{25} f(x)\,dx=0.25, \int_{25}^{100} f(x)\,dx=0.18, \int_{100}^{499} f(x)\,dx=0.021, \int_{499}^{500} f(x)\,dx=0.004, \int_{5}^{500} f(x)\,dx=1[/math] (If we're using the last one we would also have to add 0.05 cumulatively to the other values)
My thought was an exponential probability function, [math]y=me^{-mx} ; m=1/μ, μ=24, m=1/24[/math]However this doesn't work, even if we shift the equation with x->(x-5) so that we begin at 5, the integrals applied on that function do not add up to the same values. The exponential probability function cant apply if we use that last integral, since [math]\int_0^{\infin}me^{-mx}=1[/math]Unless subtracting some number makes it usable?
Thanks in advance!
As I was going about my day today I stumbled across a problem I became curious about. The context is in a video game, where an item exists that skips time forward forward for you between 5 and 500 hours. The description of the item states the following:
"Gives 24 hrs on average. Odds are: 54% for 5-10 hrs, 25% for 10-25 hrs, 18% for 25-100 hrs, 2.1% for 100-499 hrs, 0.4% for 500 hrs."
Notice it adds up to 0.995, there is some rounding.
What I want to find out is, what probability density function produces these probabilities?
The integrals I thought were then integral (eh?) were:
[math]\int_5^{10} f(x)\,dx = 0.54, \int_{10}^{25} f(x)\,dx=0.25, \int_{25}^{100} f(x)\,dx=0.18, \int_{100}^{499} f(x)\,dx=0.021, \int_{499}^{500} f(x)\,dx=0.004, \int_{5}^{500} f(x)\,dx=1[/math] (If we're using the last one we would also have to add 0.05 cumulatively to the other values)
My thought was an exponential probability function, [math]y=me^{-mx} ; m=1/μ, μ=24, m=1/24[/math]However this doesn't work, even if we shift the equation with x->(x-5) so that we begin at 5, the integrals applied on that function do not add up to the same values. The exponential probability function cant apply if we use that last integral, since [math]\int_0^{\infin}me^{-mx}=1[/math]Unless subtracting some number makes it usable?
Thanks in advance!