The fraction of a cohort of AIDS patients that survives a time t after AIDS diagnosis is given by S(t) = e^(-kt).
Suppose the mean survival for a cohort of hemophiliacs diagnosed with AIDS before 1986 was found to be T (avg) = 6.4 months.
What fraction of the cohort survived 5 years after AIDS diagnosis?
so, what percent of patients survive for 6.4 months? Is there a standard deviation? Not knowing these items, I can't figure out the value of k. However, let's say that "mean" survival means 50%. In that case,
e^-6.4t = 0.5
-6.4t = ln 0.5
t = 0.1083
So, using
S(t) = e^-0.1083t
then after 5 years (60 months) you'd expect
S(60) = e^-6.498 = 0.00150 = 1.5%
to have survived.
If I got the meaning of "mean" wrong, I'm sure you can fix the calculation.
The first part of the quesiton asked this: Suppose the fraction of a cohort of aids patients that survives a time t after aids diagnosis is given by S(t) = exp(-kt).
Show that the average survival time T average after aids diagnosis for a member of this cohort is given by T average= 1/k.
SOLVED
https://en.wikipedia.org/wiki/Exponential_decay#Solution_of_the_differential_equation
mean = 1/k = 1/6.4
s = e^-t/6.4
in months
what is that at t = 5*12 = 60 months
s = e^-(60/6.4) = e^-9.375
= .0000848
That link shows you how to get mean = 1/k
whatever
To find the fraction of the cohort that survived 5 years after AIDS diagnosis, we need to substitute the given values into the equation S(t) = e^(-kt).
Given: T (avg) = 6.4 months
We know that T (avg) is the average survival time, which is equal to the time t when the fraction of the cohort is 1/e. Therefore, we can set up the equation:
1/e = e^(-k * 6.4)
To find the value of k, we need to take the natural logarithm (ln) of both sides:
ln(1/e) = -k * 6.4
Since ln(1/e) = -1, the equation becomes:
-1 = -k * 6.4
Dividing both sides by -6.4, we can solve for k:
k = 1/6.4
Now that we have the value of k, we can find the fraction of the cohort that survived 5 years (which is equal to 60 months) after AIDS diagnosis. Plugging these values into the equation:
S(t) = e^(-k * t)
S(60) = e^(-1/6.4 * 60)
Now we can calculate this fraction using a calculator or software that supports exponentials and natural logarithms.