The fraction of a cohort of AIDS patients that survives a time t after AIDS diagnosis is given by S(t) = e^(-kt).

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

SOLVED

Answer 4

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

Answer 5

That link shows you how to get mean = 1/k

Answer 6

whatever