I am stuck with a statistics problem based on exponential distribution, can anyone help?
Hey Arjun - maybe... What's your problem?
Post the problem i might be able to help out
Thanks for the response folks. The problem is: After given an antibiotic in dogs, certain type of bacteria dies completely within six to twelve hours based on an exponential distribution; i.e. f(x) = 0.95 e ^{-x/3}., where \[6\le x \le\] . What is the probability that a dog receiving the antibiotic will be completely free of bacteria in less than eight hours?
x is greater than and equal to 6, and x is less than and equal to 12
wow hard
you could integrate the function and evaluate first on the interval 6 to 12 and then from 6 to 8. and do the following:\[\int\limits_{6}^{8}f(x) \div \int\limits_{6}^{12}f(x)\] it's just a thought, i'm not entirely sure
Thanks estebananaya, m not too completely sure but trying right now
I knew Zarkon would shoot over here. I'm posting so I can see the solution if you get it.
Well that would be great if Zarkon can figure this out..
the pdf you gave doesn't integrate to 1.
so, do you think there is a typo in question.. I tried estebananaya's way (posted above) and it came down to .25, but not sure if its right..
if \(f\) is a true pdf then it is required that \[\int_{\Omega}f(x)dx=1\] where \(\Omega\) is the support of the distribution
well, no doubts about that.. and one other way I see it, is to subtract individual probabilities, I mean calculating at 8 and 6, and subtracting.. but not sure..
I would double check the problem to make sure you have typed it correctly. If you have typed it correctly then the problem is ill posed since they didn't give you a true density function
Zarkon, in question there is an incomplete equation for x, it has to be \[6\le x \le12\]
for your \(f\) \[\int\limits_{6}^{12}f(x)dx\ne 1\]
thanks
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