posted by Craig .
An investigational drug study is being conducted. The initial screening phase needs to include enough people so that the final (approved) group has at least 10 patients. The probability of a person being approved for the study is .80. what initial sample size is needed to make sure with a .939 probability that the study will have at least 10 patients?
i don't know how to go about working this problem. i know its a discrete random variable problem and i know that i somehow have to work backwards, but i have NO idea how to set it up or work it. any and all guidance would be appreciated. thank you!!!
i don't expect you to work it for me, but i do need some help.
would this be a binomial-like should i consider it
applicable for study
not applicable for study
Yes, the binomial distribution is a discrete random variable problem.
Basically, if you screen N people, each having a success rate of 0.8. You would like to calculate the minimum candidates to have a minimum of 10 success with a probability of 0.939.
It is a very similar process to flipping coins, except the probability of success is 0.8 instead of 0.5.
The probability can be obtained by tabulation, but the a formula exists to calculate the probability of success of n times out of N candidates, with a probability of success of p.
Let say, since the proability of success is 0.8, and 10 people are required, a good starting point is to screen 10/0.8=13 candidates.
So N=13, p=0.8,
P(10,13,0.8) = 0.246
P(11,13,0.8) = 0.268
P(12,13,0.8) = 0.179
P(13,13,0.8) = 0.055
Probability of retaining 10 or more candidates
= 0.747 < 0.939 required.
So you could screen 14 candidates and calculate the probability. Keep increasing the number of candidates until you get the probability of 0.939 or higher.
Oh yes, here's a link for more explanation:
Also, in case you are not familiar with the notation,
nCr is the combination function defined as:
thank you so much! i think i got it!
what was the answer you finally got