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Central Limit Theorem

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Date: 03/08/2002 at 11:04:34
From: Stephanie Sparks
Subject: Probability of "at least"

Here's the problem:

The probability that a drug is effective on any one patient is 62%.
Find the probability that, of the next 200 patients, at least half
(or more) will survive.

I've figured that the probability of the patient dying is 38%, and out
of the next 200 patients, we want to know if at least 100 or more
people will survive, exactly 100 or more successes in 200 trials, but
I don't know what to do from there.

Thanks,
Stephanie
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Date: 03/08/2002 at 17:47:27
From: Doctor Jubal
Subject: Re: Probability of "at least"

Hi Stephanie,

Thanks for writing Dr. Math.

The most straightforward way to do this, and one that would take a lot
of time, is to recognize that the probability of achieving k successes
in N trials has a binomial distribution, which is

N!
P(k) = p^k * (1-p)^(N-k) * ----------
k!(N-k)!

where p is the probability of success on a single trial. You could use
this to calculate P(100), P(101), P(102), and so on to P(200), and
then add all these probabilities up, and you'd have the exact answer.
It would also take you far more time than you probably want to devote
to this problem.

One of the most useful theorems of statistics is the Central Limit
Theorem, which says that if you take a large number of independent
random variables and add them together, no matter what the form of the
probability distributions of the individual variables is, the sum will
have a distribution that is approximately Gaussian.

This is nice because it lets us approximate a binomial distribution
(with a "large enough" value of N) as a Gaussian one, because the
binomial distriubtion itself is really the sum of several Bernoulli
distributions (each trial has a Bernoulli distribution), so as the
number of trials gets large, the binomial distribution becomes more or
less Gaussian. Because the Gaussian distribution is continuous
(whereas the binomial distribution is discrete), it lets us do an
integral instead of the sum P(100) + P(101) + P(102) + ... + P(200).
Best of all, the values of the integral we need to do have already
been calculated and put in the back of almost any mathematical
handbook or statistics text.

The Gaussian distribution that best approximates a binomial
distribution has mean Np and variance Np(1-p), which makes a lot of
sense because Np and Np(1-p) are the mean and variance of the binomial
distribution itself.

So, since 62% of the patients survive, the mean number of survivors in
200 patients is (0.62)(200) = 124. Also, we can expect a variance in
this value of (0.62)(0.38)(200) = 47.12, for a standard deviation of
about 6.86. As you said, we want to know what the probability of at
least 100 patients surviving is, or rather what the probability of
doing no worse than 24 more deaths than the mean. 24 is
(24)/(6.86) = 3.5 times more than the standard deviation.

Now, for a Gaussian distribution, what is the probability of falling
no more than 3.5 standard deviations below the mean?

more, or if you have any other questions.

- Doctor Jubal, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Probability
High School Statistics

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