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Jennio2875 · Jun 12, 2009, 07:44 AM

A drug manufacturer states that only 5% of the patients, using a high blood pressure drug, will experience side effects. Doctors at a large university hospital use the drug to treat 200 patients. What is the probability that 15 or fewer of the 200 patients experience side effects?

galactus · Jun 12, 2009, 03:18 PM

We can do this like so:

$\sum_{k=0}^{15}C(200,k)(.05)^{k}(.95)^{200-k}$

Do you have a nice calculator to run it through?

Another way that is used when n is large as in this case (n=200), we can use the binomial this way:

${\mu}=np=200(.05)=10$

For the standard deviation, ${\sigma}=\sqrt{npq}=\sqrt{200(.05)(.95)}=3.0822$

Now, we use the z formula. But we incorporate the 'continuity correction'. Instead of 15, we use 14.5 in the formula.

$z=\frac{15.5-10}{3.0822}=1.78$

Look this up in the z table and see how close it is to the other method. It should be close

One thing about this method is that we have to make sure that np>5

In this case it is 10, so we are OK. If np<5, then we can not use it.