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?

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.