pneumonia and normal distribution

**logitech** · May 20th 2009, 08:05 PM

I am having trouble solving the following question after reading my textbook. Thank you for any help you can give me.

In adults the pneumococcus bacterium cause 70% of the pneumonia cases. In a random sample of n=84 adults who have pneumonia, let X equal the number whose pneumonia was caused by the pneumococcus bacterium. Use the normal distribution to find, approximately, $P(X \le 52)$

**matheagle** · May 20th 2009, 10:03 PM

X is a binomial rv with n=84 and p=.7.

The exact probability is $\sum_{k=0}^{52}{84\choose k}(.7)^k(.3)^{84-k}$ (or you can use the complement).

The normal approximation is...

$P(X_B\le 52) \sim P(X_N\le 52.5)$

where $X_B$ is the original binomial rv, while $X_N$ is the overlapping normal approximation to that binomial.
Next we center the normal...

$P(X_N\le 52.5)=P\biggl(Z\le {52.5-(84)(.7)\over \sqrt{(84)(.7)(.3)}}\biggr)$

**logitech** · May 28th 2009, 08:10 PM

Thank you so much for clearing that all up! I just finished your work to get $P(Z \le -1.5)=P(Z \ge 1.5)=1-0.9332=0.0668$ . Hopefully I didn't mess that little part up!

**logitech** · June 1st 2009, 06:04 PM

Can someone help me figure out where I went wrong? I got 0.1530328 for the exact probability but 0.0668 using normal distribution, but I think these numbers are too far apart. Thank you!

**logitech** · June 1st 2009, 06:55 PM

I think I got it! I calculated the real probability wrong. It should have been 0.068987

**matheagle** · June 1st 2009, 08:12 PM

With n=84, they should be quite close.
I just checked my approximation and I also $P(Z \le -1.5)$ .

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