# Thread: 5.5 help teacher is a mofe

1. ## 5.5 help teacher is a mofe

A drug company tested a new drug on 250 pigs with swine flu. Historically, 20% of pigs contracting swine flu die from the disease. Of the 250 pigs treated, 215 recovered. Estimate the probability that this result is by chance.

Please somebody help. the class average for our classes last unit test was 14%. Our teacher is such a mean sob he says its our fault...this is a question on the make-up test and i still dont get it. Also, share your horrible teacher stories, itll be hilarious stuff.

2. Originally Posted by Blockface
A drug company tested a new drug on 250 pigs with swine flu. Historically, 20% of pigs contracting swine flu die from the disease. Of the 250 pigs treated, 215 recovered. Estimate the probability that this result is by chance.
The number that die has a binomial distribution, but here we have moderately
large numbers so we will use the normal approximation to the binomial distribution.

Expected number of deaths in $\displaystyle 250$ cases is $\displaystyle \mu=0.2\times 250=50$.

The standard deviation of a binomially distributed random variable is $\displaystyle \sqrt{Np(1-p}$,
where $\displaystyle N$ is the number of trials ($\displaystyle N=250$ in this case), $\displaystyle p$ is the probability
of the outcome of interest on a single trial (in this case $\displaystyle p=0.2$). So the
standard deviation of the number of deaths is:$\displaystyle \sigma=\sqrt{250\times 0.2 \times 0.8}\approx 6.32$

So now the question is: what is the probability of 35 or fewer deaths
happening by chance assuming the treatment has no effect.

We work with 35.5 or fewer to allow for the continuous nature of the normal
approximation, and so the critical z-score we are interested in is:

$\displaystyle z=(35.5-50)/6.32 \approx -2.29$

This is looked up in a table of the cumulative standard normal distribution to
give:

$\displaystyle P(\mbox{35 or fewer deaths given no treatment})\approx 0.011$,

or about $\displaystyle 1.1 \%$

RonL