# 13% of people took a math course. What is the probability that in a 350 randomly selected sample, less than 40 people take the course?

13 % of people took a math course. What is the probability that in a 350 randomly selected sample, less than 40 people take the course?

So I have $X\sim Bin\left(350,\frac{13}{100}\right)$
Then let $Y\sim N\left(\frac{13}{100},{\sqrt{0.00032314}}^{2}\right)$
I want $P\left(X<40\right)=P\left(\stackrel{^}{p}<\frac{4}{10}\right)$
Then $P\left(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}}\right)$ which is obviously very wrong.
Where did it go wrong, and why did it go wrong?
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Octavio Barr
The number $X$ of people taking the course is $B\left(n,\phantom{\rule{thinmathspace}{0ex}}p\right)$-distributed with $n=350,\phantom{\rule{thinmathspace}{0ex}}p=0.13$. Its variance is npq,$npq,\phantom{\rule{thinmathspace}{0ex}}q:=1-p$. The $z$-score of $X$ is $Z:=\frac{X-np}{\sqrt{npq}}$, so the Normal approximation of $P\left(X<40\right)$ is
$P\left(Z<\frac{40-np}{\sqrt{npq}}\right)=P\left(Z<\frac{40-350×0.13}{\sqrt{350×0.13×0.87}}\right).$
(Depending on how you seek to discretize the Normal variable approximating $X$, you might replace 40 with e.g. 39.5.)