Jadon Melendez

2022-07-22

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 $X\sim Bin(350,\frac{13}{100})$

Then let $Y\sim N(\frac{13}{100},{\sqrt{0.00032314}}^{2})$

I want $P(X<40)=P(\hat{p}<\frac{4}{10})$

Then $P(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}})$ which is obviously very wrong.

Where did it go wrong, and why did it go wrong?

So I have $X\sim Bin(350,\frac{13}{100})$

Then let $Y\sim N(\frac{13}{100},{\sqrt{0.00032314}}^{2})$

I want $P(X<40)=P(\hat{p}<\frac{4}{10})$

Then $P(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}})$ which is obviously very wrong.

Where did it go wrong, and why did it go wrong?

Octavio Barr

Beginner2022-07-23Added 11 answers

The number $X$ of people taking the course is $B(n,\phantom{\rule{thinmathspace}{0ex}}p)$-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 $P(X<40)$ is

$P(Z<\frac{40-np}{\sqrt{npq}})=P(Z<\frac{40-350\times 0.13}{\sqrt{350\times 0.13\times 0.87}}).$

(Depending on how you seek to discretize the Normal variable approximating $X$, you might replace 40 with e.g. 39.5.)

$P(Z<\frac{40-np}{\sqrt{npq}})=P(Z<\frac{40-350\times 0.13}{\sqrt{350\times 0.13\times 0.87}}).$

(Depending on how you seek to discretize the Normal variable approximating $X$, you might replace 40 with e.g. 39.5.)

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