Finding the probability of an event with binomial distribution using a normal approximation 
A Tarheels basketball player is obsessed about practicing his free throws. It is known that he is $75\%$ free throw shooter. One morning he decides to shoot $100$ free throws. You may assume that the free throws are independent and the chance that he makes any given free throw is $0.75$. What is the chance he makes more than $80$ free throws?

I thought you would do $$80-75\sqrt{100\cdot 0.75(1-0.75)}$$ to get $3.46$, but the answer key says the answer is $0.1241$.  Can someone explain this to me?
 A: Just a calculator error I think.
$\frac{80-75}{\sqrt{75\times0.25}}=1.1547$
Now $p(Z\ge 1.1547)=0.1241$
Hope it helps
A: The distribution is binomial here, but a binomial distribution with $n$ trials and probability $q$ of success for each trial can be well-approximated by a normal distribution with mean $pn$ and variance $q(1 - q) n$, provided that $n > 5$ and
$$\frac{1}{\sqrt{n}}\left\vert\sqrt{\frac{q}{1 - q}} - \sqrt{\frac{1 - q}{q}} \right\vert < \frac{3}{10}$$
(and provided we don't work too many standard deviations away, say no more than $3$, from the mean). In our case $n = 80$ and $q = \frac{3}{4}$, and so the quantity on the left hand side is $\sim 0.1291$, and hence we're well within the domain where our distribution is effectively normal.
Now, proceed as usual, by computing the $z$-score for a mean $$\mu = qn = \frac{3}{4} \cdot 100 = 75,$$ a standard deviation $$\sigma := \sqrt{q(1 - q) n} = \sqrt{\frac{3}{4} \left(1 - \frac{3}{4}\right) (80)} \sim 4.3301,$$ and value $x = 80$:
$$z = \frac{x - \mu}{\sigma} = \frac{(80) - (75)}{(4.33)} \sim 1.1547,$$
and consulting a $z$-score table shows that $$\color{#bf0000}{p(Z \geq 1.154) \sim 0.1241}$$ as claimed.
Remark Since we're looking for the probability that the Tar Heel lands more than $80$ free throws, we should choose roughly $x = 80.5$ (as taking $x = 80$ effective counts landing $80$ as landing more than $80$ about half the time). This yields a noticeably smaller probability, $\sim 0.1020$. Using the actual (binomial) distribution rather than the normal approximation gives that the true probability is
$$\sum_{k = 81}^{100} {{100} \choose k} \left(\frac{3}{4}\right)^k \left(\frac{1}{4}\right)^k \sim 0.0995.$$
(The rule-of-thumb formula at the beginning of the answer
 comes from Box, Hunter and Hunter (1978). Statistics for Experimenters. Wiley. p. 130.)
A: A simple computer simulation using $1$ million rounds of $100$ shots is showing me $99,787$ winners ($81$+ shots made).  The highest # of shots I see being made (out of $100$) in the simulation is $94$ with $1$ occurrence and $93$ had $5$ occurrences, $92$ had $8$ occurrences....  also interestingly, $52$ shots had only $1$ occurrence.
