[Math] Finding the probability of an event with binomial distribution using a normal approximation

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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?

Best Answer

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

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