Approximation with a normal distribution Every day Alice tries the stroke playing tennis until she reaches $50$ strokes.
If each stroke is good with a probability of $0,4$, independently from others, approximately what is the probability that at least $100$ attemps are necessary to success? 
Let X be a binomial random variable with parameters $$n=100$$ and $$p=0,4$$
Since n is large we can approximate with a  normal distribution with parameters $$\mu=40$$ and $$\sigma=\sqrt{0,4*100*0,6}=\sqrt{24}$$
Appling the normal approximation
\begin{align}P(X> 49,5)&=1-P(x<49,5)\\&=1-P((X-\mu )/\sigma  < (49,5-40 )/ \sqrt{24} )\\&=1-P((X-\mu )/\sigma  < 1,939 )\\&=1-\Phi (1,939)\\&=1-0,9737\\&=0,0263\end{align}
But the solution on the book is $0,974$ (it could be $P(x<49,5)$).

Considering an approximation with a normal distribution of the binomial random variable, I think that we should calculate $$P(X<49.5)$$ because the number of attempts must be more than 100.
