The distribution of sample proportion for given population proportion and sample size

Question

If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89? (4dp)

work: $n=64$, $\hat p=0.89$, so $X=n \hat p =56.96$. $$ P(\hat p >0.89)=P( X >56.95)=1- P( X <56.95)$$ and then how to do it?

The answer is $0.6064$

More details of the solutions would be great, thanks

Answer 1

Since $np > 10$ we can apply Central Limit Theorem and use the formula $\mathbb{P}\!\left(p < \hat{p}\right)= \mathbb{P}\!\left(Z<\frac{p-\hat{p}}{s}\right)$ where $s = \sqrt{\frac{p(1-p)}{n}}$ and $Z \sim N(0,1)$.

My answer came out to be .6052 (I didn't use a standard normal table).