A case study in a exam material goes like this:
"Assume that the bank reports a daily VAR of \$100 million at the 99% level of confidence. Under the null hypothesis that the VAR model is correctly calibrated, the number of exceptions should follow a binomial distribution with expected value of $E[X] = np = 250(1 − 0.99) = 2.5$. The risk manager then has to pick a cutoff number of exceptions above which the model would be rejected. The type 1 error rate is the probability of observing higher numbers than the cutoff point. Say the risk manager chooses $n = 4$, which corresponds to a type 1 error rate or significance level of $10.8\%$. Above 4, the risk model is rejected."
I calculated the significance level corresponding to observing 4 exceptions to be 17.11%, using standard normal distribution as follows:
$$ z= (4 - np)/\sqrt{p(1-p)n} = (4-250(1-0.99))/\sqrt{0.01(1-0.01)*250}=1.5/1.573=0.95$$
$$P(-\infty < Z <0.95)=0.8289$$ $$ \alpha = (1-P)*2 = 0.1711 = 0.1811=17.11\%$$
Was there any error in my calculation?
