Estimation of probability, 70 particular events from 360 A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is

This is a GRE question.
 A: The exact probability is $$p=\frac{1}{6^{360}}\sum_{k=70}^{360}\binom{360}{k}5^{360-k}.$$
which is not so nice to compute without a computer. However for such number of trials we can use the Normal Distribution  $\cal{N}(np,np(1-p))$ as an approximation of the Binomial Distribution $\cal{B}(n,p)$ where $n=360$ and $p=1/6$.
Use the Standardized Normal Distribution: let $Z=(X-\mu)/\sigma$ where $\mu=np=60$ and $\sigma=\sqrt{np(1-p)}=\sqrt{50}$. You should compute
$$p\approx P(Z\geq (70-\mu)/\sigma)=P(Z\geq (10/\sqrt{50})=P(Z\geq 1.414)\\=0.5-P(0<Z< 1.414)=0.5-0.42073\approx 0.08.$$
wher in the last setp we used the Standard normal table. So the answer should be (C).
A: I consider a random function $\xi_i$ which is 1 if a six comes up for i-th trail and 0 otherwise.
$$\xi = \sum_{i=1}^{360} \xi_i$$
Mean of $\xi_i$ is $\frac{1}{6}$. Dispersion of $\xi_i$ is $\frac{1}{6} - \frac{1}{36} = \frac{5}{36}$
Then mean of $\xi$ is 60. Dispersion of $\xi$ is 50. Standart deviation is about 7.
Now I use normal approximation for $\xi$. The probability $53 < \xi < 67$ is about 68%. Then probability $\xi \geq 70$ is less then $\frac{100-68}{2} = 16$ %.
On the other hand probability of $60-14 < \xi < 60+14$ is about 95%. Then probability of $\xi \geq 74$ is about 2.5%. So probability of $\xi > 70$ is more then 2%.
So the answer is C.
A: Using normal approximation we get
$$P(X>70) = P(Z > \frac{10}{\sqrt{50}}) = P(Z > \sqrt{2}) = 1 - P(Z < \sqrt{2})$$
I believe we are supposed to have memorised:
$$P(Z < 2) \approx 97.5\%$$
$$P(Z < 1) \approx 84\%$$
Hence
$$84\% < P(Z < \sqrt{2}) < 97.5\%$$
$$\to 16\% > 1 - P(Z < \sqrt{2}) > 2.5\%$$
