At first glance you would think this question was easy. It's not. Maybe the answer in the back of the book is wrong. But nobody, not even my maths teacher can explain it to me.
Here is the question: an unbiased 6 sided dice is thrown 70 times. Find the probability that the mean score is less than 3.3 Answer: 0.155
My working. Mean of one dice is 3.5 variance is 35/12
Therefore the mean of 70 throws is distributed normally ~N(3.5,(35/12)/70)
P(Xmean<3.3)=P(z<-0.2/0.204)
1-p(z<0.98039)
1-0.83655
0.16345
NOT THE ANSWER???
I know this isn't the most intellectually stimulating question out there but to anyone reading this it would help me out a lot if I could have this question explained to me. Thanks
Update: Maybe I am being silly and missing something very obvious. Here is the original question. 
If the result of one throw has mean $\mu$ and variance $\sigma^2$, the sum of $70$ independent throws has mean $70 \mu$ and variance $70 \sigma^2$, and the average $X$ of these $70$ throws has mean $\mu$ and variance $\sigma^2/70$. It is wrong to say $X$ is distributed normally, but because of the CLT you can say it is distributed approximately normally. Thus $$ P(X < 3.3) \approx P\left(Z < \frac{3.3 - \mu}{\sigma/\sqrt{70}} \approx -0.9798\right) \approx .16359 $$
However, we might improve the result with the "continuity correction". An average $< 3.3$ means a sum $< 70 \times 3.3 = 231$, and this sum is an integer; since the $<$ means we don't want to include cases where the sum is exactly $231$, we might replace $231$ by $230.5$, and thus replace $3.3$ by $230.5/70$. This makes the $z$ score
$$ \frac{230.5/70 - \mu}{\sigma/\sqrt{70}} \approx -1.01479$$
and now indeed $$P(X < 3.3) \approx 0.155103 $$
By the way, the exact answer (not using any approximation) turns out to be
$$ \frac{229642652203521128605439312834749816684648611425676955}{1477602207273840622329353829895227690835664661525823488} \approx .1554157479$$