Normal distribution of the mean of a uniform distribution

I have $\bar{X}$ which is the mean of the numbers from the uniform distribution of $[0, 1]$ with $n = 100$.

I know that $\mu = \frac{1}{2}$ and $\sigma^2 = \frac{1}{1200}$, thus, $\sigma \approx 0.028 $, I need to find the probability of $\bar{X}$ having a value between $[0.47, 0.53]$.

Calculating the normal distribution, I find $1 - 2(0.3508) = 0.2984$, however the book says the answer is $0.7016$.

It is easy to see that $1 - 0.2984 = 0.7016$, and it makes me sure that I'm in the right path. I just don't know why I should subtract the value I found in the distribution from $1$, and it makes me think the book forgot one step before finishing the exercise. Could someone clarify this for me?

Hint: $P(-a<X<a)=2\Phi(a)-1$.

These are the inner (orange) areas of the graph below.

For more detailed explanation see here.

Additional:

In your case the standardized value is

$a=\frac{0.53-0.5}{\frac1{\sqrt{1200}}}=0.03\cdot \sqrt{1200}=0.6\cdot \sqrt{3}=1.03923...\approx 1.04$

This table gives $\Phi(a)=\Phi(1.04)=0.85083\approx 0.8508$

Finally we get

$$P(-1.04<X<1.04)=2\Phi(1.04)-1=2\cdot 0.8508-1=1.7016-1=0.7016$$