If $20 $ random numbers are selected independently from the interval $(0,1) $ probability that the sum of these numbers is at least $8$? 
If $20 $ random numbers are selected independently from the interval
$(0,1) $  what is the probability that the sum of these numbers is
at least $8$?

I tried to take this question https://math.stackexchange.com/questions/285362/choosing-two-random-numbers-in-0-1-what-is-the-probability-that-sum-of-them   as reference  but the step where there is a double integral, I got stuck, do I have to make 20 integrals?

Let $ \ X_i $ be the $ \ i^{th}$ number selected where $\ i= 1,2,3,4...20 $
$To $ $ calculate $
$ \ P( \sum_{i=1}^{20} X_i \ge 8 )  $
$ E(\ X_i) = \frac{(0+1)}{2} $ $ [uniform $ $ distribution ] $
$ E(\ X_i) = \frac{1}{2} $
$ E(\sum_{i=1}^{20} X_i) = 20/2 = 10 $
$ Var(\ X_i) = \frac{\ (1-0)^2}{12} $ $ [uniform $ $ distribution ] $
$ Var(\ X_i) = \frac{\ 1}{12} $
$ Var(\sum_{i=1}^{20} X_i) = 20/12 = 5/3  $
$ \ P(\frac{ \sum_{i=1}^{20} X_i  - E(\sum_{i=1}^{20} X_i)  }{\sqrt Var(\sum_{i=1}^{20} X_i)}   \ge \frac {8 -E(\sum_{i=1}^{20} X_i)}{\sqrt Var(\sum_{i=1}^{20} X_i}  )  $
$ \ P(\frac{ \sum_{i=1}^{20} X_i  - 10)  }{\sqrt {5/3}}   \ge \frac {8 -10}{\sqrt 5/3}  )  $
$
1- P(Z \le -1.55)$
= $ 0.9394 $ $ approx $
