estimate a probability

Question

Let $X_1....X_{48}$ be independent random variables, each follows a uniform probability distribution over [0,1]. What is the best way to estimate P($\Sigma_{i=1}^{48} X_i > 20)$?

Answer 1

The distribution of $ X_i$ and the normal distribution are both symmetric about the mean, which is an advantage if the Central Limit theorem (CLT) is applied. And the required condition of $n>30$ (rule of thumb) is fullfilled. We can use both distributions and see if the approximation is good enough.

• Using Irwin–Hall distribution

$$1-P\left( \sum_{i=0}^{48}X_i\leq 20\right) =1 - \frac{1}{48!} \sum_{k=0}^{20} (-1)^k *{48 \choose k }*(20-k)^{48}=0.977364\approx 97.74\%$$

• Using the normal distribution as an approximation (CLT)

$$1-P\left( \sum_{i=0}^{48}X_i\leq 20 \right) \approx 1-\Phi\left( \frac{20-\mu}{\sigma} \right)$$

$=1-\Phi\left( \frac{20-24}{\sqrt{\frac{48}{12}}} \right)$

$=1-\Phi(-2)=1-0.02275=0.97725\approx 97.73\%$

As it is expected the approximation is good enough for the most purposes.

Answer 2

The sum of iid uniform random variables is an Irwin-Hall distributed random variable:

$$P(\sum_{i=1}^{48} X_i > 20) = 1 - P(\sum_{i=1}^{48} X_i \le 20)$$

$$= 1 - \frac{1}{48!} \sum_{k=0}^{20} (-1)^k \binom{48}{k}(20-k)^{48}$$