Originally Posted by

**matheagle** I would use the normal approximation, with n=100.

By the Central Limit Theorem

$\displaystyle P\left(\sum_{i=1}^{100}X_i>300\right)=P\left(\bar X>3\right) $

$\displaystyle \approx P\left(Z>{3-\mu\over \sigma/\sqrt{100}}\right) $

NOW use your distribution..

You have a mean of 2.5 and a variance of 2.5 via the exponential distribution...

$\displaystyle \approx P\left(Z>3.16227766\right) $

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If you don't want to approximate this, then use that fact that a sum of iid exponentials is a gamma.

Either way you need to calculate $\displaystyle P\left(\sum_{i=1}^{100}X_i>300\right) $