Normal approximation to total weight above 26 kg

A cheesemaker produces $100$ blocks whose weights in grams are i.i.d. exponential with rate $\tfrac1{250}$; let $T_{100}$ be their total weight. Using the central limit theorem, $\mathbb{P}[T_{100}>26000]\approx\Phi(a)$, where $\Phi$ is the standard normal CDF. What is $a$?

$a=-2/5$. $T_{100}$ has mean $25000$ and standard deviation $\sqrt{100}\cdot250=2500$, so $26000$ is $0.4$ standard deviations above the mean. Then $\mathbb{P}\approx1-\Phi(0.4)=\Phi(-0.4)$.