Central limit theorem; Poisson distribution

Let $X_n$ be the numbers of job applications at a company in the year $1900+n,n\in\mathbb N$. Let $(X_n)_{n\in\mathbb N}$ be a sequence of independent, identically distributed random variables with the Poisson ($\lambda$) distribution, where $\lambda\in[1,144]$. Give an approximation of an upper limit of the probability $$\mathbb P(\lambda>\frac{1}{100}\sum_{n=1}^{100}X_n+1),$$ using the Central Limit Theorem.

I first rewrote this probability: $$\mathbb P(\frac{1}{100}\sum_{n=1}^{100}X_n<\lambda-1).$$ So we know that $$Z_n=\frac{S_n-n\mu}{\sqrt{\sigma^2 n}}<\frac{100(\lambda-1)-100\lambda}{\sqrt{100\lambda}}=\frac{-100}{\sqrt{100\lambda}},$$ where $S_n=X_1+\dots+X_n$.

By the Central Limit Theorem, we should have $$\mathbb P(Z_n<\frac{-100}{\sqrt{100\lambda}})\approx\int_{-\infty}^\lambda\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^2}\,\mathrm du.$$ I don't know how to continue from here on. Could someone help me?

• If you use fewer greek letters it will make the math easier. Greek letters are scary! – terrace Feb 3 '17 at 1:09
• @terrace I'm following convention, actually – Sha Vuklia Feb 3 '17 at 1:10
• You have $\mathbb{P}(Z_{n}<-10/\lambda),$ which is an increasing function in $\lambda,$ and you know that $\lambda\leq 144.$ Then an upper bound is given by $\mathbb{P}(Z_{n}<-10/144),$ which you can find using a normal probability table or statistical software. – RideTheWavelet Feb 3 '17 at 1:26