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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?

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  • $\begingroup$ If you use fewer greek letters it will make the math easier. Greek letters are scary! $\endgroup$ – terrace Feb 3 '17 at 1:09
  • $\begingroup$ @terrace I'm following convention, actually $\endgroup$ – Sha Vuklia Feb 3 '17 at 1:10
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    $\begingroup$ 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. $\endgroup$ – RideTheWavelet Feb 3 '17 at 1:26

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