Why does central limit theorem give such big x in $\phi(x)$ I am solving a problem : The number of  new customers in the mall each day follows Poisson distirbution with $\lambda = 50$.  Find approximately the probability that after one year (200 working days) the number of customers that visited was between 950 and 1100.
My idea : Day i we have $X_i \sim P_o(\lambda) $ new customers . Now we need $P(950 \leq \sum_{i=1}^{200} X_i\leq 1100 )$.
$X_i$'s are independent and equally distirbuted. We have $200 > 30$ samples and therefore Central Limit Theorem should hold:
$$ Z=\frac{S_{200}-200\lambda}{\sqrt{200\lambda} } \sim N(0,1)$$
Then :
$P(950 \leq \sum_{i=1}^{200} X_i\leq 1100 ) = P(\frac{950-200*50}{\sqrt{200*50} } \leq Z \leq \frac{1100-200*50}{\sqrt{200*50} })$
And I could actually apply continuity correction and have instead:
$$ P(\frac{949.5-200*50}{\sqrt{200*50} } \leq Z \leq \frac{1100.5-200*50}{\sqrt{200*50} }=P(-90.5 \leq Z \leq -88.9)=\Phi(-88.9) - \Phi(-90.5) = 1-\Phi(88.9) - 1 + \Phi(90.5)= \Phi(90.5)-\Phi(88.9)$$
Why are they so huge  ?  At least in theory , CLT should hold , the conditions were satisfied.
Side note : We have at average 50 people per day and also 50 deviation . So let's suppose we have 50 and 0 people/per day alternatively --> after 200 days we should have 100* 50 = 5000 so less than 1100 seems small (if this kind of thought makes sense)

The set up of the question is absurd, as originally pointed out by Michael M, in that the expected number with $200$ days is $10000$ with a standard deviation of $100$, so you are extremely unlikely to see from $950$ to $1100$ and are extremely deep in the tail - being about $90$ standard deviations away from the expectation.
The actual probability is between $10^{-2813}$ and $10^{-2812}$ while a normal approximation would give a much smaller figure, being a good absolute approximation (as one might eventually hope for from the Central Limit Theorem) of near to zero,  though poor relative approximation as often happens in the tails.
Let's instead deal with more realistic numbers away from the extreme tails and look at the probability of seeing from $9950$ to $10100$. It should not be difficult to find the exact probability, for example with R
sum(dpois(9950:10100, 10000))
# 0.5353328

The normal approximation is a continuous distribution while the Poisson is discrete on the integers, so $P(9950 \leq \sum\limits_{i=1}^{200} X_i\leq 10100 ) = P(9949 \lt \sum\limits_{i=1}^{200} X_i\leq 10101)$ in reality but not in your approximation. You could use a continuity correction and find the probability of seeing values between $9949.5$ and $10100.5$.  This will lead you to $\Phi\left(\frac{10100.5 - 10000}{\sqrt{10000}}\right) -\Phi\left(\frac{9949.5 - 10000}{\sqrt{10000}}\right) = \Phi\left(1.005\right) -\Phi\left(-0.505\right)$  which R tells us is
pnorm(1.005) - pnorm(-0.505)
# 0.5357722

and is a reasonably good approximation both in absolute and relative terms.
