Suppose that the number of traffic accidents, $N$, in a given period of time is distributed as a Poisson random variable with $E(N) = 100$. Use the normal approximation to the Poisson to find $\Delta$ such that $P(100-\Delta < N < 100 + \Delta) \approx 0.9$.
I understand the procedure on how to figure out this problem, but for some reason the way I went about figuring it out clashes with the final solution:
Attempt
Given that we have to approximate it by the normal we will be standardizing using the theory of the Central Limit Theorem. After the algebra I arrive at:
$$P(\frac{-\Delta}{10} < Z < \frac{\Delta}{10}) \approx 0.9$$
So I proceeded to treat things as follows:
$$P(\frac{-\Delta}{10} < Z < \frac{\Delta}{10})\\ \Rightarrow\ P(Z< \frac{\Delta}{10}) - P(\frac{-\Delta}{10} < Z) \\ \Rightarrow \ (1 - P(Z \leq \frac{-\Delta}{10})) \\ \Rightarrow \ 2P(Z < \frac{\Delta}{10}) - 1\ \text{(used the symmetry of the normal distribution)} \\ \Rightarrow\ 2P(Z < \frac{\Delta}{10}) - 1 \approx 0.9 \\ \Rightarrow P(Z < \frac{\Delta}{10}) = \frac{1.9}{2} = 0.2375$$
So taking the compliment of this value $(1 - 0.2375 = 0.7625)$ and using the tables of standard normal $Z$ values I arrive at an expression of:
$$0.72 = \frac{\Delta}{10} \\ 7.2 = \Delta$$
But this isn't what the solution arrived at. The solution did the following:
Which algebraically is the same thing, they instead took $\frac{-\Delta}{10}$ and worked with that while I worked with the positive side. So in theory the same result should be achieved. So what gives with what I did?
I discovered the mistake I made with regards to interpreting the question. It had to do with my interpretation of the symmetry of the normal distribution.
I was thinking that:
$$P(Z < \frac{\Delta}{10}) = P(Z < \frac{-\Delta}{10})$$
By the symmetry of the normal distribution. The only thing that is similar here is how they are written syntactically. This statement is not true. On the other hand
$$P(Z > \frac{\Delta}{10}) = P(Z < \frac{-\Delta}{10})$$
This uses the symmetry of the normal distribution. A quick look at a picture of a normal distribution will confirm this heuristically.
And as a result:
$$P(Z < \frac{\Delta}{10}) - P(Z < \frac{-\Delta}{10}) \\ \Rightarrow\ 1 - P(Z > \frac{\Delta}{10}) - P(Z < \frac{-\Delta}{10}) \\ \text{(now using the symmetry of the normal distribution)}\\ \Rightarrow 1 - P(Z < \frac{-\Delta}{10}) - P(Z < \frac{-\Delta}{10}) \\ \Rightarrow \ 1 - 2P(Z < \frac{-\Delta}{10})$$
And the rest of the algebra will follow easily.