Central Limit Theorem "find sample size" question, with modulus inequality

Question

I'm doing the below Central Limit Theorem question. I omitted sub-question a) and b) as they were unrelated.

I have included the answer below the question. I half-understand the answer but I don't understand a few specific aspects:

1. Why has the |X-u| component disappeared and replaced with just 0.5? Is it because it's an inequality, so we can use either side?

2. It seems like the modulus did not matter in the question?

3. The answer equates the LHS to 1.6449, which is the Z value for 0.05. However, 1.6449 is the z-value for only one tail of the Normal Distribution. Why do we not care about both tails and therefore 2 x 1.6449?

Question:

Answer:

Answer 1

The idea is to observe that if the fat content of a randomly selected hamburger is normally distributed with mean $\mu\%$ and standard deviation $\sigma = 2\%$, then the random variable $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1)$$ will be standard normal with mean $0$ and standard deviation $1$. This is because $\bar X = (X_1 + \cdots + X_n)/n$ will be normal with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$. Consequently, $$\Pr\left[|\bar X - \mu| < 0.5\%\right] = \Pr\left[\left|\frac{\bar X - \mu}{\sigma/\sqrt{n}}\right| < \frac{0.5}{2/\sqrt{n}}\right] = \Pr\left[|Z| < \frac{0.5}{2/\sqrt{n}}\right].$$ Now we know that for a standard normal random variable, $$\Pr[|Z| < z] = \Pr[-z < Z < z] = \Phi(z) - \Phi(-z) = 2\Phi(z) - 1,$$ where $\Phi(z)$ is the cumulative distribution function of the standard normal. So the requirement that $$\Pr\left[|Z| < \frac{0.5}{2/\sqrt{n}}\right] \ge 0.9$$ is equivalent to $$\frac{0.5}{2/\sqrt{n}} \ge \Phi^{-1}\left(\frac{0.9 + 1}{2}\right) = \Phi^{-1}(0.95) \approx 1.6449,$$ where $\Phi^{-1}$ is the quantile function of the standard normal; i.e., it gives the $z$-score for a desired cumulative probability. In this case, $\Phi^{-1}(0.95)$ is the $95^{\rm th}$ percentile of the standard normal, or in other words, $\Pr[Z \le 1.6449] \approx 0.95$.

The rest of the solution is simply an algebraic exercise in solving for $n$.