If $P(|X-19.1| \leq a) = 0.98$ then why is it necessarily $P(\frac{X-19.1}{17} \leq \frac{a}{17})= 0.99$? Suppose we have $\mu = 19.1$ and $\sigma = 1.7$.
I need to find $a \in \mathbb{R}$ such that: $P(|X-19.1|\leq a) = 0.98$
Now in the correction, it is given that:
$$P(|X-19.1|\leq a) = 0.98 \iff P\left(\frac{-a}{1.7} \leq \frac{X - 19.1}{1.7} \leq \frac{a}{1.7}\right)=0.98 \iff  P\left(\frac{X-19.1}{1.7} \leq \frac{a}{1.7}\right)=0.99$$
Can someone explain to me why is the last equality correct?
 A: This seems to be discussing some symmetric distribution - so that the 0.01 quantile ($q_{0.01}$) is as far from $19.1$ as the 0.99 quantile ($q_{0.99}$). (I've used a normal density in my image but this argument applies to symmetric distributions more generally.)
In the image below we have $P(|X-19.1|\leq a) = 0.98$ (i.e. $a=q_{0.99}-19.1$). We can see that if we drop the absolute-value part in $|X-19.1|\leq a$ (giving $X-19.1\leq a$), we will include everything below $q_{0.01}$. This adds the 1% probability that lies below that first percentile, taking us from $0.98$ to $0.99$.

So we have $P(X-19.1\leq a) = 0.99$. 
Now we can scale the left and right side of that inequality by the same constant without changing the probability (it's like going from saying "The probability that a randomly chosen basketball player's height is below $2\,\text{m}$ is $p$" to saying "The probability that half a randomly chosen basketball player's height is below $1\,\text{m}$ is $p$"). 
That is we can go from $P(X-19.1\leq a) = 0.99$ to $P(\frac{X-19.1}{17}\leq \frac{a}{17}) = 0.99$; scaling both halves of the inequality by a positive multiplier ($1/17$ in this case) doesn't change anything.
A: This assumes a normal distribution and the first relationship amounts to transform to the standard normal. You can then use 0.99 in the table of the cumulative standard normal distribution to find $a$. The change to 0.99 occurs because the left tail (all values below $-a/1.7$) is left out.
