Vrftjkry

I was doing some normal distribution questions and was stumped when I faced this question. Hope you guys can give me a hand here.

Question
If the annual precipitation $X$ in a city is a normal variable, with mean $50$cm and standard deviation $10$cm. Determine the following.
The probability that $X$ is within $5$cm from the mean annual precipitation

The answer
The probability that $X$ is within $5$cm from the mean annual precipitation is
$P(|X - 50| < 5) = P(|(X-50/10)| < 5/10).$

My Confusion
Why is $P(|(X-50/10)| < 5/10)$ not P(|(X-50**-50**/10)| < 5/10). And why divide the $5$ with $10?$

Thank you guys

We know that $frac{X-mu}{sigma} sim N(0,1)$, hence we divide by the standard deviation.

When we have an inequality $a le b$ we can divide both sides by the same positive scalar and the statements are equivalent.

If $X sim mathsf{Norm}(mu = 50, sigma=10),$ then

$$P(45 < X < 55) = Pleft(frac{45 - 50}{10}< frac{X-mu}{sigma} < frac{55 - 50}{10}right)\
= P(-0.5 < Z < 0.5) = 0.3829,$$
where $Z sim mathsf{Norm}(0, 1),$ the standard normal distribution.

The displayed equation shows the process of 'standardization', which makes it possible to find the (approximate) probability from printed tables of the standard normal distribution. For example, you can find from such tables that
$P(0 < Z < 0.5) = 0.1915,$ from which the result can be obtained by symmetry as
$P(-0.5 < Z < 0.5) = 2(0.1915) = 0.3830,$ correct to three places.

Alternatively, statistical software or statistical calculators can be used to evaluate the original expression $P(45 < X < 55) = 0.3829249$ directly. The result from
R statistical software is shown below.

diff(pnorm(c(45,55), 50, 10))

[1] 0.3829249

The left panel below shows the density function of $mathsf{Norm}(50, 10)$ and the panel at right shows the standard normal density function. In both panels, the total area beneath the density function is $1$ and the desired probability is
the area between the vertical dotted lines.