What is the probability of obtaining a defective cable? The diameter of an electric cable is normally distributed with mean $0.8$ and variance $0.0004$. Suppose that the cable is considered defective if the diameter differs from its mean by more than $0.025$. What is the probability of obtaining a defective cable?
Answer: $0.2112$
This is the question 9.3 from (Paul Meyer's "Introductory probability and Statistics", 2nd ed.)
My attempt: The question is asking for $P(|X - \mu| > 0.025)$.
$$
|X - \mu|
\begin{cases}
X - \mu, X \geq \mu \\
\mu - X, X < \mu
\end{cases}\\
P(|X - \mu| > 0.025) = P(X - \mu > 0.025) + P(\mu - X > 0.025)\\
P(|X - \mu| > 0.025) = P(X > 0.025 + \mu) + P(X < \mu - 0.025)\\
P(|X - \mu| > 0.025) = 1 - P(X \leq 0.025 + \mu) + \Phi(-0.13)\\
P(|X - \mu| > 0.025) = 1 -\Phi(0.13) + \Phi(-0.13)\\
P(|X - \mu| > 0.025) = 2 - 2\Phi(0.13)\\
P(|X - \mu| > 0.025) = 0.8966
$$
What was my mistake here?
 A: Basically it is asked for $P(X>0.825)+P(X<0.775)$. It is a good idea to standardize the random variable first: $Z=\frac{X-\mu}{\sigma}=\frac{X-0.8}{0.02}$. Then Z is standard normal distributed with $\mu=1$ and $\sigma^2=1$. So Z is symmetric distributed around the mean $0$.
For $Z$ the equation becomes  $P(X>0.825)+P(X<0.775)=P(Z<-1.25)+P(Z>1.25)$. Due the symmetry of $Z$ we get
$$=2-2\cdot P(Z<1.25)=2-2\cdot \Phi(1.25)$$
Below it is shown how the simplified term can be derived:
\begin{align}
  P(Z\le -z\text{ or }Z\ge z) &= P(|Z|\ge z)\\
                                &=\Phi(-z)+1-\Phi(z)\\
                                &= 1-\Phi(z)+1-\Phi(z)\\
                                &=2-2 \Phi(z).
\end{align}
Remark:
It seems that your main mistake is that your have $\sqrt{0.0004}=0.2$, which is not true. It is  $\sqrt{0.0004}=0.02$.
In general it is not a good idea to round to two decimal places as you did: $\frac{0.025}{0.2}=0.125\neq 0.13$. The result would be to far away from the exact value.
To summarize: You take the wrong standard deviation and additionally your rounding is not appropiate.
