Given $X \sim \mathcal{N}(6, 2)$, we are asked to calculate $P(X \leq 10)$. My professor taught me that
$$ F_Y(x) = \frac{1}{2} + \Phi(x), $$
for $Y \sim \mathcal{N}(0, 1)$. This implies two crucial facts:
$$ \lim_{x\to\pm\infty}F_Y(x) = \frac{1}{2} + \lim_{x\to\pm\infty}\Phi(x), \quad \text{and} \quad P(a < Y < b) = \Phi(b) - \Phi(a). $$
(To loosen up the formalism here, $a$ and $b$ here can also be $-\infty$ and $\infty$, meaning appopriate limits.)
Now we calculate, with transforming $X$ into a random variable $Z \sim \mathcal{N}(0, 1)$:
\begin{align*} P(X \leq 10) &= P(-\infty < X \leq 10) = P(-\infty < X < 10) = P(-\infty < Z < \frac{10 - 6}{2}) \\ &= P(-\infty < Z < 2) \\ &=\Phi(2) - \lim_{x\to-\infty}\Phi(x) = 0.9772 + 0.5 = 1.4772, \end{align*}
which is a contradiction, since a probability measure cannot exceed $1$.
Here is also the professor's manuscript:
As you can see, he input an incorrect value for $\Phi(2)$, but got the overall correct result, as $P(-\infty < Z \leq 2)$ is indeed $0.9772$.
Please, help me find out where the logical error in the calculation happens, and whether I understand the professor's calculation correctly. (He used the formulas I listed above when solving other exercises as well.) Thank you.
I found out what the problem was. It was due to a very specific and unlikely (fitting, given that we are talking about probability, lol) error in the table for $\Phi(z)$.
In my context, $\Phi(z) = \frac{1}{2} + F_Y(z)$ is true, because it is defined as
It just so happens that the table has an error at exactly $z = 2.00$. It should be $0.4772$, but is $0.9772$ instead. Taking the correct table value into account returns the correct result. I would like to thank everyone who took their time to help me with this.
Here is the table with the error: