normal distribution - random variable Please let me know if my answer is correct.  Thank you.
It’s known that a random variable X is distributed normally with E(X) = 3
and it’s also known that p(0≤X≤1)+p(5≤X≤6) = 0.6.  Find p(p(5≤X≤6).
My answer:


The normal distribution is symmetric around $\mu$, i.e. $$P(X\leq\mu) = P(\mu \leq X) = 0.5,$$ which also means that $$P(X\leq \mu-a) = P(\mu+a\leq X),$$ and consequently
\begin{aligned}
P(\mu-a\leq X \leq \mu) & = P(\mu \leq X \leq \mu+a)\\
P(\mu-a \leq X \leq \mu-b) & = P(\mu + b \leq X \leq \mu+a).
\end{aligned}
Let $\mu=3$, $a=3$ and $b=2$:
$$P(0 \leq X \leq 1) = P(5 \leq X \leq 6),$$
which means that
\begin{aligned}
P(0 \leq X \leq 1) + P(5 \leq X \leq 6) & = 0.6 \Rightarrow\\
2 P(5 \leq X \leq 6) & = 0.6 \Rightarrow\\
P(5 \leq X \leq 6) = \frac{0.6}{2} & = 0.3.
\end{aligned}
Your answer is correct.
