random variable X is distributed normally 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 $p(5 ≤ X ≤ 6)$.
Solution:
$(0 ≤  ≤ 1) = (5 ≤ ≤6)$ because intervals $(0,1)$ and $(5,6)$ are symmetric around $=3$, then $(5 ≤  ≤ 6)= 0.6/2 = 0.3$.
Your thoughts please.
 A: If we assume that the information presented is correct, then the solution is correct and well-explained.
But let us think a little more. Let the standard deviation be $\sigma$. Then $6$ is $3/\sigma$ standard deviation units up from the mean, while $5$ is $2/\sigma$ standard deviation units up.
Is there a $k$ such that $\Pr(2k\lt Z\le 3k)=0.3$, where $Z$ is standard normal? A brief look at tables shows there isn't! 
Remark: The question asks us about the cruising speed of a flying horse. If the choice of numbers was intentional, this is a mean little question. 
A: For $Z = (X-3)/\sigma \sim {\rm Normal}(0,1)$, we have $$\Pr[2/\sigma \le Z \le 3/\sigma] = \int_{z=2/\sigma}^{3/\sigma} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \, dz = \Phi(3/\sigma) - \Phi(2/\sigma).$$  Differentiating with respect to $\sigma$ gives $$3\sigma^{-2} f_Z(3/\sigma) - 2\sigma^{-2} f_Z(2/\sigma) = \frac{1}{\sqrt{2\pi}\sigma^2}e^{-9/(2\sigma^2)}(-3+2e^{5/(2\sigma^2)})$$ from which it follows that $\sigma = \sqrt{\frac{5}{\log \frac{9}{4}}} \approx 2.48309$ is a critical point for this probability.  It is not difficult to see this corresponds to a global maximum; thus the probability is at most $$\Pr[0.805447 \le Z \le 1.20817] \approx 0.09679.$$  It is not possible for any normally distributed random variable to satisfy $\Pr[3 \le X \le 5] = 0.3$.
