How would I integrate a normal distribution to find the values of k? Suppose $Z$ is the standardized normal variable and the value of $k$ needs to be found if $P(0< Z< k) = 2P(k< Z< 2k)$. I have tried to integrate the normal distribution formula $$\frac{1}{\sigma\sqrt{2\pi}}\exp((\frac{x-\mu}{\sigma})^2)$$
However I found that it was really difficult to integrate this while setting the limit between $0$ and $k$. Is there another way to approach this problem or am I doing something wrong?
 A: You can do this with the help of a standard normal distribution table.  Note that $$\Pr[0 < Z \le k] = \Pr[Z \le k] - \Pr[Z \le 0] = \Phi(k) - \frac{1}{2},$$ where $\Phi$ is the cumulative distribution function of the standard normal.  Similarly, $$\Pr[k < Z \le 2k] = \Pr[Z \le 2k] - \Pr[Z \le k] = \Phi(2k) - \Phi(k),$$ thus the condition $$\Pr[0 < Z \le k] = 2\Pr[k < Z \le 2k]$$ implies $$\Phi(k) - \frac{1}{2} = 2 \left(\Phi(2k) - \Phi(k)\right),$$ or $$g(k) = 3\Phi(k) - 2\Phi(2k) = \frac{1}{2}.$$  You would then search for $k$ in the table such that satisfy this condition.  For example, if you started with $k = 1$, then $\Phi(1) = 0.841345$ and $\Phi(2) = 0.97725$, giving $g(1) = 0.569535$.  This is too large, so try $k = 0.5$, giving $g(0.5) = 0.391698$, which is too small.  These findings would suggest that $k$ should be slightly less than $1$.  If we try $k = 0.8$, we get $$\Phi(0.8) = 0.788145, \quad \Phi(1.6) = 0.945201,$$ hence $g(0.8) = 0.474032$.  We could continue in this fashion to get something reasonably close with the table, $k \approx 0.86$.  Using a computer, a more precise value is
$$k = 0.85838209570223257181023976889293137701405020511241\ldots.$$
A: Starting from @heropup's answer, we need to find the zero of function
$$F(k)=\text{erfc}\left(\sqrt{2}
   k\right)-\frac{3}{2} \text{erfc}\left(\frac{k}{\sqrt{2}}\right)+\frac{1}{2}$$
$$F'(k)=\frac{e^{-2 k^2} \left(3 e^{\frac{3 k^2}{2}}-4\right)}{\sqrt{2 \pi }}$$
$$F''(k)=\frac{e^{-2 k^2} \left(16-3 e^{\frac{3 k^2}{2}}\right) k}{\sqrt{2 \pi }}$$ The first derivative cancels at
$$k_*=\sqrt{\frac{2}{3} \log \left(\frac{4}{3}\right)}$$ and for this value
$$F(k_*)\approx -0.111049\qquad \text{and} \qquad F''(k_*)\approx 1.42862$$ making the solution $> k_*$.
We can approximate it using a Taylor expansion to second order to get, as an estimate,
$$k_0=k_*+\sqrt{-2\frac{F(k_*)}{F''(k_*)}}\approx 0.832225$$ and, from this point, start using Newton method. The iterates would be
$$\left(
\begin{array}{cc}
 n & k_n \\
 0 & 0.832225 \\
 1 & 0.858840 \\
 2 & 0.858382
\end{array}
\right)$$ which is the solution for six significant figures.
