Let $X$ be normal random variable with $P(X>15)$ and $\sigma^{2}=4$. find $\mu$ let $X$ be normal random variable with $P(X>15)=0.75$ and $\sigma^{2}=4$. find $\mu$
MY WORKING:
Now I am using $pdf$  of normal random variable with $\sigma=2$ and $P(X>15)=0.75$ after plugging these values I get the following expression:
$$0.75=\int_{15}^\infty \frac{1}{\sqrt{8\pi}} e^{-\frac{1}{8}(x-\mu)^2} \,dx= \left.-\frac{1}{\sqrt{8\pi}}e^{-\frac{1}{8}(x-\mu)^2} \frac{1}{4}(x-\mu)\right|_{15}^\infty$$
Now from here when I evaluate the limit I get the expression $0\times\infty$ in the first term. What should I do next? Is there a mistake. Kindly guide me.
 A: \begin{align}
0.75 = {} & \Pr( X>15) \\[8pt]
= {} & \Pr\left( \frac{X-\mu} \sigma > \frac{15-\mu} 2 \right) \\[8pt]
= {} & \Pr\left( Z> \frac{15-\mu} 2 \right)
\end{align}
The question now is: For what value of $z$ is $\Pr(Z>z)$ equal to $0.75$? This is the inverse of the standard normal cdf evaluated at $1-0.75.$ Standard software tells me this is $-0.6744898$ (rounded, of course). Thus we have
$$
\frac{15-\mu} 2 = -0.6744898
$$
This is to be solved for $\mu.$
If you have no suitable software you can use this table: https://freakonometrics.hypotheses.org/files/2013/10/Capture-d%E2%80%99e%CC%81cran-2013-10-15-a%CC%80-14.22.40.png
This table says $\Pr(Z<0.67) \approx 0.7486$ and $\Pr(Z<0.68)\approx 0.7517.$ So the number you seek is between those. Naive linear interpolation ("naive" = not thinking about how those numbers were rounded) yields${}\approx0.6745161.$
A: $$\int e^{-\frac{1}{8}(x-\mu)^2}\mathrm dx\neq -\frac{1}{4}(x-\mu)e^{-a(x-\mu)^2}$$
To see this just take the derivative on both sides. We need to use the complementary error function,
$$\int_a^\infty e^{-b(x-\mu)^2}\mathrm{d}x=\frac{\sqrt{\pi/b}}{2}\operatorname{erfc}(\sqrt{b}~(a-\mu))$$
It is related to the normal error function simply by $\operatorname{erfc}(z)=1-\operatorname{erf}(z)$
To compute the complementary error function, you can simply go to Wolfram|Alpha. In our case we want to find a number $\mu$ such that $$\frac{1}{2}\operatorname{erfc}\left(\sqrt{\frac{1}{8}}\left(15-\mu\right)\right)=0.75$$
To do this we can use any kind of root finding algorithm we wish. Wolfram finds
$$\mu\approx 16.349$$
