Given that $X$~$N(10, 0.5^2)$, find $c$ such that: $$P(12.6-2c < x < 11.9+c)=0.2$$
I can solve simpler problems of this sort, for example I can solve:
$P(10-c < x < 10+c)=0.2$ but only because the two end areas are equal. I wanted to know if there was an easy way to calculate $c$ in my first example.
I am aware that you can integrate the equation of the normal curve using the error function, but that is way beyond my understanding at the moment.
This problem is easier than it appears at first glance.
With a little algebra, $$12.6-2c < x < 11.9+c$$ if and only if $$5.2-4c < \frac{x-10}{0.5} < 3.8+2c$$ and $(x-10)/0.5$ has a $N(0,1)$ distribution. So we want to solve $$\Phi(3.8+2c)-\Phi(5.2-4c) = 0.2$$ for $c$, where $\Phi(x)$ is the cumulative standard normal distribution.
At $c=0$, the left-hand side of the equation is essentially $1-1=0$, while for large $c$ it becomes $1-0=1$, so we expect a positive root. But for $c>0$, $\Phi(3.8+2c) \approx 1$, so we can replace our equation with $$1-\Phi(5.2-4c) = 0.2$$ I'm guessing you can probably take it from here.