How do I approximate the normal distribution?

Question

$G\underset{\sigma,µ} {(x)} = \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{(x - µ)^2}{2\sigma^2}} $

$q * G\underset{\sigma,µ} {(µ \pm a*\sigma)} \lt G\underset{\sigma,µ} {(µ)} $

Calculate the factor a.

I have no idea what I am supposed to do, I think I should calculate the factor a to approximate the normal distribution. I know this already to be $3\sigma$, but I have no idea how I would do this. I am also not sure what the purpose of q is.

I started by substitution

$q * G\underset{\sigma,µ} {(µ \pm a*\sigma)} = q * \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{((µ \pm a*\sigma) - µ)^2}{2\sigma^2}} $

$ = q * \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{(\pm a*\sigma)^2}{2\sigma^2}} $

$ = q * \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{a^2 *\sigma^2}{2\sigma^2}} $

$ = q * \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{a^2}{2}} $

$G\underset{\sigma,µ} {(µ)} = \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{(µ - µ)^2}{2\sigma^2}} = \frac{1}{ \sigma \sqrt{2\pi}}$

$q * G\underset{\sigma,µ} {(µ \pm a*\sigma)} \lt G\underset{\sigma,µ} {(µ)} = q * \frac{1}{ \sigma \sqrt{2\pi}}e^{-\frac{a^2}{2}} \lt \frac{1}{ \sigma \sqrt{2\pi}}$

$ = q * e^{-\frac{a^2}{2}} \lt 1$

How does this help me? I am not sure how I would solve for a and I have still another variable q.

Answer 1

I suspect that the question is supposed to be this: given $q > 1$, find what positive numbers $a$ make $q G(\mu \pm a \sigma) < G(\mu)$.

As you noted, this inequality is equivalent to $$ q e^{-a^2/2} < 1$$ Hint: multiply both sides by $e^{a^2/2}$ and take logarithms.