To find $\sigma$ of a normal distribution

Question

Given $X \sim \mathcal{N}( n, \sigma^2)$.

The question told me $\mathbb{P}(X\lt 3) = \mathbb{P}(X\gt 7)$

So I found $n$ which is $5$.

I'm also given $2\mathbb{P}(X\lt 2) = \mathbb{P}(X\lt 8)$. So how do I find $\sigma^2$?

Answer 1

Hint: By standardizing, you have $\;2\,\mathsf P (Z< -\frac 3\sigma) = \mathsf P(Z<\frac 3\sigma)$ for $Z\sim \mathcal N(0,1^2)$

Further, a standard normal distribution is symmetric around the zero value, so $\mathsf P(Z<-\frac 3 \sigma) = \mathsf P(Z>\frac 3 \sigma) \quad = 1 -\mathsf P(Z<\frac 3 \sigma)$