Let $X$ be a random variable with a normal distribution such that $E(X) = µ$ and $Var(X) = σ^2$
From my understanding I can write the p.d.f. of a normal distribution as
$f(x) = \frac {e^{ \frac {-(x-µ)^2} {2 \sigma^2}}} {\sigma \sqrt {2\pi}}$ ,or equivalently (?) $f(w) = \frac {e^{\frac{-w}{2}}} {\sigma \sqrt {2\pi}}$ where $w= \frac {-(x-µ)^2} {\sigma^2}$
A question in my probability textbook arises to compute $P(\frac {(X-µ)^2} {\sigma^2} < 2.706)$ . Is this equivalent to asking when $P(W<2.706)$ i.e. the integral $\int_{- \infty}^{2.706} \frac {e^{\frac{-w}{2}}} {\sigma \sqrt {2\pi}} dw$ or perhaps I am confused. Any insights or clarifications much appreciated.