Let $Y ∼ N(−8, 4)$. Compute each of the following, in terms of the function use Table D.2 (or software) to evaluate these probabilities numerically
(a) $P(Y ≤ −5)$
attempt:
this is $\frac{Y+8}{4} ∼ N(0,1)$ once normalized
$P(Y \leq -5) = \phi(\frac{-5+8}{4}) = \phi(0.75) = 0.7734$
The solution gets $0.933$
Is the soluton wrong?
Differing notations for the distribution seem to be in play.
Their answer is for $\mathcal N(\mu{:=}{-8}, \sigma^2{:=}4)$. You have used $\mathcal N(\mu{:=}{-8}, \sigma{:=}4)$.
$$\phi\left(\tfrac{-5+8}{2}\right)\approx 0.933$$