Normal Distribution problem using the table

Question

the problem goes like this Y has a normal distribution with mean 1 and standart deviation 2. determine P(Y^2 < 9)

so i rewrote like this P(Y< sq root 9)=P(P<3)= norm dist ((3-1)/2)=norm dist of 1 =.8413

however the book says its supposed to be .8185, and according to table that means my values should be not 1 but .91, so where did i make mistake.

Answer 1

$P(Y^2 < 9) = P(-3 < Y < 3) = P(-3 -1 < Y-1 < 3 - 1) = P((-3-1)/2 < (Y-1)/2 < (3-1)/2) = P(-2 < Z < 1) = \phi(1) - \phi(-2)$ where $Z$ is standard normal, $\phi$ is the normal CDF.

Answer 2

Hint: $\Phi (-z)=1-\Phi(z)\quad \Longrightarrow \quad \Phi (1)-\Phi(-2) \ = \ \Phi(1)-(1-\Phi (2)) \ = \\ \Phi(1)+\Phi (2)-1$