Assume $X$ is a normal distributed random variable with mean $2$ and variance $4$. Determine the conditional probability $P(1 \le X \le 3|0 \le X \le 4)$
What I did: $$Z_0 = \frac{0-2}{2}=-1$$ $$Z_1 = \frac{1-2}{2}=-\frac12$$ $$Z_3 = \frac{3-2}{2}=\frac12$$ $$Z_4 = \frac{4-2}{2}=1$$ using z-score table... $$P(1 \le X \le 3) = .6915 - .3085 = .383$$ $$P(0 \le X \le 4) = .8413 -.1587 = .6826$$ $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$ $$P(1 \le X \le 3|0 \le X \le 4) = .383$$
I know the correct answer is $ 0.560907$. What am I doing wrong?
You did it all right, except for the final fraction. The joint event is:
$$A \cap B \equiv (1 \le X \le 3) \cap (0 \le X \le 4)\equiv (1 \le X \le 3)\equiv A$$
(Notice that this holds because here $A \subset B$)
Hence
$$\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{.383}{.6826}=0.561$$