Probability and using z-table in Normal Distribution

Question

There's a medication whose target dosage is $\mu$. Due to external errors, the actual dosage follows a normal distribution with mean $\mu$ and standard deviation $\sigma$. I'm supposed to find the probability that the dose administered differs from the mean $\mu$ by less than $\sigma$.

We're supposed to use the z-table in this problem. Suppose a random variable X denotes the dosage of medicine. We want $P(X-\mu<\sigma)$ if $X>\mu$ and $P(\mu-X<\sigma)$ if $X<\mu$. I took $z=\frac{X-\mu}{\sigma}$ ,and then I had to replace X by z and find the corresponding probability value from the table. It turns out $P(z<1)$ which is equal to $0.8413$. If I do for the other part, I just get $1-0.8413$.

Now, I'm not sure whether I'm doing it correctly. For one reason, the given answer is $0.6826$. I'm not given any values, only $\mu$ and $\sigma$. What am I doing wrong?

Answer 1

You don't want $P(X-\mu < \sigma)$ if $X>\mu$ and $P(\mu-X < \sigma)$ if $X < \sigma$, you want $P(|X-\mu|<\sigma) = P(|z|<1) = P(z<1) - P(z<-1)$.

Answer 2

What "nullUser" said is correct. Looking at the table you find P(z<-1) = .16 and p(z<1) = .84 roughly. You can solve then by just subtracting .84-.16 to get the probability in-between. If you just calculate p(z<1) like it seems you did, it overlaps with z<-1 so there is no way to discern the correct answer unless you "delete" the overlapping portion.