Given that $$ is standard normal, there is a $68\%$ probability of it lying within one S.D. of its mean. Determine the probability in which $$ is within one standard deviation of its mean.
If $X = 4Z - 1$, I understand that mean = $-1$ and S.D. = $4$
Therefore I have to find $P(X≤(-1+4)) = P(X≤3)$ using:
$Z = \frac{(x+1)}{4}$ and therefore obtaining $P(Z≤1) = 0.5 + 0.3413$ (from $Z$-table) = $0.8413$
Am I on the right track? Because i am aware that I have not factored in the $68\%$ probability of it lying within one S.D. of the mean. If not, how do I move on from here?
Further to my comments above, we need to find $P(-1-4\le X\le - 1+4)=P(-1\le Z\le 1)=P(Z\le 1) - P(Z\le - 1)=0.68$. No need to look at table!