# Normal distribution interval

I am trying to find $c$ so that $P(|X - 5| < c) = .95$ with $\mu = 5, \sigma^2 = 4$.

I came up with: $P( \frac{-c}{2} < Z < \frac{c}{2}) = .95$

Attempting to solve for $c$, the $z$-score for $c$ is $c = 1.65$, but I know this is not the correct answer. Can anybody tell me what I am doing wrong?

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You want probability $0.025$ in the right tail, same in the left tail. Look for $1-0.025=0.975$ in the body of the table. This corresponds (approximately) to $1.96$. So you want $c/2\approx 1.96$. –  André Nicolas Mar 9 '13 at 18:55
@AndréNicolas: Thanks for your input. Why am I looking for .025 in the right tail? –  Doug Ramsey Mar 9 '13 at 18:58
@Doug Ramsey:$P( \frac{-c}{2} < Z < \frac{c}{2}) = 2P( 0 < Z < \frac{c}{2})=.95$ then$$P( 0 < Z < \frac{c}{2})=\frac{.95}{2}$$ you find value of $\frac{c}{2}$ Standard normal table –  Maisam Hedyelloo Mar 9 '13 at 18:59
You need to have a picture of the standard normal in your head, or in front of you. We want the number $d$, which is $c/2$, be such that $\Pr(-d\le Z\le d)=0.95$. So the part between $-d$ and $d$ should have area (probability) $0.95$. That means that the part beyond $d$, and the part before $-d$, should have combined area $0.05$. By symmetry, the part after $d$ has area $0.025$, so the part before $d$ has area $0.975$. Look this up in the table. You get $d=1.96$. (Continued) –  André Nicolas Mar 10 '13 at 1:23
Another way: Between $-d$ and $d$ is $0.95$. So by symmetry the part between $0$ and $d$ has area half of that, which is $0.475$. So the area from $-\infty$ to $d$ is $0.5+0.475=0.975$. Look for $0.975$ in the body of the normal table to find $d$. –  André Nicolas Mar 10 '13 at 1:28