If I toss a coin 100 times, what's the probability of getting greater than 40 heads? I found one answer that says, use the binomial distribution. Since p = 0.50 and N = 100, the mean is 50, the variance is 25 and the standard deviation is 5. So 40 corresponds to a one sided 2 std move. A two sided 2 std move contains 95% of the area, so a 1 sided 2 std move contains 95% + 2.5% = 97.5%.
I would have thought it was just two standard deviations away so it's 95%. Where does the additional 2.5% come from? Isn't a move usually to one side, and what would I do if it was one standard deviation, would I do 68% + 2.5% as well?
 A: There's an about 95% probability of getting between 40 and 60 heads.
By symmetry, the remaining 5% is split evenly between "fewer than 40" and "more than 60", that is, 2.5% for each.
Your "more than 40" event is the disjoint union of "between 40 and 60" and "more than 60", so you get 95% + 2.5%.
A: You're asking for the probability that the number of head is greater than $40$—that is,
$$
P(40<H)\ ,
$$
where $\ H\ $ is the number of heads.  The two-sided confidence interval that has a probability of $95$% is the event $\ \{\ 40<H<60\ \}\ $. But
\begin{align}
P(40<H)&=P(\{\ 40<H<60\ \}\cup\{60\le H\})\\
&=P( 40<H<60)+P(60\le H)\\
&=0.95+P(60\le H)\ ,
\end{align}
because the events are mutually exclusive.  Thus the extra $2.5$% comes from the addition of $\ P(60\le H)\ $. Because the distribution is symmetric,
\begin{align}
P(40< H)&=1-P(H\le40)\\
&=1-P(H\ge60)\ ,
\end{align}
so the identity above gives us
$$
1-P(H\ge60)=0.95+P(60\le H)\ ,
$$
or
$$
1-0.95=0.05=2P(60\le H)\ ,
$$
which gives $\ P(60\le H)=0.025\ $.
For one standard deivation, what you add is not just $2.5$%, but $\ \frac{1-0.68}{2}$$=0.16\ $—that is, $16$%—to get $84$%.
