Calculating two standard deviations above the mean I have a variable a need to find data points which are two standard deviations above the mean. 
Example, let say we have: 2, 3, 4, 120, 5.
Sample mean=26.11 Stan.deviation=52.11
I have been calculating something like: 2*52.11+26.11=131.02
I think I am missing something since the 120 is outside of normal distribution curve but if I use this calculation method the 120 is smaller than 131.02, which means that it is normally distributed. 
Could anyone explain that to me?
 A: I think the criterion of having an observation exceed $\bar X + 2S$ is intended
as a way to highlight an observation as an "outlier." This criterion does
not work very well because the large observation 120 itself leads to
inflating both $\bar X$ and $S$ so that $\bar X + 2S$  is larger than it "should" be.
A better criterion might be to delete 120 before computing mean $\bar X^\prime$ and $S^\prime.$ Then you'd have $120 > 6.082.$ [Computation using R.]
x.d = c(2,3,4,5);  mean(x.d) + 2*sd(x.d)
[1] 6.081989

Another method is to use the interquartile range (IQR) as a 'robust' measure
of variability (which does not depend so crucially on any one observation).
That is the idea behind using boxplots to suggest possible outliers.
x = c(2, 3, 4, 120, 5)
boxplot(x, horizontal=T)


The boxplot 'outlier' rule is to suggest any observation above
$Q_3 + 1.5\text{IQR}$ as a high outlier, where $Q_3$ is the third quartile.
Similarly, any observation below $Q_1 - 1.5\text{IQR}$ is suggested
as a possible low outlier.
All outlier criteria need to be used with great caution because
some distributions routinely have 'outliers'--especially in large
samples. Almost all samples of size 100 from an exponential distribution
have 'outliers'. More specifically, for the simulation below about 99%  of
exponential samples of size 100 had at least one outlier and the average
number of outliers per sample was nearly 5.
nr.out = replicate(10^5, length(boxplot.stats(rexp(100))$out))
mean(nr.out)
[1] 4.84265
mean(nr.out > 0)
[1] 0.99023

Even among samples of size 100 from a normal distribution, over 90%
have at least one 'boxplot outlier'--but usually only one or two of them
per sample. Generating such a sample with today's date as seed, I happened to get three 'outliers'.
set.seed(1116); x = rnorm(100)
boxplot(x, horizontal=T)


