Sample and population standard deviations in a confidence interval Sample deviation is calculated as population standard deviation divided by square root of sample size $σ_x=\left(\frac{σ}{\sqrt x}\right)$. Then why in an example like this below, is the answer $5.086$ ? I understand that it comes from $4.2+1.96 \times \left(\frac{1.5}{\sqrt{11}}\right)$, but why are we dividing with $\sqrt {11}$ when going from sample to population, when according to the formula we should multiply with it?
The total amount of vegetation held by the earth’s forests is called the “biomass”. A random sample of $11$ one-square-metre plots was taken in Canada’s northern forests; the sample mean biomass was $4.2$ kilograms per square metre, and the sample standard deviation was $1.5$ kilograms per square metre. What is the upper limit of a $95%$ confidence interval for the mean biomass (kilograms per square metre) in all of Canada’s northern forests?
 A: Your goal is to get a 95% confidence interval for the mean $\mu$ of a normal population. Let's begin by pretending we know the population variance. 
A point estimate of $\mu$ is the sample mean $\bar X$ of the data.
In this situation, can show that $E(\bar X) = \mu$ and that
$Var(\bar X) = \sigma^2/n.$ Also, that the distribution of
$\bar X$ is given by $\bar X \sim Norm(\mu, \sigma/\sqrt{n}),$
where the second argument is the standard deviation.
Standardizing, we see that $Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$ is standard normal. Because 95% of the area under the standard normal curve lies between -1.96 and 1.96, we have
$$P(-1.96 < Z < 1.96) = P\left(-1.96 < \frac{\bar X - \mu}{\sigma/\sqrt{n}} < 1.96 \right) = 0.95.$$
By manipulating the inequality on the right, we can see that
$$P(\bar X -  1.96\sigma/\sqrt{n} < \mu < \bar X -  1.96\sigma/\sqrt{n}) = 0.95.$$ 
If $\sigma$ is known, this relationship is the basis to say that
a 95% confidence interval for $\mu$ has endpoints $\bar X \pm  1.96\sigma/\sqrt{n}.$
In this estimation context, $\sigma/\sqrt{n}$ is called the 'standard error' of the mean $\bar X.$ This is a way of saying that $SD(\bar X) = \sigma/\sqrt{n}.$ (The terminology 'standard error' is usually
used only with estimators.)
The denominator $\sqrt{n}$ indicates that the variability of the
mean of a sample of size $n$ gets smaller as the sample gets larger.
This is a different thing than the population standard deviation
$\sigma$. For a single randomly chosen observation $X_i$, one
could write $SD(X_i) = \sigma.$
In your specific problem, if $\sigma = 1.5,$ then the upper confidence limit is
$$\bar X +  1.96\sigma/\sqrt{n} = 4.2 + 1.96(1.5)/\sqrt{11},$$
based on the information provided.
However, in case $\sigma$ is not known, it is usually estimated by
the sample standard deviation $S = \sqrt{\frac{\sum_{i=1}^n(X_i - \bar X)^2}{n-1}}.$ In that case, $T = \frac{\bar X - \mu}{S/\sqrt{n}}$ 
has a Student t distribution with $n-1$ degrees of freedom, and
a number from printed tables or software must be
substituted for 1.96 above. For example, with $n = 11,$ your 95% CI would have the upper limit $\bar X +  2.228 S/\sqrt{n}.$ The R code
below shows the computation of 2.228.
 qt(.975, 10)
 ## 2.228139

