If you add two independent random variables, what is the standard deviation of the combined distribution, if the standard deviations of the two original distributions were, for example $7$ and $5$?

Hint: To find the standard deviation of two independent random variables, we cannot just add the standard deviations. First, we have to calculate variances. Standard deviation is defined as the square root of the variance and we can also say that variance is the square of standard deviation. So, first we square individual standard deviations and add them to get the variance then we take the square root to get total standard deviation.

Complete step by step answer:
Standard deviation is defined as the measurement of the dispersion of the data set from its mean value. It is always measured in arithmetic mean value. Standard deviation is always positive and denoted by the symbol $\sigma $ (sigma). Standard deviation is calculated as the square root of variance by determining each data- point’s deviation related to the arithmetic mean.
Variance is defined as the difference between the mean and its data points or it can be defined as how far a set of data points from their mean value. It is represented as ${\sigma ^2}$.
Example:
We have standard deviation of two original distribution i.e., $7$ and $5$
So, first we square the individual standard deviation. We get,
$ \Rightarrow {(7)^2} = 49 $
$ \Rightarrow {(5)^2} = 25 $
We add these together to get the total variance.
So, total variance$ = {(7)^2} + {(5)^2} = 49 + 25 = 74$
We take the square root of the total variance to get total standard deviation.
Therefore, Total standard deviation $ = \sqrt[2]{{74}} = 8.60$

Note:
We cannot just add standard deviation first we have to calculate variance. Standard deviation is only used in measuring dispersion around the mean value of the data set. Standard deviation can never be negative however, it can be zero when the values of a particular data set are the same. Without standard deviation we can’t compare the two sets of data effectively.