How do you find the standard deviation from the probability distribution?

Hint: In this question, we have to find the standard deviation. As we know, a probability distribution is a mathematical function that gives all the possible values of a random variable within a given range, which implies it gives the probabilities of different possible outcomes of an experiment. A standard deviation means a measure of dispersion that is the absolute variability of a distribution. So, we simply take the general form of probability distribution and then find its mean, and after that variance. Then, we take the square root of the variance, to get the required result of the problem.

Complete step by step answer:
According to the problem, we have to find the standard deviation.
As we know, the probability distribution is a statistical function that takes all possible values of a random variable. It is denoted by $ f(x) $ , therefore the pdf of a random variable is
 $ pdf=\int_{-\infty }^{\infty }{f(x)dx} $
Now, we know that the mean of a probability distribution, is the product of the variable and its pdf. It is denoted by $ \overline{x} $ or E(x) , therefore we get
 $ mean=\int_{-\infty }^{\infty }{x.f(x)dx} $
So, we will now find the variance of the probability distribution. As we know, the variance is the variability of the probability distribution. It is denoted by $ \sigma _{X}^{2} $ , therefore its formula is the sum of the mean of $ {{x}^{2}} $ and square of the mean, that is
 $ \sigma _{X}^{2}=E({{x}^{2}})-{{\left( E(x) \right)}^{2}} $
Therefore, we get
 $ \text{variance=}{{\int_{-\infty }^{\infty }{{{x}^{2}}.f(x)dx-\left( \int_{-\infty }^{\infty }{x.f(x)dx} \right)}}^{2}} $
Since the standard deviation is the square root of the variance, it is denoted by $ {{\sigma }_{X}} $ , therefore we get
 $ \begin{align}
  & SD=\sqrt{\sigma _{X}^{2}} \\
 & {{\sigma }_{X}}=\sqrt{E({{x}^{2}})-{{\left( E(x) \right)}^{2}}} \\
\end{align} $
Therefore, we get
 $ \text{SD=}\sqrt{{{\int_{-\infty }^{\infty }{{{x}^{2}}.f(x)dx-\left( \int_{-\infty }^{\infty }{x.f(x)dx} \right)}}^{2}}} $
Thus, for the probability distribution, its standard deviation is $ \sqrt{{{\int_{-\infty }^{\infty }{{{x}^{2}}.f(x)dx-\left( \int_{-\infty }^{\infty }{x.f(x)dx} \right)}}^{2}}} $ or $ \sqrt{E({{x}^{2}})-{{\left( E(x) \right)}^{2}}} $.

Note:
While finding the standard deviation, keep in mind the definitions of all the terms. Do not get confused about the variance and the standard deviation, because standard deviation is the square root of the variance. Do mention the symbols used to represent all the terms.