Question 1204182
<font color=black size=3>
Review the formulas mentioned here
<a href = "https://stats.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame/MATH_345__-_Probability_(Kuter)/4%3A_Continuous_Random_Variables/4.2%3A_Expected_Value_and_Variance_of_Continuous_Random_Variables">https://stats.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame/MATH_345__-_Probability_(Kuter)/4%3A_Continuous_Random_Variables/4.2%3A_Expected_Value_and_Variance_of_Continuous_Random_Variables</a>


E[X] = expected value of X = mean


The formula on that page says:
*[tex \Large E(X) = \int_{-\infty}^{\infty}\text{x}f(\text{x})d\text{x}]
However, we do not have to worry about the entire real number line.
Instead, we can shrink the domain to focus on the interval 0 < x < 1, since f(x) = 0 otherwise.


You need to calculate 
*[tex \Large E(X) = \int_{0}^{1}\text{x}f(\text{x})d\text{x} = \int_{0}^{1}2\text{x}(1-\text{x})d\text{x}]


I'll skip steps
Integrating should be fairly trivial because 2x(1-x) = 2x-2x^2 is a polynomial


You should find that
E[X] = 1/3
in other words,
mu = mean = 1/3


-----------------------------


The variance is defined as
*[tex \Large Var(X) = E(X^2) - \mu^2]
i.e.
*[tex \Large Var(X) = E(X^2) - (E(X))^2]


So we'll need to calculate E(X^2)
*[tex \Large E(X^2) = \int_{0}^{1}\text{x}^2f(\text{x})d\text{x} = \int_{0}^{1}2\text{x}^2(1-x)d\text{x}]
Once again I'll skip steps. 


The result you should get is E[X^2] = 1/6


-----------------------------


Therefore,
Var(X) = variance
Var(X) = E[X^2] - ( E[X] )^2
Var(X) = 1/6 - ( 1/3 )^2
Var(X) = 1/18


And,
SD(X) = standard deviation
SD(X) = sqrt( Var(X) )
SD(X) = sqrt( 1/18 )
SD(X) = 1/sqrt(18)
SD(X) = sqrt(18)/18
</font>