SOLUTION: For a laboratory assignment, if the equipment is
working, the density function of the observed outcome
X is
f(x) = {2(1 − x), 0 < x < 1,
{0, otherwise.
Find the varia
Algebra.Com
Question 1204182: For a laboratory assignment, if the equipment is
working, the density function of the observed outcome
X is
f(x) = {2(1 − x), 0 < x < 1,
{0, otherwise.
Find the variance and standard deviation of X.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
.
Hello, sorry,
I mistakenly pressed wrong button.
Please re-post your problem again.
Please do not post it to me in person.
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
Review the formulas mentioned here
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
E[X] = expected value of X = mean
The formula on that page says:
 = \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
 = \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
 = E(X^2) - \mu^2)
i.e.
 = E(X^2) - (E(X))^2)
So we'll need to calculate 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
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