Determine the mean and variance of the random variable with the following probability mass function. f(x) = (125/31)(1/5)^x ........ x = 1,2,3

LAST QUESTION I SWEAR!!!

Respuesta :

[tex]f(x)=\begin{cases}\dfrac{125}{31}&\text{for }x\in\{1,2,3\}\\\\0&\text{otherwise}\end{cases}[/tex]

The mean is

[tex]\mathbb E(X)=\displaystyle\frac{125}{31}\sum_{x=1}^3x\left(\frac15\right)^x=\dfrac{38}{31}[/tex]

You then have

[tex]\mathbb E(X^2)=\displaystyle\frac{125}{31}\sum_{x=1}^3x^2\left(\frac15\right)^x=\dfrac{54}{31}\approx1.74[/tex]

so the variance is

[tex]\mathbb V(X)=\dfrac{54}{31}-\left(\dfrac{38}{31}\right)^2=\dfrac{230}{961}\approx0.239[/tex]