Given a probability density function:
$$\boxed{\begin{array}{c|r:r:r:r:r} x & -10& -5& 0& 5& 10\\ \hline P(x)& 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \end{array}}$$
1) Find the Standard Deviation:
Sorry the makeshift table I made up there might be confusing. But this is how I found the standard deviation:
$\newcommand{\Var}{\mathsf{Var}}\newcommand{\E}{\mathsf {E}}\Var(X) = \E(X^2) - (\E(X))^2$
$(\E(X))^2 = ((-10\cdot 0.1) + (-5\cdot 0.2) + (0\cdot 0.3) + (5\cdot 0.2) + (10\cdot 0.2))^2 = 1$
$\E(X^2) = ((-10^2\cdot 0.1) + (-5^2\cdot 0.2) + (0^2\cdot 0.3) + (5^2\cdot 0.2) + (10^2\cdot 0.2)) = 40$
$\Var(X) = \E(X^2) - (\E(X))^2 = 40 -1^2 = 39$
$\newcommand{\SD}{\mathsf{SD}}\SD(X) = \sqrt{\Var(X)} = \sqrt{39}$
2) If we have 100 independent replications of this experiment with measurements $X_i$ and the average of these 100 random variables is $\bar X$, then calculate $\Var(\bar X)$.
This is where it gets tricky for me. My professor said it was along lines of
$\Var(\bar X) = \Var(\tfrac 1 n \cdot \sum X)$
$\Var(\tfrac 1 n \sum X) = \tfrac 1{n^2}\Var(X_1 + X_2 + \cdots + X_n)$
So do I just multiply the variance I got for 1) by $\tfrac 1{100^2}$?
In other words does $\Var(X_1 + X_2 + \cdots + X_n)$ equal the $\Var(X)$ I found in 1)?
3) Use Chebyshev’s inequality to bound the probability. $\mathsf P\{\operatorname{abs}(\bar X − 1) > 2\}$
I'm completely lost at this one.
1) Looks technically correct, though I have not checked the calculations.$\newcommand{\Var}{\mathsf{Var}}\newcommand{\E}{\mathsf {E}}$
2) Independence means all covariances are zero, so $\Var(\sum X_k)$ equals $\sum\Var(X_k)$ , and identicallity means that equals: $n\Var(X)$ .
3) Chebyshev’s inequality is: $~\mathsf P\Big\{\big(X-\E(X)\big)^2\geq k^2\Big\}~{\large\leq}~\dfrac{\Var(X)}{k^2}$
Alternatively stated (for $k\geqslant 0$) as: $~\mathsf P\Big\{\big\lvert X-\E(X)\big\rvert \geq k\Big\}~{\large\leq}~\dfrac{\Var(X)}{k^2}$
So...