Let we have $X_{1},.....,X_{100}$ be iid random variable from $U(-0.5,0.5)$. Then, prove using the chebychev inequality
$P(T^2\geq25)\leq\frac{1}{3}$
Where $T = X_{1}+....+X_{100}$
My approach
The inequality can be written as:
$1-P(T^2\leq25)\leq\frac{1}{3}$ implies that $P(T^2\leq25)\geq\frac{2}{3}$ $P(-5\leq T\leq 5)\geq \frac{2}{3}$
I can make the T as sum of Uniform random variable with parameters $U(0,1)$. After that, we can use Irwin hall distribution of sum of iid $U(0,1)$ random variables.
After adjusting terms,
$P(-5-50\leq T-50\leq 5-50)\geq \frac{2}{3}$ $P(-55\leq T-50\leq -45)\geq \frac{2}{3}$
$T-50$ will be sum of 100 $U(0,1)$ iid random variables with mean $50$ and variance $\frac{100}{12}$.
But now, chebychev inequality doesn't seems to be applicable. From here, I am not able to proceed.
Any help?
It seems that you just have to use the chebychev inequality. You have already found that $$P(T^2\geq25)=P(|T|\geq5)$$
In combination with the chebychev inequality we get
$$P(|T-\mu|\geq k\sigma)\leq \frac{1}{k^2} \Rightarrow P(|T|\geq k\sigma)\leq \frac{1}{k^2}$$
Comparing $P(|T|>5)$ and P$(|T|\geq k\sigma)$ we see that $5=k\sigma$
$5=k\cdot \frac{5}{\sqrt 3}$ It follows that $k=\sqrt 3$ Thus $P(|T|\geq 5)\leq \frac{1}{3}$. Equivalently $P(T^2\geq 25)\leq \frac13$