Rolling a die with Chebyshev's inequality

Question

We roll a die $360$ times. We call an event a success when a face with even number of dots appears. Use Chebyshev's inequality to calculate interval in which the probability of having an obtained number of successes equals at least $0.9$.

My attempt: $$Pr(|X-\mu|\geq t\sigma)\leq \frac{1}{t^2}$$ $$1-Pr(|X-\mu|\geq t\sigma)\geq1- \frac{1}{t^2}$$ $$Pr(|X-\mu|\leq t\sigma)\geq1- \frac{1}{t^2}$$ We have $$1-\frac{1}{t^2}=0.9$$ $$\frac{1}{t^2}=0.1 \Rightarrow t=\sqrt{10}$$

The probability of obtaining an even numbered face in a single roll is $$p=0.5$$ This is where I'm getting stuck. How do I calculate $\mu$ and $\sigma$?

Answer 1

$μ=1/2$ and $σ^2=1/4$?

That´s right.

But that's for a single roll, let's say $X_i$ and $X=X_1+...+X_{360}$. Are $μ$ and $σ$ the same for $X$ or do I need to divide/multiply them by 360?

The expected value of the sum of the die rolls is equal to the sum of the extected values: $$\mathbb E\left(\sum\limits_{i=1}^{360} X_i \right)=360\cdot \mu_{x_1} $$

And the variance of the sum of the die rolls is equal to the sum of the variances, due independence of the rolls: $$Var\left(\sum\limits_{i=1}^{360} X_i \right)=360\cdot \sigma^2_{x_1}$$