# Find Chebyshev's upper bound for this probability

We are given that $$X$$ is $$Uniform[4,10]$$ with mean $$7$$ and variance $$3$$.

We are first asked to figure out $$P(X \leq 5 \text { or } X \geq 9$$)

Well this is a uniform graph with height = $$1/(10-4)=1/6$$, and we can do the following:

$$P(X \leq 5 \text { or } X \geq 9) = P(X \leq 5) + P(X\geq 9)=(5-4)/6+(10-9)/6=2/6=1/3$$

Now I am asking the following:

Let $$X_1, X_2, X_3, \dots, X_{16}$$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $$P(\overline{X} \leq 5 \text { or } \overline{X} \geq 9)$$ where $$\overline{X}=\dfrac{\sum_{i=1}^{16}X_i}{16}$$

Again, we can break up the probability to be as follows:

$$P(\overline{X} \leq 5) + P(\overline{X} \geq 9)\leq \text{upper bound}$$

Chebyshev Inequality is the following: $$P(|X-\mu_x| \geq a)\leq \dfrac{var(x)}{a^2}$$

So, for $$P(\overline{X}\leq 5)$$, we have $$P(|\overline{X}-7|\leq5)=1-\dfrac{3}{25}$$

For $$P(\overline{X}\geq 9)$$, we have $$P(|\overline{X}-7|\geq 9)=\dfrac{3}{9^2}$$

So in total, the upper bound must be $$1-\dfrac{3}{25}+\dfrac{3}{9^2}=0.917$$

Is this correct? I don't see where we used $$\overline{X}$$ in this. There would be nothing different if I considered $$X$$, instead of $$\overline{X}$$, so I think I am missing something.

Maybe the $$var(x)$$ should actually be $$\sigma^2/n=3/16$$. But in my notes, I have that this is only the case if $$X_1, X_2, \dots, X_n$$ are $$N(\mu, \sigma^2)$$. Here we have that they are uniform, so I am not sure.

• In general, $\text{var}(\bar{X}) = \frac{1}{n^2} \text{var}(\sum_{i=1}^n X_i) = \frac{1}{n} \text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $\bar{X}$. – angryavian Dec 5 '18 at 21:27

$$P(X \leq 5 \text { or } X \geq 9) = P(|X-7| \geq 2)$$
• Yes, because your calculations steps are not correct. $P(\overline{X }\leq 5)$, doesn't mean $P(|\overline{X}-7| \leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive. – karakfa Dec 5 '18 at 22:45
• What is the correct way to solve the $P(\overline{X} \lt 5)$ ? – K Split X Dec 5 '18 at 22:50
• If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not? – K Split X Dec 6 '18 at 0:27