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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.

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  • $\begingroup$ 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 $\bar{X}$. $\endgroup$ – angryavian Dec 5 '18 at 21:27
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Hint:

$P(X \leq 5 \text { or } X \geq 9) = P(|X-7| \geq 2) $

two away from mean on both sides..

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  • $\begingroup$ This would give a different upper bound then what I got $\endgroup$ – K Split X Dec 5 '18 at 22:40
  • $\begingroup$ 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. $\endgroup$ – karakfa Dec 5 '18 at 22:45
  • $\begingroup$ What is the correct way to solve the $P(\overline{X} \lt 5)$ ? $\endgroup$ – K Split X Dec 5 '18 at 22:50
  • $\begingroup$ Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed. $\endgroup$ – karakfa Dec 5 '18 at 22:53
  • $\begingroup$ 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? $\endgroup$ – K Split X Dec 6 '18 at 0:27

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