# Chebyshev's Hoeffding's inequalities

Distribution of Student's Grades ( 25 points) A student submits 7 assignments graded on the $0-100$ scale. We assume that each assignment is an independent sample of his/her knowledge of the material and all scores are sampled from the same distribution. Let $X_{1}, \ldots, X_{7}$ denote the scores and $\hat{Z}=\frac{1}{7} \sum_{i=1}^{7} X_{i}$ their average. Let $p$ denote the unknown expected score, so that $\mathbb{E}\left[X_{i}\right]=p$ for all $i$. What is the maximal value $z$, such that the probability of observing $\hat{Z} \leq z$ when $p=60$ is at most $\delta=0.05$ ?

2. Use Chebyshev's inequality to answer the question. (You can use the fact that for a random variable $X \in[a, b]$ and a random variable $Y \in\{a, b\}$ with $\mathbb{E}[X]=\mathbb{E}[Y]$ we have $\operatorname{Var}[X] \leq \operatorname{Var}[Y]$. In words, the variance of a random variable taking values in a bounded interval is maximized when the distribution is concentrated on the boundaries of the interval. You should determine what should be the values of $\mathbb{P}(Y=a)$ and $\mathbb{P}(Y=b)$ in order to get the right expectation and then you can obtain a bound on the variance.)

3. Use Hoeffding's inequality to answer the question.

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There seems to be something off about part 2. Based on the given information the maximized variance is 18.51 which is too large to get something meaningful from Chebyshev's inequality. -

actually maximum standard deviation is 18.51, and not the variance.

Double check the question.

this is all the info that we got. I think we have to show that the answer dosent make sense.

yeah, the problem doesn't seem to make sense. I did the computations and it did not get to anything meaningful.

You may want to repost and mention this in your new post.