# 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.

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.