Estimate the sample distribution given mean and root mean square

Question: Given a dataset $\{x_k\}_{k=1}^n$ with $n=500000$ points which have an average $\frac1n\sum_{k=1}^nx_n=13.06$ and a square root mean $\sqrt{\frac1n\sum_{k=1}^nx_n^2}=13.67$. Under this condition, at most how many data points could have a value greater than 17.1? please give a nontrivial upper bound.

A first glimpse of this question reminded me of something like Chebyshev inequality. But the question asks for the distribution of the sample points, so I think such results in probability theory may fail to work here. So I tried to use some elementary algebra approaches, but I don’t know where to start with.

Many thanks.

• Considering you need only the right tail of the distribution, you're better off using Cantelli's inequality. – Raskolnikov Apr 15 '18 at 6:00
• @Raskolnikov hi, thank you for your reply! This inequality gives an upper bound of probability, but how can I use it to estimate the distribution of the original sample? – WallTi Apr 15 '18 at 6:05
• With such a large $n,$ you have enough information to get a (fairly short) z confidence interval. But that will not give you a true upper bound. So this is a probability problem, not a statistics problem. – BruceET Apr 15 '18 at 6:05
• Not sure I understand your question, are you in possession of the dataset, or only of the average and square root mean? If the latter, the best you can do is some Chebyshev type inequality. If the former, then you just count how many data points have a value greater than 17.1. – Raskolnikov Apr 15 '18 at 6:08
• @Raskolnikov it is an exercise rather than a practical problem. And yes, only a average and square root mean is given. – WallTi Apr 15 '18 at 6:13

Whether you call it Cantelli's inequality or a one-sided Chebyshev inequality, you need to find the variance $13.67^2 - 13.06^2 = 16.3053$ and then look at $$500000 \times \left(\dfrac{16.3053}{16.3053+(17.1-13.06)^2}\right) \approx 249875.1$$ but counts have to be integers so the maximum possible number of observations over $17.1$ is $249875$, almost half of the sample size; this is not much of a surprise as the difference between $17.1$ and the mean is just over one standard deviation
So for example if $249875$ observations were actually all $17.100001665 > 17.1$ and $250125$ observations were actually all $9.024036317$, then you would recover the $13.06$ and $13.67$ of the question to $9$ decimal places