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Suppose that a data set $ \{x_n: n = 1,\dots,N\} $, with $ N = 500,000 $ has average \begin{equation} <x> = \frac{1}{N}\sum\limits_{n=1}^{N}x_n=13.06 \end{equation} and root mean square \begin{equation} \sigma = \sqrt{\frac{1}{N}\sum\limits_{n=1}^{N}x_n^2}=13.67 \end{equation} Using this information, derive the best upper bound you can, for the number of measurements that are greater than $ 17.1 $.


Intuition says that we should determine such $ k \in \mathbb{N}, l_0 \in \mathbb{R} $ so that exactly k measurements would be equal to $ 17.1 $, and exactly $ N - k $ measurements would be equal to $ l_0 $ (so that $ <x> $ and $ \sigma $ would be as given - obviously, it is very easy). But I can't even prove that $ k $ $ is $ an upper boundary, let alone the best one. can anybody hint, is there a well-known formula for this case?

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If you're being asked to solve this problem, you have probably seen a few inequalities named after Russians. –  Michael Lugo Aug 22 '13 at 3:49
    
Markov & Chebyshev? –  user85663 Aug 22 '13 at 3:52
    
They seem related, but not too much. We are dealing with a sample here, not with random variables and probabilities. –  user85663 Aug 22 '13 at 3:56
    
In the context of your problem, what you are calling the "sample" is actually a population. You need to view it as a probability mass function. And the inequality does apply, as seen below. –  soakley Aug 24 '13 at 14:29

3 Answers 3

up vote 2 down vote accepted

The one-sided Chebyshev inequality states $$P[X \geq \mu + k\sigma] \leq {{1} \over {1+k^2}} ,$$ where $\sigma$ is the population standard deviation (different from your definition above). For your data, I find the population standard deviation to be $\sigma \approx 4.04,$ which leads to $k=1.$

Using this in the inequality gives $$P[X \geq \mu + \sigma] =P[X \geq 17.1] \leq {{1} \over {2}}.$$

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$$ \approx 500,000\ \times\ {1 \over \sqrt{\,2\pi\,}\,\sigma\, }\ \int_{17.1}^{\infty} \exp\left(-\,{\left\lbrack x - \left\langle x\right\rangle\right\rbrack^{2} \over 2\sigma^{2}}\right) \,{\rm d}x $$

assuming that $\displaystyle{\int_{x\ <\ 0} \approx 0}$.

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While the sample size is large enough that the mean and variance of the sample will behave as a normal distribution due to the Central Limit Theorem, it is not the case that the sample SET behaves as if it were normally distributed, so we are stuck using the Chebyesev-Cantelli inequality as mentioned above.

To flesh out answer 1 a bit more, since the root mean square is 13.67, this means that $E(X^2) = 13.67^2$ so the standard deviation becomes: $$ \sigma = \sqrt{\sigma^2} = \sqrt{E(X^2) - E(X)^2} \approx \sqrt{13.67^2 - 13.06^2} \approx 4.038 $$ Therefore, as mentioned above the best upper bound without knowing anything more about the shape of the distribution is 50%.

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