Chebyshev inequality in terms of RMS I'm self studying the book Introduction to Applied Linear Algebra – Vectors, Matrices, and Least Squares
In page 48, the author write: "It says,for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5."
I need more explain about it. Especially about why the factor is 5?
 A: According to Chebyshev's_inequality, the probability of a value to deviate more than $k=5$ standard deviations from the mean is at most $1/k^2$.
When applied to vectors in your specific case, and following the book you cited, let $k$ be the number of elements of the vector $\vec{x}=(x_1,\ldots,x_n)$ such that $||x_i|| \geq a > 0$.
Hence $\|\vec{x}\|^2 = \sum x_i^2 \geq k a^2 + (n - k) \times 0$, which means that we have $k$ values larger than $a^2$ and the others $n-k$ values are at least zero.
Since the root mean square value is $\operatorname{rms}(\vec{x}) = \sqrt{\frac{\|\vec{x}\|^2}{n}}$, it follows that $\operatorname{rms}(\vec{x})^2 = \frac{\|\vec{x}\|^2}{n} \geq \frac {k a^2}{n}$.
Therefore, we get the final expression that says 
$$\frac {k}{n} \leq \left( \frac{\operatorname{rms}(\vec{x})}{a} \right) ^2$$
So, following the example, where $a = 5 \operatorname{rms}(\vec{x})$, we have that $\frac {k}{n} \leq \left( \frac{1}{5} \right) ^2 = 4 \%$, so, the fraction of elements of the vector larger (in absolute value) than $5\operatorname{rms}$ is at most $4\%$.
If we chose another number, say $a = 2 \operatorname{rms}(\vec{x})$, we would have that $\frac {k}{n} \leq \left( \frac{1}{2} \right) ^2 = 25 \%$.
