14. Intern's Mistake (from 8th grade, 3 points). At a bakery, a dosing machine measures out portions of dough weighing 400 g. The rule is: if the standard deviation of ten random measurements exceeds 5% of the nominal portion mass, the machine requires repair.

To check the machine, young intern Ivanov weighed 10 random portions dispensed by the machine, with an accuracy of one gram, and recorded in the table how many grams the portions differ from the nominal mass of 400 g, either way.

| Portion Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Deviation from 400 g | 12 | 32 | 18 | 19 | 28 | 15 | 7 | 22 | 12 | 21 |

Young engineer Petrov got angry and said that all the work was in vain, since the signs of the deviations are unknown, and he cannot calculate either the standard deviation or even the arithmetic mean.

Experienced master Sidorov just shook his head and said that everything was not so bad. Of course, finding the standard deviation is impossible, but it can be mathematically proven that the machine does not require repair. How could Master Sidorov have reasoned?
Solution. The master could reason as follows.

The root mean square deviation from 400 g is

$$
\sqrt{\frac{12^{2}+32^{2}+18^{2}+\ldots+21^{2}}{10}}=\sqrt{398}
$$

The standard deviation, which is the root mean square deviation of the numbers in the array from the arithmetic mean, is no greater than the root mean square deviation from any other number. Therefore, the standard deviation of the test mass in the control batch is no more than $\sqrt{398}$ g $<20$ g, which is less than 5% of the nominal mass of 400 g.

In the reasoning, a known fact was used: the standard deviation of any numerical array $X=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ does not exceed the root mean square deviation of this array from any number. Let's prove this. It is sufficient to show that the variance does not exceed the mean square deviation from any number.

Write down the mean square deviation from an arbitrary number $y$:

$$
\begin{aligned}
& \frac{\left(x_{1}-y\right)^{2}+\left(x_{2}-y\right)^{2}+\left(x_{3}-y\right)^{2}+\ldots+\left(x_{n}-y\right)^{2}}{n}= \\
& =y^{2}-\frac{2}{n}\left(x_{1}+x_{2}+\ldots+x_{n}\right) y+\frac{1}{n}\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)
\end{aligned}
$$

On the right side, we have a quadratic trinomial in the variable $y$. It takes its minimum value at the point

$$
y_{0}=-\frac{1}{2} \cdot\left(-\frac{2}{n}\left(x_{1}+x_{2}+\ldots+x_{n}\right)\right)=\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}=\bar{x}
$$

Thus, when $y=\bar{x}$, the mean square of the deviations from $y$ reaches its minimum value, which is equal to the variance of the array.