18. Weather Diary (8th grade. 5 points). Throughout March, 7th-grader Ivanov kept a weather observation diary. Every day at exactly 3 PM, he would peek out the kitchen window, look at the outdoor thermometer, and record the air temperature in his diary. After statistical processing, the results were:

- Average temperature for March: $\bar{x}=0$ (degrees Celsius).
- Median: $m=4$ (degrees Celsius).
- Variance: $S^{2}=15.917$ (likely square degrees or degrees squared Celsius, but this did not concern Ivanov).

The geography teacher was very pleased and gave Ivanov a five. The math teacher was also pleased with Ivanov and told him to go and find the mistake. Why did the math teacher think Ivanov had made a mistake?
Solution. Probably, the math teacher knew that for any set of numbers, the median is not more than one standard deviation away from the arithmetic mean:

$$
|\bar{x}-m| \leq \sqrt{S^{2}} \text {, hence }(\bar{x}-m)^{2} \leq S^{2} \text {. }
$$

Let's prove this. Suppose the set consists of numbers $x_{1}, x_{2}, \ldots, x_{n}$. Then

$$
|\bar{x}-m|=\left|\frac{x_{1}-m}{n}+\frac{x_{2}-m}{n}+\ldots+\frac{x_{n}-m}{n}\right| \leq \frac{\left|x_{1}-m\right|+\left|x_{2}-m\right|+\ldots+\left|x_{n}-m\right|}{n}
$$

This follows from the triangle inequality - the modulus of the sum of numbers is not greater than the sum of their moduli, regardless of these numbers. The average modulus of deviation from the median, which stands on the right side, is not greater than the average modulus of deviation from any other number, for example, from the arithmetic mean $\bar{x}$. The proof of this fact can be found in the solution to problem 14 of the 2010 olympiad (http://www.mccme.ru/terver/Solutions2010.pdf). We obtain:

$$
|\bar{x}-m| \leq \frac{\left|x_{1}-m\right|+\left|x_{2}-m\right|+\ldots+\left|x_{2}-m\right|}{n} \leq \frac{\left|x_{1}-\bar{x}\right|+\left|x_{2}-\bar{x}\right|+\ldots+\left|x_{2}-\bar{x}\right|}{n}
$$

It remains to apply the inequality between the arithmetic mean and the quadratic mean (sometimes called the Cauchy inequality):

$$
|\bar{x}-m| \leq \frac{\left|x_{1}-\bar{x}\right|+\left|x_{2}-\bar{x}\right|+\ldots+\left|x_{n}-\bar{x}\right|}{n} \leq \sqrt{\frac{\left(x_{1}-\bar{x}\right)^{2}+\left(x_{2}-\bar{x}\right)^{2}+\ldots+\left(x_{n}-\bar{x}\right)^{2}}{n}}=S
$$

Therefore, $(\bar{x}-m)^{2} \leq S^{2}$.

Substituting the known numbers, we get: $(0-4)^{2}<15.917$. This is false. Therefore, Ivanov made a mistake somewhere.