Example 6. Determine how many measurements of the cross-sectional diameter of trees need to be made on a large plot so that the average diameter of the trees differs from the true value $a$ by no more than 2 cm with a probability of at least 0.95. The standard deviation of the cross-sectional diameter of the trees does not exceed $10 \mathrm{~cm}$ and measurements are taken without error.
Solution. We will consider the selection of trees for measurements such that the measurement results can be regarded as independent random variables. Let $X_{i}$ denote the measurement result of the cross-section of the $i$-th tree. According to the problem, $\sigma\left(X_{i}\right)=\sqrt{D\left(X_{i}\right)} \leq 10$, hence, $D\left(X_{i}\right) \leq 100$.

Assuming in inequality (4.2.3.) $\varepsilon=2, C=100$, we get

$$
P\left(\left|\frac{1}{n} \sum_{i=1}^{n} X_{i}-a\right|<2\right) \geq 1-\frac{100}{n \cdot 2^{2}} \geq 0.95
$$

Since

$$
1-\frac{100}{n \cdot 2^{2}} \geq 0.95, \text { then } 0.05 \geq \frac{100}{n \cdot 4}, n \geq \frac{100}{4 \cdot 0.05}=500
$$

Therefore, it is sufficient to perform 500 measurements of the cross-section of the trees.