457. As a result of five measurements of the length of a rod with one instrument (without systematic errors), the following results (in mm) were obtained: $92 ; 94 ; 103 ; 105 ; 106$. Find: a) the sample mean length of the rod; b) the sample and corrected variances of the instrument errors.
Solution. a) Let's find the sample mean:

$$
\bar{x}_{\mathrm{B}}=92+(0+2+11+13+14) / 5=92+8=100
$$

b) Let's find the sample variance:

$$
\begin{gathered}
D_{\mathrm{B}}=\frac{\sum\left(x_{i}-\bar{x}_{\mathrm{B}}\right)^{2}}{n}=\left[(92-100)^{2}+(94-100)^{2}+(103-100)^{2}\right] / 5+ \\
+\left[(105-100)^{2}+(106-100)^{2}\right] / 5=34 .
\end{gathered}
$$

Let's find the corrected variance:

$$
s^{2}=\frac{n}{n-1} \cdot D_{B}=\frac{5}{4} \cdot 34=42.5
$$