Task 1. Given a statistical series - the number of days missed due to illness by employees of a laboratory.

| Number of days | 0 | 2 | 3 | 4 | 5 | 7 | 10 | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of employees | 7 | 3 | 5 | 2 | 5 | 6 | 2 | 30 |

Determine the average number of days missed due to illness per person and the dispersion of this number around the average.
Solution. Let's determine the sample mean of size $n=30(k=7)$ using formula (1):

$$
\bar{x}_{3}=\frac{1}{30}(7 \cdot 0+3 \cdot 2+5 \cdot 3+2 \cdot 4+5 \cdot 5+6 \cdot 7+2 \cdot 10)=\frac{116}{30}=3.87
$$

We will calculate the variance and standard deviation using formula (2):

$$
\begin{aligned}
\overline{x_{\mathrm{B}}^{2}}=\frac{1}{30}\left(7 \cdot 0^{2}+3 \cdot 2^{2}+5 \cdot 3^{2}+2 \cdot 4^{2}+5 \cdot 5^{2}+6 \cdot 7^{2}+2 \cdot\right. & \left.10^{2}\right)= \\
= & \frac{692}{30}=23.1
\end{aligned}
$$

We have $D_{\mathrm{B}}=\overline{x_{\mathrm{B}}^{2}}-\left(\bar{x}_{\mathrm{B}}\right)^{2}=23.1-3.87^{2}=8.3 ; \sigma=\sqrt{D}=2.8$. Rounding to the nearest whole number, we get that the average number (expected value) of absences in a year per person is 4 days, with a spread of 3 days.

Problem 2. The sample provided in the following table corresponds to the service time (in minutes) for a customer in a store:

| Time Interval | $34 \div 38$ | $38 \div 42$ | $42 \div 46$ | $46 \div 50$ | $50 \div 54$ | $54 \div 58$ | $\Sigma$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of Customers | 2 | 3 | 15 | 18 | 10 | 2 | 50 |

Provide a forecast of the service time by the store and the standard deviation of this time. Find the mode of the variation series.

Solution. We transition from the interval table to the variation series by replacing the time interval with the center of this interval:

$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x_{i} & 36 & 40 & 44 & 48 & 52 & 56 & \Sigma \\
\hline n_{i} & 2 & 3 & 15 & 18 & 10 & 2 & 50 \\
\hline
\end{array}
$$

The mode is the variant with the maximum frequency; $M o=48$.
The numbers in the variation series are not convenient for calculations because they are large, so we transition to conditional variants with a new center $c=48$ and step (difference) $h=4$. Let $u_{i}=\frac{x_{i}-c}{h}=\frac{x_{i}-48}{4}$. We will calculate the sample mean and variance for the new variant. For convenience, we provide a table with intermediate calculations.

| $i$ | $x_{2}$ | $n_{2}$ | $u_{i}$ | $u_{2} n_{2}$ | $n_{i} u_{2}^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 36 | 2 | -3 | -6 | 18 |
| 2 | 40 | 3 | -2 | -6 | 12 |
| 3 | 44 | 15 | -1 | -15 | 15 |
| 4 | 48 | 18 | 0 | 0 | 0 |
| 5 | 52 | 10 | 1 | 10 | 10 |
| 6 | 56 | 2 | 2 | 4 | 8 |
| $\Sigma$ | - | 50 | -3 | -13 | 63 |

Using the sums in the last row of the table, we calculate: $\bar{u}_{\mathrm{B}}=-\frac{13}{50}=-0.26, \overline{u_{\mathrm{B}}^{2}}=\frac{63}{50}=1.26, \quad D_{\mathrm{B}}=1.26-0.0676=1.1924$, $\sigma(x)=1.1$.

The reverse transition to the variants $x$ is carried out using formulas that follow from the formulas in point $8^{\circ}$:

$$
\bar{x}_{\mathrm{B}}=c+\bar{u}_{\mathrm{B}} \cdot h, \quad D_{\mathrm{B}}(x)=D_{\mathrm{B}}(u) h^{2}, \quad \sigma(x)=\sigma(u) h
$$

We get: $\bar{x}_{\mathrm{B}}=48-4 \cdot 0.26=46.96 ; D_{\mathrm{B}}(x)=19.08 ; \sigma(x)=4.4$.

Forecast of service time: $\approx 47$ min, spread $\approx 4.4$ min.