1775. To determine the average amount of deposits in a savings bank with 2200 depositors, a sample survey (sampling without replacement) of 111 deposits was conducted, which yielded the following results:

| Deposit amount, rubles | $10-30$ | $30-50$ | $50-70$ | $70-90$ | $90-110$ | $110-130$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of deposits | 1 | 3 | 10 | 30 | 60 | 7 |

Using these data, find the confidence limits for the population mean that could be guaranteed with a probability of 0.96.
Solution. First, we find the sample mean and sample variance:

| $X_{i}^{\prime}-X_{i}^{\prime \prime}$ | $x_{i}$ | $m_{i}$ | $u_{i}=\frac{x_{i}-80}{20}$ | $m_{i} u_{i}$ | $m_{i} u_{i}^{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $10-30$ | 20 | 1 | -3 | -3 | 9 |
| $30-50$ | 40 | 3 | -2 | -6 | 12 |
| $50-70$ | 60 | 10 | -1 | -10 | 10 |
| $70-90$ | 80 | 30 | 0 | 0 | 0 |
| $90-110$ | 100 | 60 | 1 | 60 | 60 |
| $110-130$ | 120 | 7 | 2 | 14 | 28 |
| Sum |  | 111 |  | 55 | 119 |

$$
\begin{gathered}
\bar{U}=\frac{55}{111}=0.4955 ; \overline{U^{2}}=\frac{119}{111}=1.0721 \\
\sigma_{U}^{2}=\overline{U^{2}}-(\bar{U})^{2}=1.0721-0.2455=0.8266 \\
\breve{X}=80+20 \cdot 0.4955=89.91 \\
\sigma^{2}=20^{2} \cdot 0.8266=330.64
\end{gathered}
$$

Therefore, $\mu \approx \sqrt{\frac{\sigma^{2}}{n-1}\left(1-\frac{n}{N}\right)}=\sqrt{\frac{330.64}{111-1}\left(1-\frac{111}{2200}\right)}=1.69$.

Since $\Phi(t)=0.96$ when $t=2.05$, the maximum error $\varepsilon=t \mu=2.05 \cdot 1.69=3.46$ and the confidence limits for the population mean will be

$$
\tilde{X}-\varepsilon=89.91-3.46=86.45 \text { and } \tilde{X}+\varepsilon=89.91+3.46=93.35
$$