Solution. If the probability that a part made by the $i$-th on a given machine will be of the highest grade is denoted by $p_{l}$, then

$$\begin{gathered} p_{1}=0.9 ; p_{2}=0.9-0.9 \cdot 0.01=0.9 \cdot 0.99 ; p_{3}=0.9 \cdot 0.99- \\ -0.9 \cdot 0.99 \cdot 0.01=0.9 \cdot(0.99)^{2} ; \ldots ; p_{i}=0.9 \cdot(0.99)^{i-1} ; \ldots ; p_{100}= \\ =0.9 \cdot(0.99)^{99} ; \\ \bar{p}=\frac{1}{n} \sum_{i=1}^{n} p_{i}=\frac{1}{100} \sum_{i=1}^{100} 0.9 \cdot(0.99)^{i-1}=\frac{0.9}{100} \cdot \frac{1-\left(0.999^{100}\right.}{1-0.99}=\frac{0.9}{100} \times \\ \times \frac{1-0.3665}{0.01}=0.570 ; \\ \sum_{i=1}^{n} p_{i}\left(1-p_{i}\right)=\sum_{i=1}^{100} 0.9 \cdot(0.99)^{i-1} \cdot\left[1-0.9 \cdot(0.99)^{i-1}\right]= \\ =\sum_{i=1}^{n} 0.9 \cdot(0.99)^{i-1}-\sum_{i=1}^{100}(0.9)^{2}(0.99)^{2(i-1)}=0.9 \cdot \frac{1-(0.99)^{100}}{1-0.99}- \\ -(0.9)^{2} \cdot \frac{1-(0.99)^{200}}{1-(0.99)^{2}}=21.39 \end{gathered}$$

Therefore, in our case, the inequality of the Poisson theorem takes the form

$$P\left(\left|\frac{m}{n}-0.570\right|<\varepsilon\right) \geqslant 1-\frac{21.39}{\varepsilon^{2} \cdot 100^{2}}$$

If we choose $\varepsilon$ such that $1-\frac{21.39}{\varepsilon^{2} \cdot 100^{2}}=0.8$, then the probability $P\left(\left|\frac{m}{n}-0.570\right|<\varepsilon\right)$ will be no less than 0.8. From the equation above, we find:

$$0.2=\frac{21.39}{100^{2} \cdot \varepsilon^{2}} ; \quad \varepsilon^{2}=\frac{21.39}{0.2 \cdot 100^{2}}=0.0107 ; \varepsilon=0.103$$

Thus, with a probability of no less than 0.8, we can assert that

$$\left|\frac{m}{n}-0.57\right|<0.103 ;|m-57|<10.3 ; 46.7<m<67.3$$

i.e., the number of parts of the highest grade is between 47 and 67.