The solution is as follows. Given $n=900, p=0.9, q=0.1$. Therefore,

$$
2 \Phi(\varepsilon \sqrt{900 /(0.9 \cdot 0.1)})=0.95, \text { or } \Phi(100 \varepsilon)=0.475
$$

From the table in Appendix 2, we find $\Phi(1.96)=0.475$, so $100 \varepsilon=1.96$. Hence, $\varepsilon \approx 0.02$.

Thus, with a probability of 0.95, the deviation of the relative frequency of the number of standard parts from the probability of 0.9 satisfies the inequality

$$
|m / 900 - 0.9| \leqslant 0.02, \quad \text { or } \quad 0.88 \leqslant m / 900 < 0.92 \text {. }
$$

From this, the number $m$ of standard parts among 900 checked, with a probability of 0.95, is within the following bounds: $792 \leqslant m \leqslant 828$.
142. The quality control department checks 900 parts for standardization. The probability that a part is standard is 0.9. Find, with a probability of 0.95, the boundaries within which the number $m$ of standard parts among those checked will be contained.