Solution. From the condition, it follows that $n=500, p=0.02$, therefore $q=1-p=0.98, n p=500 \cdot 0.02=10, n p q=500 \cdot 0.02 \cdot 0.98=9.8$.

According to formula (4.3.2), we find

$$
P_{500}(10)=\frac{1}{\sqrt{9.8}} \varphi\left(\frac{10-10}{\sqrt{9.8}}\right)=\frac{1}{\sqrt{9.8}} \varphi(0)=0.319 \cdot 0.399=0.127
$$

In accordance with formula (4.3.7), we obtain

$$
\Phi(10 \leq k \leq 20)=\Phi\left(\frac{20-10}{\sqrt{9.8}}\right)-\Phi\left(\frac{10-10}{\sqrt{9.8}}\right)=\Phi(3.2)-\Phi(0)=0.499
$$
Example 16. 500 items of the product manufactured by the enterprise are inspected, where the defect rate is $2 \%$. Find the probabilities that: 1) among them there will be exactly 10 defective items; 2) the number of defective items is between 10 and 20.