Example 6. Production gives $1 \%$ of defects. What is the probability that out of 1100 items taken for research, no more than $17$ will be defective?
Solution. From the condition, it follows that $n=1100, p=0.01, q=0.99$, $0 \leq k \leq 17$. Using formulas (4.3.6), we determine

$$
\begin{gathered}
x_{1}=\frac{0-1100 \cdot 0.01}{\sqrt{1100 \cdot 0.01 \cdot 0.99}}=\frac{-11}{\sqrt{10.89}}=-\frac{11}{3.3} \approx-3.33 \\
x_{2}=\frac{17-11}{3.3}=\frac{6}{3.3} \approx 1.82
\end{gathered}
$$

Using the table of values of the Laplace function (see appendix), we find

$$
\begin{gathered}
\Phi\left(x_{1}\right)=\Phi(-3.33)=-\Phi(3.33) \approx-0.4995 \\
\Phi\left(x_{2}\right)=\Phi(1.82) \approx 0.4656
\end{gathered}
$$

In accordance with formula (4.3.7), we obtain

$$
P_{1100}(0 ; 17)=0.4656-(-0.4995)=0.9651
$$