1692. A power station serves a network with 18000 lamps, the probability of each being turned on in the winter evening is 0.9. What is the probability that the number of lamps turned on in the network in the winter evening differs from its expected value in absolute value by no more than 200?
Solution. Since the number of turned-on lamps $X$ is a random variable distributed according to the binomial law, the mathematical expectation of this random variable is equal to the product of the number of all independent trials ($n=18000$) by the probability of the event occurring in a single trial $(p=0.9)$, i.e., $E X=n p=18000 \cdot 0.9=16200$. The variance of $X$ is found using the formula $D X=n p q=18000 \cdot 0.9 \cdot 0.1=1620(q=1-p)$.

Chebyshev's inequality provides a lower bound for the probability that the inequality $|X-E X| \leqslant 200$ holds, i.e.,

$$
P(|X-E X| \leqslant 200) \geqslant 1-\frac{D X}{200^{2}}=1-\frac{1620}{40000}=1-0.045=0.955
$$

The exact value of this probability is found using the integral theorem of Laplace:

$$
P(|X-E X| \leqslant 200)=\Phi\left(\frac{200}{\sqrt{1620}}\right)=\Phi(4.97)=1
$$

(here $\Phi(t)$ is the Laplace function).