101. One of the most important characteristics of human blood is the Rh factor. The "Rh positive" gene is dominant over the "Rh negative" gene.

In one of the surveyed populations, the probability that a person has a positive Rh factor is 0.91. For the random variable $X$ - the number of Rh-negative people among 1000 people in this population, find:

a) the expected value, variance, and standard deviation;

b) the probability that the absolute value of the deviation of the number of Rh-negative people from its expected value does not exceed a given positive number $\delta$;

c) the interval, symmetric about the expected value, in which the possible values of $X$ will be contained with a probability of 0.9722.
The random variable $X$ under consideration - the number of Rh-negative individuals among 1000 people in the examined population, is distributed according to the binomial law.

The total number of trials $n=1000$. In each trial, event $A$ - a person is Rh-negative, may or may not occur. The opposite event $\bar{A}$ - a person is Rh-positive. According to the condition $P(\bar{A})=q=0.91$, therefore $p=1-q=0.09$.

a) The mathematical expectation, variance, and standard deviation of the random variable $X$ will be found respectively by formulas (44), (45), and (46):

$$
\begin{aligned}
& M(X)=n p=1000 \cdot 0.09=90 \\
& D(X)=n p q=1000 \cdot 0.09 \cdot 0.91=81.9 \\
& \sigma(X)=\sqrt{n p q}=\sqrt{81.9} \approx 9.05
\end{aligned}
$$

b) The possible values of the random variable $X$ are $x_{i}=m$, where $m$ is the number of occurrences of event $A$ in $n$ independent trials. The mathematical expectation of the variable $X$ is $M(X)=n p$. The required probability is

$$
P(|m-n p| \leq \delta)=P(n p-\delta \leq m \leq n p+\delta)
$$

According to the integral formula of Laplace (22) for $m_{1}=n p-\delta$, $m_{2}=n p+\delta$, we find

$$
\begin{aligned}
& P(n p-\delta \leq m \leq n p+\delta)=P\left(m_{1} \leq m \leq m_{2}\right) \approx \\
& \approx \Phi\left(\frac{n p+\delta-n p}{\sqrt{n p q}}\right)-\Phi\left(\frac{n p-\delta-n p}{\sqrt{n p q}}\right)=2 \Phi\left(\frac{\delta}{\sqrt{n p q}}\right)
\end{aligned}
$$

Thus, we obtain

$$
P(|m-n p| \leq \delta) \approx 2 \Phi\left(\frac{\delta}{\sqrt{n p q}}\right)
$$

c) The boundaries of the required interval are determined by an integer $\delta>0$. Let's find this number.

According to the condition $P(|m-n p| \leq \delta)=0.9722$. According to formula (*), we have

$$
2 \Phi\left(\frac{\delta}{\sqrt{n p q}}\right) \approx 0.9722, \quad \Phi\left(\frac{\delta}{\sqrt{n p q}}\right) \approx 0.4861
$$

According to the table of values of the Laplace function (see table 3 in the appendix), we find

$\frac{\delta}{\sqrt{n p q}}=2.20$. Therefore, $\delta=2.2 \sqrt{n p q} \approx 2.2 \cdot 9.05=19.91$. Since the number $\delta$ must be an integer, then $\delta=20$.

Thus, $70 \leq m \leq 110$, i.e., $x_{i} \in [70 ; 110]$.

## 4.4. Poisson's Law

Suppose a discrete random variable $X$ can take only non-negative integer values: $x_{i}=0,1,2, \ldots, m, \ldots$, and this sequence of values is theoretically unlimited.

If the probability that $X$ will take a certain value $m$ is expressed by the formula

$$
P(X=m)=P_{m}=\frac{\lambda^{m}}{m!} e^{-\lambda}, \quad(m=0,1,2, \ldots)
$$

where $\lambda>0$ is the parameter of the distribution, then this random variable $X$ is distributed according to Poisson's law.

A random variable $X$ distributed according to Poisson's law is characterized by the following distribution series (Table 4):

Table 4

| $x_{i}$ | 0 | 1 | 2 | $\cdots$ | $\boldsymbol{m}$ | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p_{i}$ | $e^{-\lambda}$ | $\frac{\lambda}{1!} \cdot e^{-\lambda}$ | $\frac{\lambda^{2}}{2!} \cdot e^{-\lambda}$ | $\cdots$ | $\frac{\lambda^{m}}{m!} \cdot e^{-\lambda}$ | $\cdots$ |

The mathematical expectation $M(X)$ and variance $D(X)$ of a random variable $X$ distributed according to Poisson's law are equal to

$$
M(X)=D(X)=\lambda
$$