76. When crossing two varieties of peas (with yellow and green seeds), G. Mendel found that in the second generation, the probability of green seeds appearing is 0.25. How many pea seeds need to be taken to expect with a probability of 0.9770 that the relative frequency of green seeds will deviate (in absolute value) from their probability of occurrence by no more than 0.02?
Solution. According to the condition $p=0.25 ; \varepsilon=0.02 ;$ we find $q=0.75$. It is known that $P\left(\left|\frac{m}{n}-0.25\right| \leq 0.02\right)=0.997$. We need to find the number of seeds $n$.

In accordance with formula (23) we have

$$
P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right) \approx 2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)=0.997
$$

From this equality, we find $\Phi(x)=0.4985$, where $x=\varepsilon \sqrt{\frac{n}{p q}}$.

Using the table of values of the Laplace function (see table 3 in the appendix), we find $x=2.96$. Since

$$
\varepsilon \sqrt{\frac{n}{p q}}=0.02 \cdot \sqrt{\frac{n}{0.25 \cdot 0.75}}
$$

we obtain the following equality:

$$
0.02 \sqrt{\frac{n}{0.25 \cdot 0.75}}=2.96
$$

From the last equality, we determine $n=4107$. Therefore, we need to take 4107 seeds.

Chapter 4

## RANDOM VARIABLES

## AND THEIR DISTRIBUTION LAWS

I asked the cuckoo, how many years I will live... The tops of the pines trembled, a yellow ray sank into the grass, but there was no sound in the fresh air... I go home, and the cool wind caresses my hot forehead.

Anna Akhmatova

## ***

You don't often appear to me in dreams, my native

Golden my, bittersweet era. HOME,

But everything that will happen to me later, -

It all starts from here!

Alexander Galich

## 4.1. Series, frequency polygon, and distribution function of a discrete random variable

The distribution law of a discrete random variable can be represented by a series of distribution, a frequency polygon, and a distribution function.

A discrete random variable $X$, having a finite set of possible values $x_{i}$ with corresponding probabilities $p_{i}=P\left(X=x_{i}\right)(i=1,2, \ldots, n)$, can be defined by a series of distribution of the following form (Table 1):

Table 1

| $x_{i}$ | $x_{1}$ | $x_{2}$ | $\cdots$ | $x_{n}$ |
| :---: | :---: | :---: | :---: | :---: |
| $p_{i}$ | $p_{1}$ | $p_{2}$ | $\cdots$ | $p_{n}$ |

For the probabilities $p_{i}$ given in Table 1, the condition must be satisfied:

$$
\sum_{i=1}^{n} p_{i}=1
$$

A discrete random variable $X$, having an infinite (countable) set of possible values $x_{i}$ with corresponding probabilities $p_{i}=P\left(X=x_{i}\right)(i=1,2, \ldots, n, \ldots)$, can be defined by a series of distribution of the following form (Table 2):

Table 2

| $x_{i}$ | $x_{1}$ | $x_{2}$ | $\cdots$ | $x_{n}$ | $\cdots$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $p_{i}$ | $p_{1}$ | $p_{2}$ | $\cdots$ | $p_{n}$ | $\cdots$ |

For the probabilities $p_{i}$ given in Table 2, the conditions must be satisfied: the series

$$
p_{1}+p_{2}+\ldots+p_{n}+\ldots
$$

converges and

$$
\sum_{i=1}^{\infty} p_{i}=1
$$

Graphically, the distribution of a discrete random variable $X$ can be represented by a frequency polygon. To construct a frequency polygon, points $\left(x_{i}, p_{i}\right)$ are plotted on a plane in the coordinate system $x O y$ and connected by a broken line. Figure 12 shows the frequency polygon of a random variable with $n=6$ possible values.

A discrete random variable $X$ can be defined by the distribution function $F(x)$, also called the integral distribution function:

For a discrete random variable defined by a series of distribution (Tables 1 and 2), the values of the distribution function are found using the formula

$$
F(x)=\sum_{x_{i}<x} p_{i}
$$

where the summation of probabilities $p_{i}$ is performed over all values of $i$ for which $x_{i}<x$.

$$
F(x)=P(X<x)
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-096.jpg?height=410&width=700&top_left_y=418&top_left_x=269)

Fig. 12

The probability that a random variable $X$ will take a value $\check{x} \in[\alpha, \beta)$ is expressed through the distribution function by the formula

$$
P(\alpha \leq X<\beta)=F(\beta)-F(\alpha)
$$