89. As a result of processing data from long-term observations, distributions of random variables $X$ and $Y$ have been obtained - the number of households in each of two districts of the region where the yield of spring cereals can exceed 35 centners/ha.

For the first district of the region:

$$
X: \begin{array}{cccc}
x_{i} & 1 & 2 & 3 \\
p\left(x_{i}\right) & 0.1 & 0.6 & 0.3
\end{array}
$$

For the second district of the region:

$$
Y: \begin{array}{ccc}
y_{j} & 0 & 1 \\
p\left(y_{j}\right) & 0.2 & 0.8
\end{array}
$$

Find the mathematical expectation $M(Z)$ and variance $D(Z)$ of the random variable $Z=X+Y$ in two ways:

a) based on the distribution law of $Z$;

b) using the properties of mathematical expectation and variance reflected in formulas (33) and (42).

Verify that in the conditions of this problem, these properties of numerical characteristics of independent random variables are satisfied.
Solution. a) Let's find all possible values of the random variable $Z=X+Y$ and the probabilities $P\left(Z=z_{k}\right)$ of these values, using reasoning similar to that presented in problems 79 and 80. As a result, we will obtain the following distribution of the variable $Z$:

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-114.jpg?height=113&width=508&top_left_y=321&top_left_x=417)

The expected value and variance of $Z$ will be found using formulas (30) and (38):

$$
\begin{aligned}
& M(Z)=1 \cdot 0.02 + 2 \cdot 0.20 + 3 \cdot 0.54 + 4 \cdot 0.24 = 3 \\
& D(Z)=1^{2} \cdot 0.02 + 2^{2} \cdot 0.20 + 3^{2} \cdot 0.54 + 4^{2} \cdot 0.24 - 3^{2} = 0.52
\end{aligned}
$$

b) Let's find the expected value and variance of the random variables $X$ and $Y$:

$$
\begin{aligned}
& M(X)=1 \cdot 0.1 + 2 \cdot 0.6 + 3 \cdot 0.3 = 2.2 \\
& D(X)=1^{2} \cdot 0.1 + 2^{2} \cdot 0.6 + 3^{2} \cdot 0.3 - 2.2^{2} = 0.36 \\
& M(Y)=0 \cdot 0.2 + 1 \cdot 0.8 = 0.8 \\
& D(Y)=0.8 - 0.8^{2} = 0.16
\end{aligned}
$$

Using the properties of the expected value and variance reflected in formulas (33) and (42), we get:

$$
\begin{aligned}
& M(Z)=M(X+Y)=M(X)+M(Y)=2.2+0.8=3 \\
& D(Z)=D(X+Y)=D(X)+D(Y)=0.36+0.16=0.52
\end{aligned}
$$

By comparing the values of $M(Z)$ and $D(Z)$ obtained in part a) with the corresponding values obtained in part b), we confirm that the same numerical characteristics of $Z$, found by two different methods, are equal to each other.