Task 1. Calculate the expected value of a random variable $X$, distributed according to the binomial law with parameters $p=0.6$ and $n=5$.
Solution. Let's form the distribution series of $X$:

$$
p(X=k)=p_{5}(k)=C_{5}^{k} \cdot(0.6)^{k} \cdot(0.4)^{5-k}, \quad k=0,1,2,3,4,5
$$

We will place the values of $x_{k}$ and $p_{k}$ in a table (p. 305), to which we will add another row for the products $x_{k} \cdot p_{k}$. The sum of the numbers in this row constitutes $M(X)$.

| $x$ | 0 | 1 | 2 | 3 | 4 | 5 | $\Sigma$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p$ | 0.01024 | 0.0768 | 0.2304 | 0.3456 | 0.2592 | 0.07776 | 1.0 |
| $p x$ | 0 | 0.0768 | 0.4608 | 1.0368 | 1.0368 | 0.3888 | 3.0 |

Note that $M(X)=3.0=5 \cdot 0.6=p n$.

Answer. 3.0

Problem 2 Calculate the variance and standard deviation of the random variable $X$, distributed according to the Bernoulli law with parameters $p=0.6$ and $n=5$.

Solution. The distribution series was obtained in Problem 1. We will use the formula $D(X)=M\left(X^{2}\right)-[M(X)]^{2}$. To calculate $M\left(X^{2}\right)$, we will form a row for $p x^{2}$, i.e., multiply the third row of the table from Problem 1 element-wise by the row $X$.

We have $p x^{2}: 0 ; 0.0768 ; 0.9216 ; 3.1104 ; 4.1472 ; 1.944$

The sum of all numbers in this row is 10.2, i.e., $M\left(X^{2}\right)=10.2$. Then $D(X)=10.2-3^{2}=1.2 ; \sigma(X)=\sqrt{1.2}=1.1$.

Note that $D(X)=1.2=5 \cdot 0.6 \cdot 0.4=n p q$.

Problem 3. The random variable $X$ is given by the distribution law

$$
\begin{array}{|l|c|c|c|c|c|}
\hline x & 0.3 & 0.7 & 1.2 & 2.1 & 3.5 \\
\hline p & 0.15 & 0.25 & 0.3 & 0.2 & 0.1 \\
\hline
\end{array}
$$

The distribution polygon is shown in Fig. 7.9. Calculate the first four initial and central moments of the variable $X$ and provide the corresponding interpretation.

Solution. It is convenient to perform the calculations in the columns of the table (p. 306). The sums of the numbers in the corresponding columns are indicated in the bottom row.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-154.jpg?height=369&width=503&top_left_y=1324&top_left_x=1956)

We obtained (rounding to two decimal places): $M_{1}=1.35$; $M_{2}=2.68 ; M_{3}=6.75 ; M_{4}=19.58$. Using the formulas of point $6^{\circ}$, we will calculate the central moments: $m_{1}=0 ; m_{2}=0.9, m_{3}=0.81 ; m_{4}=2.44$. We have

| $x$ | $p$ | $p x$ | $p x^{2}$ | $p x^{3}$ | $p x^{4}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0.3 | 0.15 | 0.045 | 0.0135 | 0.00405 | 0.001215 |
| 0.7 | 0.25 | 0.175 | 0.1225 | 0.08575 | 0.060025 |
| 1.2 | 0.3 | 0.36 | 0.432 | 0.5184 | 0.62208 |
| 2.1 | 0.2 | 0.42 | 0.882 | 1.8522 | 3.88962 |
| 3.5 | 0.1 | 1.35 | 1.225 | 4.2875 | 15.00625 |
| $\Sigma$ | 1.0 | 1.350 | 2.675 | 6.7479 | 19.57919 |

We also have $\sigma=\sqrt{D}=0.9 ; A s=0.81: 0.95^{3}=0.94 ; E k=2.44: 0.95^{4}-3=$ $=-0.004$.

Answer. The expected value $M(X)=1.35$. The variance $D(X)=0.9$; the standard deviation, i.e., the average deviation $\sigma=0.95$; the skewness $A s=0.94$, indicating a right-skewed distribution graph; the kurtosis $E k(X)=-0.004$, indicating a slightly flat-topped distribution.