We are solving a probability problem based on the given discrete random variable $X$ with its probability distribution. The task is to calculate:

1. $M(X-3)$ (expected value of $X-3$),
2. $D(X-3)$ (variance of $X-3$),
3. $P(4 \leq X < 9)$ (probability that $X$ falls within the interval $[4, 9)$),
4. Draw the probability polygon for the distribution.

Given Data:

Step 1: Calculate $M(X-3)$

The expected value of $X-3$ is given by:

$M(X-3) = M(X) - 3$

First, calculate $M(X)$ (the expected value of $X$) using the formula:

$M(X) = \sum (X_i \cdot p_i)$

Substitute values from the table:

$M(X) = (1 \cdot 0.1) + (4 \cdot 0.3) + (8 \cdot 0.3) + (9 \cdot 0.2) + (11 \cdot 0.1)$

$M(X) = 0.1 + 1.2 + 2.4 + 1.8 + 1.1 = 6.6$

Now, subtract 3:

$M(X-3) = 6.6 - 3 = 3.6$

Step 2: Calculate $D(X-3)$

The variance of $X-3$ is equal to the variance of $X$, since subtracting a constant does not affect variance:

$D(X-3) = D(X)$

The variance $D(X)$ is calculated as:

$D(X) = M(X^2) - \left(M(X)\right)^2$

Step 2.1: Calculate $M(X^2)$

$M(X^2) = \sum (X_i^2 \cdot p_i)$

Substitute values from the table:

$M(X^2) = (1^2 \cdot 0.1) + (4^2 \cdot 0.3) + (8^2 \cdot 0.3) + (9^2 \cdot 0.2) + (11^2 \cdot 0.1)$

$M(X^2) = (1 \cdot 0.1) + (16 \cdot 0.3) + (64 \cdot 0.3) + (81 \cdot 0.2) + (121 \cdot 0.1)$

$M(X^2) = 0.1 + 4.8 + 19.2 + 16.2 + 12.1 = 52.4$

Step 2.2: Calculate $D(X)$

$D(X) = 52.4 - (6.6)^2 = 52.4 - 43.56 = 8.84$

Thus:

$D(X-3) = 8.84$

Step 3: Calculate $P(4 \leq X < 9)$

We sum the probabilities for $X$ values within the range $[4, 9)$, which includes $X = 4$ and $X = 8$:

$P(4 \leq X < 9) = P(X=4) + P(X=8)$

Substitute the probabilities from the table:

$P(4 \leq X < 9) = 0.3 + 0.3 = 0.6$

Step 4: Draw the Probability Polygon

The polygon represents the probability distribution. Each $X$ value is plotted on the x-axis, and the corresponding probabilities ($p$) on the y-axis:

Points to plot:

• $(1, 0.1)$
• $(4, 0.3)$
• $(8, 0.3)$
• $(9, 0.2)$
• $(11, 0.1)$

The polygon is formed by connecting these points linearly.

Let me know if you'd like me to generate the graph for you.

Final Results:

1. $M(X-3) = 3.6$
2. $D(X-3) = 8.84$
3. $P(4 \leq X < 9) = 0.6$

Do you want the distribution polygon or further explanations?
Here are five related questions to deepen your understanding:

1. How does subtracting a constant affect the mean and variance of a random variable?
2. How would $D(aX+b)$ (variance after scaling and shifting) differ from $D(X)$?
3. What is the difference between a probability polygon and a probability histogram?
4. How would adding an additional $X$-value affect the mean and variance?
5. Can you explain why the variance is unaffected by subtracting a constant?

Tip: Remember that the expected value is a weighted average of possible outcomes, and variance measures the spread of these outcomes around the mean.