9. Steel doors (from 9th grade. 2 points). At the factory named after Seaman Zheleznyak, they manufacture rectangles with a length of 2 m and a width of 1 m. The length is measured by worker Ivanov, and the width, independently of Ivanov, is measured by worker Petrov. Both have an average error of zero, but Ivanov has a standard measurement error (standard deviation of length) of 3 mm, and Petrov has a standard measurement error of 2 mm.

a) (from 9th grade. 1 point). Find the expected value of the area of the resulting rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_164205c0ca2288d5a185g-13.jpg?height=582&width=272&top_left_y=240&top_left_x=1623)
of the resulting rectangle.

b) (from 9th grade. 2 points). Find the standard deviation of the area of the resulting rectangle in square centimeters.
# Solution.

a) Let $X$ be the width and $Y$ be the length of the cut rectangle in meters. According to the problem, $\mathrm{E} X=2$, $\mathrm{E} Y=1$. Since the measurements are independent, $\mathrm{E}(X Y)=\mathrm{E} X \cdot \mathrm{E} Y=2$ (sq.m.).

b) From the condition, it follows that $\mathrm{D} X=0.003^{2}=9 \cdot 10^{-6}$ and $\mathrm{D} Y=4 \cdot 10^{-6}$.

$$
\mathrm{D}(X Y)=\mathrm{E}(X Y)^{2}-\mathrm{E}^{2}(X Y)=\mathrm{E} X^{2} \cdot \mathrm{E} Y^{2}-4
$$

Using the equality $\mathrm{D} X=\mathrm{E} X^{2}-\mathrm{E}^{2} X$, we can express $\mathrm{E} X^{2}$ and substitute it into the obtained expression. We will do the same for $\mathrm{E} Y^{2}$:

$$
\mathrm{D}(X Y)=\left(\mathrm{D} X+\mathrm{E}^{2} X\right)\left(\mathrm{D} Y+\mathrm{E}^{2} Y\right)-4=\left(4 \cdot 10^{-6}+1\right)\left(9 \cdot 10^{-6}+4\right)-4=36 \cdot 10^{-12}+25 \cdot 10^{-6}
$$

Compared to the second term, the first term is so small that it can be neglected. We get: $\sqrt{\mathrm{D}(X Y)} \approx \sqrt{25 \cdot 10^{-6}}=0.005$ (sq.m), which gives 50 sq.cm.

Answer: a) 2 sq.m; b) approx. 50 sq.cm.