At the factory named after Sailor 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 error of 2 mm.

a) Find the expected value of the area of the resulting rectangle.

b) Find the standard deviation of the area of the resulting rectangle in square centimeters.
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,

$\mathrm{E}(X Y)=\mathrm{E} X \cdot \mathrm{E} Y=2\left(\mathrm{M}^{2}\right)$.

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

From the equality $\mathrm{D} X=\mathrm{E} X^{2}-\mathrm{E}^{2} X$, we express $\mathrm{E} X^{2}$ and substitute it into the obtained expression. Similarly, we handle $\mathrm{E} Y^{2}$: $\mathrm{D}(X Y)=\left(\mathrm{D} X+\mathrm{E}^{2} X\right)\left(\mathrm{D} Y+\mathrm{E}^{2} Y\right)-4=\left(4 \cdot 10^{-6}+1\right)\left(9 \cdot 10^{-6}+4\right)-4=36 \cdot 10^{-12}+25 \cdot 10^{-6}$.

Compared to the second term, the first term is so small that it can be neglected. We get:

$\sqrt{\mathrm{D}(X Y)} \approx \sqrt{25 \cdot 10^{-6}}=0.005\left(\mathrm{M}^{2}\right)$, which gives 50 cm $^{2}$.

## Answer

a) $2 \mathrm{M}^{2} ;$ b) $\approx 50 \mathrm{CM}^{2}$.

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