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Question

# What are the mean and standard deviation of the difference Y - X between the readings? (The readings X and Y are independent.)

Solution

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Given:

$\mu_X=2.000$

$\sigma_X=0.002$

$\mu_Y=2.001$

$\sigma_Y=0.001$

For the linear combination $W=aX_1+bX_2$, the mean, variance, and standard deviation are as follows:

$\mu_W=a\mu_1+b\mu_2$

$\sigma_W^2=a^2\sigma_1^2+b^2\sigma_2^2\text{ (If X_1 and X_2 are independent)}$

$\sigma_W=\sqrt{a^2\sigma_1^2+b^2\sigma_2^2}\text{ (If X_1 and X_2 are independent)}$

The mean and standard deviation of the difference $Y-X$ are then:

$\mu_{Y-X}=\mu_Y-\mu_X=2.001-2.000=0.001$

$\sigma_{Y-X}=\sqrt{1^2\sigma_Y^2+(-1)^2\sigma_X^2}=\sqrt{\sigma_X^2+\sigma_Y^2}=\sqrt{0.001^2+0.002^2}=\sqrt{0.000005}\approx 0.002236$

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