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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:

μX=2.000\mu_X=2.000

σX=0.002\sigma_X=0.002

μY=2.001\mu_Y=2.001

σY=0.001\sigma_Y=0.001

For the linear combination W=aX1+bX2W=aX_1+bX_2, the mean, variance, and standard deviation are as follows:

μW=aμ1+bμ2\mu_W=a\mu_1+b\mu_2

σW2=a2σ12+b2σ22 (If X1 and X2 are independent)\sigma_W^2=a^2\sigma_1^2+b^2\sigma_2^2\text{ (If $X_1$ and $X_2$ are independent)}

σW=a2σ12+b2σ22 (If X1 and X2 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 YXY-X are then:

μYX=μYμX=2.0012.000=0.001\mu_{Y-X}=\mu_Y-\mu_X=2.001-2.000=0.001

σYX=12σY2+(1)2σX2=σX2+σY2=0.0012+0.0022=0.0000050.002236\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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