Mean, and standard deviation question.

[faust9]

Here's the question:

In the expression for uniaxial strain e = d/l, the elongation is specified as
d ~ N(0.0015, 0.000092) in. and the length l ~ N(2.0000, 0.0081) in.
What are the mean, standard deviation, and coefficient of variation of the corresponding strain e?

so, I said the mean (X) of delta is 0.0015 and the standard deviation (S) of delta is 0.000092

$$X_d=0.0015, S_d=0.000092$$
$$X_l=2.000, S_l=0.0081$$

I said $$Z=d/l\ thus\ X_z=X_d/X_l$$ and the $$S_z^2=(C_d^2+C_l^2)/X_z^2$$


So, I did the following: $$X_e=X_d/X_l=0.0015/2.000=0.00075$$

and $$S_z=\sqrt{((0.000092/0.0015)^2+(0.0081/2.000)^2)/0.00075^2}=81.956$$

The last part doesn't seem right though. Any thoughts?

Thanks.

 

[ehild]

Here's the question:

In the expression for uniaxial strain e = d/l, the elongation is specified as
d ~ N(0.0015, 0.000092) in. and the length l ~ N(2.0000, 0.0081) in.
What are the mean, standard deviation, and coefficient of variation of the corresponding strain e?


$$X_d=0.0015, S_d=0.000092$$
$$X_l=2.000, S_l=0.0081$$

I said $$Z=d/l\ thus\ X_z=X_d/X_l$$ and the $$S_z^2=(C_d^2+C_l^2)/X_z^2$$

If you denoted the relative errors by C

$$S_z^2=(C_d^2+C_l^2)*X_z^2$$ instead of what you used.

 

[wais]

Your calculation for the mean and standard deviation of the strain e looks correct. However, the coefficient of variation (CV) is usually calculated as the ratio of the standard deviation to the mean, so in this case it would be 0.000092/0.0015 = 0.0613 or 6.13%. The value you calculated for the standard deviation of e (81.956) is quite large compared to the mean (0.00075), which indicates that the data may not be normally distributed. It might be worth checking the data and making sure it follows a normal distribution before using these calculations.