- #1
faust9
- 692
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Here's the question:
so, I said the mean (X) of delta is 0.0015 and the standard deviation (S) of delta is 0.000092
[tex]X_d=0.0015, S_d=0.000092[/tex]
[tex]X_l=2.000, S_l=0.0081[/tex]
I said [tex]Z=d/l\ thus\ X_z=X_d/X_l[/tex] and the [tex]S_z^2=(C_d^2+C_l^2)/X_z^2[/tex]
So, I did the following: [tex]X_e=X_d/X_l=0.0015/2.000=0.00075[/tex]
and [tex]S_z=\sqrt{((0.000092/0.0015)^2+(0.0081/2.000)^2)/0.00075^2}=81.956[/tex]
The last part doesn't seem right though. Any thoughts?
Thanks.
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
[tex]X_d=0.0015, S_d=0.000092[/tex]
[tex]X_l=2.000, S_l=0.0081[/tex]
I said [tex]Z=d/l\ thus\ X_z=X_d/X_l[/tex] and the [tex]S_z^2=(C_d^2+C_l^2)/X_z^2[/tex]
So, I did the following: [tex]X_e=X_d/X_l=0.0015/2.000=0.00075[/tex]
and [tex]S_z=\sqrt{((0.000092/0.0015)^2+(0.0081/2.000)^2)/0.00075^2}=81.956[/tex]
The last part doesn't seem right though. Any thoughts?
Thanks.