- #1
Tiberious
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I've managed to work out the below for the sample standard deviation. Any error's that I should be aware of?
The ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows.
711 N 〖mm〗^(-2),732 N 〖mm〗^(-2),759 N 〖mm〗^(-2),670 N 〖mm〗^(-2),701 N 〖mm〗^(-2),
765 N 〖mm〗^(-2),743 N 〖mm〗^(-2),755 N 〖mm〗^(-2),715 N 〖mm〗^(-2),713 N 〖mm〗^(-2).
[/B]
(a) Determine the mean standard deviation of these results.
(a) Express the values found in (a) in GPa.
s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )
(a) As we are referring to a 'sample' we apply the below formula for Standard Deviation:
s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )
Calculating the mean value x ̅
(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10
x ̅= 726.4 〖N mm〗^(-2)
Determining the (x_i-x ̅)^2
(711-726.4 )^2= 237.16
(732-726.4 )^2= 31.36
(759-726.4 )^2= 1062.76
(670-726.4 )^2= 3180.95
(701-726.4 )^2= 625.16
(765-726.4 )^2= 1489.96
(743-726.4 )^2= 275.56
(755-726.4 )^2= 817.96
(715-726.4 )^2= 129.96
(713-726.4 )^2= 179.56
Determining the sum of the above
∑▒237.16+31.36 +1062.76 +3180.95 +625.16 +1489.96 +275.56 +817.96 +129.96 +179.56
= 8,030.29 〖N mm〗^(-2)
Divide by N-1
(1/5)∙8030.29= 1606.06 N 〖mm〗^(-2)Determining the sample standard deviation σ
σ= √((1606.06)=40.08 N 〖mm〗^(-2)=4.008 MPa
(b)4.008 MPa=0.004008 GPa
Or in standard form, 4.008∙10^(-3) GPa.
Homework Statement
The ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows.
711 N 〖mm〗^(-2),732 N 〖mm〗^(-2),759 N 〖mm〗^(-2),670 N 〖mm〗^(-2),701 N 〖mm〗^(-2),
765 N 〖mm〗^(-2),743 N 〖mm〗^(-2),755 N 〖mm〗^(-2),715 N 〖mm〗^(-2),713 N 〖mm〗^(-2).
[/B]
(a) Determine the mean standard deviation of these results.
(a) Express the values found in (a) in GPa.
Homework Equations
s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )
The Attempt at a Solution
(a) As we are referring to a 'sample' we apply the below formula for Standard Deviation:
s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )
Calculating the mean value x ̅
(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10
x ̅= 726.4 〖N mm〗^(-2)
Determining the (x_i-x ̅)^2
(711-726.4 )^2= 237.16
(732-726.4 )^2= 31.36
(759-726.4 )^2= 1062.76
(670-726.4 )^2= 3180.95
(701-726.4 )^2= 625.16
(765-726.4 )^2= 1489.96
(743-726.4 )^2= 275.56
(755-726.4 )^2= 817.96
(715-726.4 )^2= 129.96
(713-726.4 )^2= 179.56
Determining the sum of the above
∑▒237.16+31.36 +1062.76 +3180.95 +625.16 +1489.96 +275.56 +817.96 +129.96 +179.56
= 8,030.29 〖N mm〗^(-2)
Divide by N-1
(1/5)∙8030.29= 1606.06 N 〖mm〗^(-2)Determining the sample standard deviation σ
σ= √((1606.06)=40.08 N 〖mm〗^(-2)=4.008 MPa
(b)4.008 MPa=0.004008 GPa
Or in standard form, 4.008∙10^(-3) GPa.