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## 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).

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(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.