Determine the standard deviation of these results

In summary: Should it be this then ? σ= √((805.02)=29.09 N mm^(-2)= 0.02909 MpaConverting to Gpa yields: - 2.909*10^-5 GpaYes, your latest answer is correct. It is just a matter of how you estimate the mean. See my last post for details.In summary, the mean standard deviation for the ultimate tensile strength of a material tested using 10 samples is 0.02837 Gpa, with a standard deviation of 2.837*10^-5 Gpa. This is calculated using the formula for sample standard deviation and dividing by N-1 instead of N, as the mean is
  • #1
Tiberious
73
3
I've managed to work out the below for the sample standard deviation. Any error's that I should be aware of?

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.
 
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  • #2
##\sigma=40 ## N/mm^2=40 MPa=.040 GPa if my arithmetic is correct. ## \\ ## And you don't need to keep all the extra sig figs on the standard deviation. ## \\ ## And I think they want the mean and standard deviation. You didn't give the value of the mean in GPa.
 
  • #3
Tiberious said:
Determine the mean standard deviation of these results.
There's a mean and there's a standard deviation. I'm not aware of a term called mean standard deviation, but there is a standard error, where you look at the variation of a collection of samples.
 
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  • #4
There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
 
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  • #5
FactChecker said:
There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
 
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  • #6
Charles Link said:
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
They are all very standard things to ask for. But I agree that the sample mean and sample standard deviation are the first two things to calculate.
 
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  • #7
Yes - it does seem my OP was missing the word 'and' apologies for the confusion. Just to confirm;

The answer for standard deviation is 0.040 Gpa ?

So, would the mean be 0.016Gpa ?
 
  • #8
In your original post, why are you dividing by 5 instead of 9 for the sample variance?

PS. small error: (701-726.4 )^2=645.16, not 625.16.
 
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  • #9
So, amending (701-726.4 )^2 from 625.16 to 645.16 and amending the following would lead to 805.02N mm^-2

Amendments being: -

(701-726.4 )^2 = 645.16

(1/10)∙8050.29= 805.02 N mm^-2

This leads to: -

σ= √((805.02)=28.37 N mm^(-2)= 0.02837 Mpa

Converting to Gpa yields: -

2.837*10^-5 Gpa

Looking any better ?
 
  • #10
Tiberious said:
(1/10)∙8050.29= 805.02 N mm^-2
Why are you dividing by 10 instead of N-1=10-1=9?

If you know the true mean, ##\mu##, you should use that in the formula for the sample standard deviation and divide by N. But if you estimate the mean with ##\bar x ##, your results of ##\sum_{i=1}^{i=N} (x_i-\bar x)^2## will be smaller and you should divide by ##N-1##.
 
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  • #11
Ah! thanks, brief moment of confusion. I assume this would amend the overall error ?
 

1. What is the standard deviation and why is it important?

The standard deviation is a measure of how spread out a set of data is from its mean. It is important because it allows us to understand how much variability there is in the data, which can help us make comparisons between different sets of data. It also helps us identify outliers and understand the overall distribution of the data.

2. How do you calculate the standard deviation?

The standard deviation is calculated by taking the square root of the variance. The variance is calculated by finding the average of the squared differences between each data point and the mean. This value is then used to determine the standard deviation, which is a more easily interpretable measure of spread.

3. Can the standard deviation ever be negative?

No, the standard deviation cannot be negative. It is always a positive value, as it represents the distance from the mean. However, a value of 0 for the standard deviation indicates that there is no variability in the data and all values are the same.

4. How does the standard deviation relate to the normal distribution?

The standard deviation is used to describe the distribution of data, and it plays an important role in the normal distribution. In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

5. How is the standard deviation affected by outliers?

Outliers can greatly affect the standard deviation. If there are extreme values in the data, the standard deviation will be larger, as it takes into account the distance of each data point from the mean. Removing outliers can decrease the standard deviation and provide a more accurate representation of the spread of the data.

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