Quick question about finding standard deviation

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SUMMARY

The discussion clarifies the correct formulas for calculating standard deviation in different contexts. The formula \(\sigma=\sqrt{\frac{1}{N}\sum_{i=1}^{N}(x-\overline{x})^2}\) is used when the mean is the true mean, while \(\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x-\overline{x})^2}\) is appropriate for experimental data. Additionally, the coefficient of variation is defined as \(c_v = \frac{\sigma}{\overline{x}}\). Understanding the distinction between these formulas is crucial for accurate statistical analysis in fields such as chemistry.

PREREQUISITES
  • Understanding of basic statistics, including mean and variance
  • Familiarity with standard deviation calculations
  • Knowledge of the difference between population and sample statistics
  • Basic algebra skills for manipulating equations
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  • Research the differences between population variance and sample variance
  • Learn about the implications of using \(N\) versus \(N-1\) in statistical calculations
  • Explore the concept of the coefficient of variation and its applications
  • Study practical examples of standard deviation calculations in experimental data analysis
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[SOLVED] Quick question about finding standard deviation

Now I know that \sigma=\sqrt{var(x)}

which simplifies to this expression : \sigma=\sqrt{\frac{1}{N}\sum_{i=1}^{N}(x-\overline{x})^2} can someone show me how they got such an expression?

and in chemistry I have to use a standard deviation calculation to get out a problem. Now normally I would use the above equation but my notes tell me to use this equation:

\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x-\overline{x})^2}


Which one is correct to use? and can someone tell me if this is correct c_v =\frac{\sigma}{\overline{x}} where c_v is the coefficient of variation
 
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The definition of the st. dev. depends on whether or not the mean is the true mean or the average of the experimental data. For the true mean use 1/N, for the experimental average use 1/(N-1). If you take the average of the variance with the experimental mean and compare it to the average of the variance with the true mean, you will see they are equal.
 
Oh I see now,thank you
 

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