Today I taught an introductory class of statistics and a student came up to me with a question, which I rephrase here as: "Why is the standard deviation defined as sqrt of variance and not as the sqrt of sum of squares over N?"

We define population variance: $\sigma^2=\frac{1}{N}\sum{(x_i-\mu)^2}$

And standard deviation: $\sigma=\sqrt{\sigma^2}=\frac{1}{\sqrt{N}}\sqrt{\sum{(x_i-\mu)^2}}$.

The interpretation we may give to $\sigma$ is that it gives the average deviation of units in the population from the population mean of $X$.

However, in the definition of the s.d. we divide the sqrt of the sum of squares through $\sqrt{N}$. The question the student raises is why we do not divide the sqrt of the sume of squares by $N$ instead. Thus we come to competing formula: $$\sigma_{new}=\frac{1}{N}\sqrt{\sum{(x_i-\mu)^2}}.$$ The student argued that this formula looks more like an "average" deviation from the mean than when dividing through $\sqrt{N}$ as in $\sigma$.

I thought this question is not stupid. I would like to give an answer to the student that goes further than saying that the s.d. is defined as sqrt of the variance which is the average squared deviaton. Put differently, why should the student use the correct formula and not follow her idea?

This question relates to an older thread and answers provided here. Answers there go in three directions:

1. $\sigma$ is the root-mean-squared (RMS) deviation, not the "typical" deviation from the mean (i.e., $\sigma_{new}$). Thus, it is defined differently.
2. It has nice mathematical properties.
3. Furthermore, the sqrt would bring back "units" to their original scale. However, this would also be the case for $\sigma_{new}$, which divides by $N$ instead.

Both of points 1 and 2 are arguments in favour of the s.d. as RMS, but I do not see an argument against the use of $\sigma_{new}$. What would be the good arguments to convince introductory level students of the use of the average RMS distance $\sigma$ from the mean?

Today I taught an introductory class of statistics and a student came up to me with a question, which I rephrase here as: "Why is the standard deviation defined as sqrt of variance and not as the sqrt of sum of squares over N?"

We define population variance: $\sigma^2=\frac{1}{N}\sum{(x_i-\mu)^2}$

And standard deviation: $\sigma=\sqrt{\sigma^2}=\frac{1}{\sqrt{N}}\sqrt{\sum{(x_i-\mu)^2}}$.

The interpretation we may give to $\sigma$ is that it gives the average deviation of units in the population from the population mean of $X$.

However, in the definition of the s.d. we divide the sqrt of the sum of squares through $\sqrt{N}$. The question the student raises is why we do not divide the sqrt of the sume of squares by $N$ instead. Thus we come to competing formula: $$\sigma_{new}=\frac{1}{N}\sqrt{\sum{(x_i-\mu)^2}}.$$ The student argued that this formula looks more like an "average" deviation from the mean than when dividing through $\sqrt{N}$ as in $\sigma$.

I thought this question is not stupid. I would like to give an answer to the student that goes further than saying that the s.d. is defined as sqrt of the variance which is the average squared deviaton. Put differently, why should the student use the correct formula and not follow her idea?

This question relates to an older thread and answers provided here. Answers there go in three directions:

1. $\sigma$ is the root-mean-squared (RMS) deviation, not the "typical" deviation from the mean (i.e., $\sigma_{new}$). Thus, it is defined differently.
2. It has nice mathematical properties.
3. Furthermore, the sqrt would bring back "units" to their original scale. However, this would also be the case for $\sigma_{new}$, which divides by $N$ instead.

Both of points 1 and 2 are arguments in favour of the s.d. as RMS, but I do not see an argument against the use of $\sigma_{new}$. What would be the good arguments to convince introductory level students of the use of the average RMS distance $\sigma$ from the mean?

Today I taught an introductory class of statistics and a student came up to me with a question, which I rephrase here as: "Why is the standard deviation defined as sqrt of variance and not as average of the sqrt of sum of squares?"

So we define population variance $\sigma^2=\frac{1}{N}\sum{(x_i-\mu)^2}$ and standard deviation $\sigma=\sqrt{\sigma^2}$. The interpretation we may give to $\sigma$ is that it gives the average deviation of units in the population from the population mean of $X$. However, in the definition of the s.d. we divide the sqrt of the sum of squares through $\sqrt{N}$. The question is why we do not divide the by $N$ instead.

I thought this question is not stupid. I would like to give an answer to the student that goes further than saying that the s.d. is defined as sqrt of the variance which is the average squared deviaton. Put differently, why should the student use the correct formula and not follow her idea?

This question relates to an older thread and answers provided here. Answers there go in three directions:

1. It is the root-mean-squared (RMS) deviation, not the "typical" deviation from the mean. Thus, it is defined differently.
2. It has nice mathematical properties.
3. Furthermore, the sqrt would bring back "units" to their original scale. However, this would also be the case for the "typical" deviation, which divides by $N$ instead.

Both of points 1 and 2 are arguments in favour of the s.d. as RMS, but I do not see an argument against the use of the "typical" deviation. What would be the good arguments to convince introductory level students of the use of the RMS distance from the mean?

Today I taught an introductory class of statistics and a student came up to me with a question, which I rephrase here as: "Why is the standard deviation defined as sqrt of variance and not as the sqrt of sum of squares over N?"

We define population variance: $\sigma^2=\frac{1}{N}\sum{(x_i-\mu)^2}$

And standard deviation: $\sigma=\sqrt{\sigma^2}=\frac{1}{\sqrt{N}}\sqrt{\sum{(x_i-\mu)^2}}$. 

The interpretation we may give to $\sigma$ is that it gives the average deviation of units in the population from the population mean of $X$. 

However, in the definition of the s.d. we divide the sqrt of the sum of squares through $\sqrt{N}$. The question the student raises is why we do not divide the sqrt of the sume of squares by $N$ instead. Thus we come to competing formula: $$\sigma_{new}=\frac{1}{N}\sqrt{\sum{(x_i-\mu)^2}}.$$ The student argued that this formula looks more like an "average" deviation from the mean than when dividing through $\sqrt{N}$ as in $\sigma$.

I thought this question is not stupid. I would like to give an answer to the student that goes further than saying that the s.d. is defined as sqrt of the variance which is the average squared deviaton. Put differently, why should the student use the correct formula and not follow her idea?

This question relates to an older thread and answers provided here. Answers there go in three directions:

1. $\sigma$ is the root-mean-squared (RMS) deviation, not the "typical" deviation from the mean (i.e., $\sigma_{new}$). Thus, it is defined differently.
2. It has nice mathematical properties.
3. Furthermore, the sqrt would bring back "units" to their original scale. However, this would also be the case for $\sigma_{new}$, which divides by $N$ instead.

Both of points 1 and 2 are arguments in favour of the s.d. as RMS, but I do not see an argument against the use of $\sigma_{new}$. What would be the good arguments to convince introductory level students of the use of the average RMS distance $\sigma$ from the mean?