# Calculation of Standard Deviation of coordinates?

Hello, I am currently in high school and we are learning about Standard Deviation. My teacher says that the applications of SD can be found in calculating the marks/population etc. But what I'm wondering is: can we calculate the SD of co-ordinates. For example, the cosine curve has a particular shape to it. But if a child draws it freehand, then it will not be as perfect as the cosine curve plotted by a calculator. So the curve drawn by the child deviates from the normal curve (and hence his curve has different co-ordinates.)

Now can we calculate the standard deviation for such a problem? What are your opinions? Note by Namrata Haribal
7 years ago

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- 7 years ago

Yes, you should be able to. Take several y-values of the child's curve and subtract them from the normal curve. From there you can average these and get the "standard deviation." It's basically like a normal standard deviation, except that the mean is not fixed, but it is the normal curve at the different sampled y-values.

- 7 years ago

hey, thanks so much!! If its possible, can you give a more detailed explanation, please? what do you mean by "subtract them from the normal curve" ?

- 7 years ago

Let's have the true curve be $f(x)$. Call the estimation of the curve $f_0(x)$. Sample several points $(x_i,f_0(x_i))$ from the estimation of the curve. We also know the points $(x_i,f(x_i))$ from the true curve. You can find the deviation of each of these points on the estimated curve from the true curve by subtracting: $f_0(x_i)-f(x_i)$. Averaging these up will give you an approximate standard deviation. Of course, sampling more points will give you a better approximation. Thus $s=\displaystyle \sum \frac{f_0(x_i)-f(x_i)}{n}$

- 7 years ago

You're awesome! Thanks a LOT.

- 7 years ago

Thanks! :) This is just my guesswork though. This is the same general idea as finding the standard deviation from the best fit line in a linear regression situation. Also, because standard deviation is weird, you might want to try using $n-1$ instead of $n$.

- 7 years ago