Standard deviation of weighted data

AI Thread Summary
The discussion focuses on calculating the standard deviation of weighted data from resistance measurements of 50 resistors. The user initially computes the mean resistance but struggles with the standard deviation, mistakenly calculating a very low variance. Clarifications are provided regarding the correct formula for variance with weighted data, emphasizing that the standard deviation is the square root of the variance. The conversation highlights the importance of properly applying the formula for weighted data and distinguishes between variance and standard deviation. Ultimately, the user arrives at a variance of 0.392 and a standard deviation of 0.6096.
MadmanMurray
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Homework Statement


The resistances of 50 resistors are measured and the results recorded are as follows:

Resistance x Frequency
1.) 5 x 17,
2.) 5.5 x 12,
3.) 6 x 10,
4.) 6.5 x 6,
5.) 7 x 5)


Calculate standard deviation of the measurements



Homework Equations





The Attempt at a Solution


Added up the frequences and got n = 50

Multiplied the resistance by its weight or frequency and got the following:
1.) 86
2.) 66
3.) 60
4.) 39
5.) 35


then I added the values up and divided the result by the sum of the frequencies (n) to get the mean (cant do an x bar so I'll denote it as M)
M = 5.7

If I follow the standard deviation formula and add up all results of (x - m)2 I get a really low number so if I then divide that really low number by n which is 50 I get something like 0.0049.

What am I doing wrong? Is the x of the formula resistance x frequency in this case since its weighted data I am dealing with?
 
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Hmm I can only comment on your procedure in calculating the standard deviation. I think what you have is the variance. The standard deviation is the square root of that quantity you calculated. But like you said, the data that you're dealing with might require a different calculation.
 
Yes I know that the standard deviation is the variance squared but what I'm asking is how I get the variance with weighted data. I know I have to use the formula variance = f(x - M)2/n but what does that mean? Do I subtract M from x then square the result then multiply by f or what?
 
Well first of all, the standard deviation is the square root of the variance. I'm not sure what f is in your formula even though I suspect that it is a function, not a variable. The formula for sample variance is the [sum of (x - M)^2 over all x] / [n-1]. The standard deviation is the square root of this quantity.

Like I said, this is just the very basic sample variance formula.
 
That's not what I get. I get 0.392 for the variance and 0.6096 for the standard deviation.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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