# Standard deviation of weighted data

1. Jan 10, 2009

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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 Im dealing with?

2. Jan 10, 2009

### snipez90

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.

3. Jan 10, 2009

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?

4. Jan 10, 2009

### snipez90

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.

5. Jan 10, 2009

### HallsofIvy

Staff Emeritus
That's not what I get. I get 0.392 for the variance and 0.6096 for the standard deviation.