Standard deviation from grades distribution how am I wrong?

AI Thread Summary
The discussion revolves around calculating the standard deviation of a given grades distribution, with participants attempting to identify errors in their calculations. The average of the grades is confirmed as 7, but discrepancies arise in the standard deviation results. One participant calculates a standard deviation of 1.658 using the formula for sample standard deviation, while the answer book claims it should be 1.563. It is clarified that the difference stems from using the population versus sample standard deviation formulas, with the correct calculation yielding approximately 1.564. The conversation highlights the importance of using the appropriate formula based on the context of the data set.
Femme_physics
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Homework Statement



The grades distribution in class is

10, 8, 8, 7, 7, 7, 6, 6, 4

What's the standard deviation?



The Attempt at a Solution



I found the average is 7 (which is true) and used the formula for standard deviation. I checked it twice on the calculator but I'm still getting the same score, which is wrong. Anyone have a clue?

http://img135.imageshack.us/img135/6773/standarddev.jpg
 
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Hey :smile:
Femme_physics said:
I found the average is 7 (which is true) and used the formula for standard deviation. I checked it twice on the calculator but I'm still getting the same score, which is wrong. Anyone have a clue?

What formula do you have for the standard deviation?

You need to divide the sum of squares by the number of grades minus one, before you draw the square root.
 
Maybe I have the wrong formula. The formula is what I used that you see. Each grade minus the average squared, all under a square root.

The sum of the squares is 22.
The sum of all grades minus one is 8
The square root of 22/8 is 1.658

The answer book is going for 1.563
 
Also, there's not a single standard deviation. There's the population standard deviation and there's the sample standard deviation. The formulas for these are different, with sample standard deviation always being a bit larger. The one that I like Serena alluded to is the sample standard deviation.
 
Femme_physics said:
Maybe I have the wrong formula. The formula is what I used that you see. Each grade minus the average squared, all under a square root.

The sum of the squares is 22.
The sum of all grades minus one is 8
The square root of 22/8 is 1.658

The answer book is going for 1.563

You will have the wrong formula then.

What the answer book will have done, is to divide the sum of squares by the number of grades.
[edit] That is the square root of 22/9 [/edit]

This is actually the wrong answer, but it'll do for now, until we get to the fine points of standard deviations.
 
Femme_physics said:
Maybe I have the wrong formula. The formula is what I used that you see. Each grade minus the average squared, all under a square root.

The sum of the squares is 22.
The sum of all grades minus one is 8
The square root of 22/8 is 1.658
That's what I'm getting, too. I don't see anything wrong in what you did. You might check to make sure that the data you show is the same as given in the problem. If it is, I suspect a wrong answer in the book.
Femme_physics said:
The answer book is going for 1.563
 
Good enough for me :) Thanks, great helpers!
 
Femme_physics said:
Maybe I have the wrong formula. The formula is what I used that you see. Each grade minus the average squared, all under a square root.

The sum of the squares is 22.
The sum of all grades minus one is 8
The square root of 22/8 is 1.658

The answer book is going for 1.563

std = sqrt(v), where v = variance. We have v = sum[(x-m)^2,i=1..n)]/n, where m = mean = sum(x,i=1..n)/n. In your case n = 9, and calculation gives m = 7, so v = 22/9, hence std = sqrt(22/9) = 1.564, approx.

RGV
 

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