Standard deviation from grades distribution how am I wrong?

So in summary, the person asking for help found the average to be 7 and used a formula for standard deviation, but is still getting the wrong answer. They then asked for help and received a suggestion to divide the sum of squares by the number of grades minus one before taking the square root. However, they later discovered that this was the wrong formula and the correct formula is to divide the sum of squares by the number of grades. The suggested formula gave a result of 1.658, while the correct formula gives a result of 1.563. The person ends the conversation by thanking those who helped them and summarizing the final correct formula for standard deviation.
  • #1
Femme_physics
Gold Member
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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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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
 
  • #7
Good enough for me :) Thanks, great helpers!
 
  • #8
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
 

Related to Standard deviation from grades distribution how am I wrong?

1. What is standard deviation and how is it calculated?

Standard deviation is a measure of how spread out a set of data is from its mean. It is calculated by taking the square root of the variance, which is the average of the squared differences from the mean.

2. Why is standard deviation important in grade distributions?

Standard deviation helps to understand the variability of grades in a distribution. A higher standard deviation means that the grades are more spread out, while a lower standard deviation means that the grades are clustered closer to the mean. This information can be useful for identifying patterns and outliers in the data.

3. How does standard deviation relate to my grades?

Standard deviation can tell you how your grades compare to the rest of the class. If your grade falls within one standard deviation of the mean, it is considered average. Grades that fall more than one standard deviation above or below the mean can be considered above or below average, respectively.

4. Can standard deviation be used to determine the overall performance of a class?

Standard deviation alone cannot determine the overall performance of a class. It is just one measure of variability in the data. Other factors such as the class size, grading system, and assignment difficulty should also be taken into consideration.

5. How can I use standard deviation to improve my grades?

Standard deviation can help you identify areas where you may need to improve. If your grades consistently fall below the mean by more than one standard deviation, it may be a sign that you need to focus more on those topics or seek extra help. Additionally, comparing your standard deviation to the class average can give you an idea of how you are performing compared to your peers.

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