Test Scores: Variance & Maximum Mark

Ted123
Messages
428
Reaction score
0
Say we have a sample of test scores, all marked between 0 and 100.

Does the sample variance have to be less than or equal to the maximum mark 100 or can it exceed this?
 
Physics news on Phys.org
It can exceed it: say your test scores are 0 and 100. Then the sample mean is 50. The sample variance equals

\frac{1}{n-1}\sum_{i=1}^n{(X_i-\overline{X_n})^2}=5000.

You probably mean the sample standard deviation... In this example, it is 70. I don't think it can exceed 100...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
7
Views
1K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
5
Views
1K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
2
Views
2K
Back
Top