Solving Deviation Questions: +8 or -8?

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Can anybody tell me how to solve or solve it?

Scores from a statistics exam are reported as deviation (difference from the mean) scores. Which of the following deviation scores indicates that the student has a higher position in the class distribution?

A). cannot determine without more information
B). +8
C). 0
D). -8
 
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Welcome to Physics Forums! :wink:
Show your attempt first.Also use the template provided when posting.So that we can know your exact problem.
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I guest I remove myself from this forum
 
kevweb said:
I guest I remove myself from this forum
You can't. :-p

Anyway,Show a line of your work and we will help you.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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