Prove linear function of a variable

Nyasha
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Homework Statement



In the data table they are noon-hour temperatures of a certain week.

datatable.jpg



I calculated the mean which is 25 and l also calculated the standard deviation which is 3.74. Now they want me to show that :

if y=ax+b then \bar{y}=a\bar{x}+b and s_{y}= \left| a\right|s_{x}






The Attempt at a Solution




I don't know where they get this y from. Can you give me hints on how to solve this question ?
 
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Hi Nyasha! :smile:
Nyasha said:
… if y=ax+b then \bar{y}=a\bar{x}+b and s_{y}= \left| a\right|s_{x}

I don't know where they get this y from. Can you give me hints on how to solve this question ?

For example, x could be in degrees Celsius, while y could be (9/5)x + 32, which is degrees Fahrenheit.

It's just a change in scale. :wink:
 
They are not "getting" y from anywhere. They are defining y to be this linear function of x. The point is to show that if y is a linear function of x then the mean of y is that same linear function of the mean of x and the standard deviation of y is a multiple of the standard deviation of x.
 
Okay guys is this correct for the other part which says show that :
s_{y}= \left| a\right|s_{x}

Attempt to solution:

s^2_{y}=(a^2)\cdot(s^2_{x})
\sqrt{(s^2_{y})}=\sqrt{(a^2)\cdot(s^2_{x})
s_{y}= \left| a\right|s_{x}
 
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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