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**[SOLVED]Mean of a probability distribution**

**1. Homework Statement**

**2. Homework Equations**

[tex]\int^{b}_{a}p(x)dx=1[/tex]

[tex]V=M_2-\bar{x}^2[/tex]

[tex]\bar{x}^2=\int^{b}_{a}xp(x)dx[/tex]

[tex]M_2=\int^{b}_{a} x^2p(x)dx[/tex]

**3. The Attempt at a Solution**

I found that [tex]c=\frac{1}{b}[/tex] which is a right answer.

What I did next was:

[tex]

\bar{x}=\int^{b}_{-b}xp(x)dx[/tex]

[tex]=\int^{0}_{-b}x(\frac{cx}{b}+c)dx\ + \int^{b}_{0}x(\frac{-cx}{b}+c)dx[/tex]

[tex]= \int^{0}_{-b}\frac{cx^2}{b}\ +\ c\ dx\ + \int^{b}_{0}\frac{-cx^2}{b}\ +\ c\ dx[/tex]

[tex]=\left[ \frac{cx^3}{3b}+cx \right]_{-b}^{0}+\left[ \frac{-cx^3}{3b}+cx \right]_{0}^{b}[/tex]

[tex]=\frac{-2cb^3}{3b}+{2cb}[/tex]

[tex]=\frac{-2b^2}{3b}+\frac{2b}{b}[/tex]

[tex]=\frac{-2b}{3}+2

[/tex]

But the answer says that [tex]\bar{x}=0[/tex]

If I can manage to get x-bar, I can manage to get the variance and SD.

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