# Homework Help: Mean of a probability distribution

1. Mar 31, 2008

### Snazzy

[SOLVED]Mean of a probability distribution

1. The problem statement, all variables and given/known data

2. Relevant equations
$$\int^{b}_{a}p(x)dx=1$$

$$V=M_2-\bar{x}^2$$

$$\bar{x}^2=\int^{b}_{a}xp(x)dx$$

$$M_2=\int^{b}_{a} x^2p(x)dx$$

3. The attempt at a solution

I found that $$c=\frac{1}{b}$$ which is a right answer.

What I did next was:

$$\bar{x}=\int^{b}_{-b}xp(x)dx$$

$$=\int^{0}_{-b}x(\frac{cx}{b}+c)dx\ + \int^{b}_{0}x(\frac{-cx}{b}+c)dx$$

$$= \int^{0}_{-b}\frac{cx^2}{b}\ +\ c\ dx\ + \int^{b}_{0}\frac{-cx^2}{b}\ +\ c\ dx$$

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

$$=\frac{-2cb^3}{3b}+{2cb}$$

$$=\frac{-2b^2}{3b}+\frac{2b}{b}$$

$$=\frac{-2b}{3}+2$$

But the answer says that $$\bar{x}=0$$

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

Last edited: Mar 31, 2008
2. Mar 31, 2008

### Dick

i) you are integrating cx, that should give you an antiderivative of cx^2/2. ii) put in the top limit and subtract the bottom limit. You are making sign errors. The cx^3/3b terms also cancel.

3. Mar 31, 2008

### Snazzy

What an amateur mistake. I feel foolish. Cheers, Dick.