# Math Help - Standard Deviation

1. ## Standard Deviation

Let X be a random variable with probability density function:

f(x)= (3/4)(1-x^2), -1 < x < 1
0, otherwise

What is the standard deviaton of x?

Let X be a random variable with probability density function:

f(x)= (3/4)(1-x^2), -1 < x < 1
0, otherwise

What is the standard deviaton of x?
You need to calculate E(X) and E(X^2):

$E(X) = \int_{-\infty}^{\infty} x\, f(x) \, dx = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = .....$

$E(X^2) = \int_{-\infty}^{\infty} x^2\, f(x) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = .....$

Then variance $= E(X^2) - [E(X)]^2 = .....$

3. Okay, so $
E(X) = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = \frac{3}{8} \, \int_{-1}^{1} u \, du
$
where $u = (1-x^2)$
Thus $
E(X) = \frac{12}{8}
$

Is that right?
Then $
E(X^2) = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 \, dx + \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx = 3 + \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx
$
but I am not sure how to find that last integral?

Okay, so $
E(X) = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = \frac{3}{8} \, \int_{-1}^{1} u \, du
$
where $u = (1-x^2)$
Thus $
E(X) = \frac{12}{8}
$

Is that right?
Then $
E(X^2) = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 \, dx + \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx = 3 + \frac{3}{4} \, \int_{-1}^{1} (1 - x^2) \, dx
$
but I am not sure how to find that last integral?
When you're working with pdf's it's assumed you can integrate. So you need to thoroughly revise basic integration techniques.

$E(X) = \frac{3}{4} \, \int_{-1}^{1} x \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x - x^3 \, dx = \frac{3}{4} \left[ \frac{x^2}{2} - \frac{x^4}{4}\right]_{-1}^{1} = 0$. This result could be anticipated since you're integrating an odd function .....

$E(X^2) = \frac{3}{4} \, \int_{-1}^{1} x^2 \, (1 - x^2) \, dx = \frac{3}{4} \, \int_{-1}^{1} x^2 - x^4 \, dx = \frac{3}{4} \left[ \frac{x^3}{3} - \frac{x^5}{5}\right]_{-1}^{1}$

$= \frac{3}{4} \left( \, \left( \frac{1}{3} - \frac{1}{5}\right) - \left( -\frac{1}{3} + \frac{1}{5} \right) \, \right) = \frac{2}{3} - \frac{2}{5} = \frac{4}{15}$.