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Find the mean and variance for the following probability density functions $f(x) = \left\{ \begin{array}{l l} \frac {1}{24} ,& \quad \text{-12$\leq$$ x$$\leq $$12$}\\ 0, & \quad \text{otherwise} \end{array} \right.$

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Toolbox:
  • Let X be a continuous random variable with probability density function f(x). Then the mathematical expectation of X is defined as $E(X)=\int_{-\infty}^\infty x f(x)dx$
  • $E(\phi (X))=\int_{\infty}^{\infty}\phi(x) f(x)dx$
  • Var$(X)=E(X^2)-[E(X)]^2$
  • $E(c)=c$
  • $E(aX\pm b)=aE (X)\pm b$
Step 1:
$E(X)=\int_{-\infty}^\infty xf(x) dx$
$\qquad=\int_{-12}^{12} \large\frac{x}{24}$$dx$
$\qquad=0$(odd function)
Mean=0
Step 2:
$E(X^2)=\int_{\infty}^\infty x^2f(x) dx$
$\qquad=\int_{-12}^{12}\large\frac{x^2}{24}$$dx$
$\qquad=\large\frac{2}{24}$$\int_0^{12} x^2 dx$
$\qquad=\large\frac{1}{12}\big[\frac{x^3}{3}\big]_0^{12}$
$\qquad=48$
Step 3:
$Var(X)=E(X^2)-[E(X)]^2$
$\qquad\;\;\;\;=48-0$
$\qquad\;\;\;\;=48$
answered Sep 17, 2013 by sreemathi.v
 

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