The expected value of a continuous random variable with the given probability density function (PDF) is calculated using two integrals due to the piecewise nature of the function. The PDF is defined as f(x) = 1/12 for 0 < x <= 3 and f(x) = x/18 for 3 < x <= 6. The expected value E(X) is computed as E(X) = (1/12) ∫(0 to 3) x dx + (1/18) ∫(3 to 6) x² dx. This approach effectively handles the two intervals of the PDF.
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rayne said:Given the PDF:
f(x) = 1/12 , 0 < x <= 3
x/18, 3 < x <= 6
0, otherwise
find the expected value, E(x).
I know how to find the expected value if there was only one interval, but don't how to do it for two.
chisigma said:The integral can be devided in two integrals as follows...
$\displaystyle E \{X \} = \frac{1}{12}\ \int_{0}^{3} x\ d x + \frac{1}{18}\ \int_{3}^{6} x^{2}\ dx\ (1)$
Kind regards
$\chi$ $\sigma$