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Sagot :
Solution :
Given :
[tex]f(x) = \left\{\begin{matrix}\frac{1}{450}(x^2-x) & \text{if } 6 < x < 12 \\ 0 & \text{otherwise}\end{matrix}\right.[/tex]
1. Cumulative distribution function
[tex]$P(X \leq x) = \int_{- \infty}^x f(x) \ dx$[/tex]
[tex]$=\int_{- \infty}^6 f(x) dx + \int_{6}^x f(x) dx $[/tex]
[tex]$=0+\int_6^x \frac{1}{450}(x^2-x) \ dx$[/tex]
[tex]$=\frac{1}{450} \int_6^x (x^2-x) \ dx$[/tex]
[tex]$=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$[/tex]
[tex]$=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right] $[/tex]
[tex]$=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$[/tex]
2. Mean [tex]$E(x) = \int_{- \infty}^{\infty} \ x \ f(x) \ dx$[/tex]
[tex]$=\int_{6}^{12}x . \left( \frac{1}{450} \ (x^2-x)\right)\ dx$[/tex]
[tex]$=\frac{1}{450} \int_6^{12} \ (x^3 - x^2) \ dx$[/tex]
[tex]$=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$[/tex]
[tex]$=\frac{1}{450} \left[ \left(\frac{(12)^4}{4} - \frac{(12)^3}{3} \right) - \left(\frac{(6)^4}{4} - \frac{(6)^3}{3} \right) $[/tex]
[tex]$=\frac{1}{450} [4608 - 252]$[/tex]
= 17.2857
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