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The probability density function of a random variable $x$ is $f(x) = \left\{ \begin{array}{l l} kx^{\alpha-1}e^{-\beta\;x^{\alpha}}, & \quad \text{x,$\alpha$$,\beta$>$0$}\\ 0 ,& \quad \text{elsewhere} \end{array} \right.$, Find $p(x$>$10)$

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1 Answer

  • The probability density function (continuous probability function $f(x)$ satisfies the following properties :
  • (i) $P(a\leq x\leq b)=\int_a^b f(x) dx$
  • (ii) $f(x)$ is non-negative for all real $x$
  • (iii) $\int_{-\infty}^\infty f(x) dx=1$
  • Also $P(x=a)=0$
  • $P(a\leq x\leq b)=P(a\leq x\leq b)$=P(a < x < b)
Step 1:
$f(x)=\left\{\begin{array}{1 1}kx^{\alpha-1}e^{-\beta x^\alpha},&x,\alpha,\beta>0\\0 ,&elsewhere\end{array}\right.$ is a probability density function.
$\therefore f(x) \geq 0$ for all x and $\int_{-\infty}^\infty f(x) dx=1$
Step 2:
$\int_{-\infty}^\infty f(x) dx=1$
$\int_{-\infty}^0 kx^{\alpha-1}e^{-\beta x^\alpha} dx=1$
Now $\large\frac{d}{dx}$$e^{-\beta x^\alpha}=-\beta e^{-\beta x^\alpha}.\alpha x^{\alpha-1}$
$\qquad\qquad\;\;\;\;\;=-\alpha \beta x^{\alpha -1}e^{-\beta x^\alpha}$
$\therefore -\large\frac{1}{\alpha\beta}$$d(e^{-\beta x^\alpha})=x^{\alpha-1}e^{-\beta x^\alpha}dx$
$\therefore k\int_0^{\infty}x^{\alpha -1}e^{-\beta x^\alpha} dx=1$
$\Rightarrow \large\frac{-k}{\alpha \beta}$$\int_0^{\infty}d(e^{-\beta x^{\alpha}})dx=1$
$\Rightarrow \large\frac{-k}{\alpha \beta}$$(e^{-\beta x^{\alpha}}\big)_0^\infty=1$
$\Rightarrow \large\frac{-k}{\alpha \beta}$$(0-1)=1$
$\Rightarrow k=\alpha \beta$
Step 3:
$P(x >10)=\int_{10}^\infty f(x) dx$
$\qquad\qquad=-\int_{10}^\infty \alpha \beta x^{\alpha-1}e^{-\beta x^\alpha}dx$
$\qquad\qquad=\alpha \beta \large\frac{e^{-\beta x^\alpha}}{-\alpha \beta}\bigg]_{10}^\infty$
$\qquad\qquad=e^{-\beta 10^\alpha}-0$
$\qquad\qquad=e^{\large -\beta (10^\alpha)}$
answered Sep 16, 2013 by sreemathi.v

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