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For the distribution function given by $f(x) = \left\{ \begin{array}{l l} 0&\quad\text{x<0}\\ x^{2} & \quad \text{$0$$\leq$$x$$\leq$$1$}\\ 1 & \quad \text{x>1} \end{array} \right.$\[\] Find the density function. Also evaluate $p(x\leq0.5)$

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Toolbox:
  • 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)
  • Cumulative distribution function of a continuous random variable $X$ is given by
  • $f(x)=P(X\leq x)=\int_{-\infty}^x f(t)dt$ for $-\infty$ < x >$\infty$
  • Where $f(t)dt$ is the probability density function of $X$
  • Properties :
  • (1) $F(x)$ is a non decreasing function of $x$
  • (2) $0 \leq F(x) \leq 1,-\infty < x <\infty$
  • (3) $F(-\infty)=\lim\limits_{x\to -\infty}\int_{-\infty}^x f(x) dx=0$
  • (4) $F(\infty)=\lim\limits_{x\to \infty}\int_{-\infty}^x f(x) dx=1$
  • (5) $P(a\leq x\leq b)=F(b)-F(a)$
  • (6) $f(x)=F'(x)$
Step 1:
$F(x)=\left\{\begin{array}{1 1} 0 & x<0\\x^2 & 0\leq x\leq 1\\1&x>1\end{array}\right.$
Now $f(x)=F'(x)=\left\{\begin{array}{1 1} 0 & x<0\\x^2 & 0\leq x\leq 1\\1&x>1\end{array}\right.$
$\therefore $ the probability density function is $f(x)=\left\{\begin{array}{1 1}2x,& 0\leq x\leq 1\\0,&else\;where\end{array}\right.$
Step 2:
$P(x\leq 0.5)=\int_{-\infty}^{0.5} f(x)dx$
$\qquad\qquad\;=\int_0^{0.5}2xdx$
$\qquad\qquad\;=x^2\bigg]_0^{0.5}$
$\qquad\qquad\;=0.25$
answered Sep 17, 2013 by sreemathi.v
 

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