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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(0.5< x <0.75)

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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(0.5 < x < 0.75)=$\int_{0.5}^{0.75} f(x) dx$
$\qquad\qquad\qquad=\int_{0.5}^{0.75} 2xdx$
$\qquad\qquad\qquad=x^2\bigg]_{0.5}^{0.75}$
$\qquad\qquad\qquad=(0.75)^2-(0.5)^2$
$\qquad\qquad\qquad=(0.75+0.5)(0.75-0.5)$
$\qquad\qquad\qquad=0.25$
answered Sep 16, 2013 by sreemathi.v
 

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