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For the p.d.f $f(x) = \left\{ \begin{array}{l l} cx(1-x)^{3}, & \quad \text{0 $<$ x$<$1}\\ 0 & \quad \text{elsewhere} \end{array} \right.$ \[\]Find the constant $c$

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  • 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) $\int_{-\infty}^\infty f(x) dx=1$
  • $P(a\leq x\leq b)=P(a\leq x\leq b)$=P(a < x < b)
Step 1:
$f(x)=\left\{\begin{array}{1 1}cx(1-x)^3,& 0< x<1\\0 ,&elsewhere\end{array}\right.$ is a probability density function.
$\therefore f(x) \geq 0$ for all x and $\int_{-\infty}^\infty f(x) dx=1$
Step 2:
$\int_{-\infty}^\infty f(x) dx=1$
$\int_0^1 cx(1-x)^3dx=1$
(since $f(x)=0$ elsewhere)
$\Rightarrow c\bigg[\large\frac{x(1-x)^4}{4}-\large\frac{1}{-4}\frac{(1-x)^5}{-5}\bigg]_0^1$$=1$
$\Rightarrow \large\frac{-c(1-x)^4}{4}\bigg[x+\large\frac{1-x}{5}\bigg]_0^1$$=1$
$\Rightarrow -\bigg[0-\large\frac{c}{4}(\frac{1}{5})\bigg]$$=1$
$\Rightarrow \large\frac{c}{20}$$=1$
answered Sep 16, 2013 by sreemathi.v

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