For the probability distribution function (p.d.f) $\;$ $f\left(x\right)= \begin{cases}
c x \left(1 - x\right)^3, & 0 < x < 1 \\
& \\
0, & \text{elsewhere}
\end{cases}$ $\;$ find
- the constant $c$
- $P \left(x < \dfrac{1}{2}\right)$
- By definition of p.d.f, $\;$ $\int \limits_{0}^{1} f \left(x\right) dx = 1$
i.e. $\;$ $\int \limits_{0}^{1} cx \left(1 - x\right)^3 dx = 1$
i.e. $\;$ $\int \limits_{0}^{1} c \left(1 - x\right) \left[1 - \left(1 - x\right)\right]^3 dx = 1$ $\; \left[\because \; \int \limits_{0}^{a} f \left(x\right) dx = \int \limits_{0}^{a} f \left(a - x\right) dx\right]$
i.e. $\;$ $\int \limits_{0}^{1} c \left(1 - x\right) x^3 dx = 1$
i.e. $\;$ $c \left\{\int \limits_{0}^{1} x^3 dx - \int \limits_{0}^{1} x^4 dx \right\} = 1$
i.e. $\;$ $c \left\{\left[\dfrac{x^4}{4}\right]_{0}^{1} - \left[\dfrac{x^5}{5}\right]_{0}^{1} \right\} = 1$
i.e. $\;$ $c \left(\dfrac{1}{4} - \dfrac{1}{5}\right) = 1$
$\implies$ $c = 20$
- $P \left(x < \dfrac{1}{2}\right) = \int \limits_{0}^{1/2} f \left(x\right) dx$
Here, $f \left(x\right) = \begin{cases} 20x \left(1 - x\right)^3, & 0 < x < 1 \\ & \\ 0, & \text{elsewhere} \end{cases}$
$\begin{aligned} \therefore \; P \left(x < \dfrac{1}{2}\right) & = 20 \int \limits_{0}^{1/2} x \left(1 - x\right)^3 dx \\\\ & = 20 \int \limits_{0}^{1/2} x \left(1 - 3x + 3x^2 -x^3\right) dx \\\\ & = 20 \int \limits_{0}^{1/2} \left(x - 3x^2 + 3x^3 -x^4\right) dx \\\\ & = 20 \left[\dfrac{x^2}{2} - 3 \left(\dfrac{x^3}{3}\right) + 3 \left(\dfrac{x^4}{4}\right) - \dfrac{x^5}{5}\right]_{0}^{1/2} \\\\ & = 20 \left[\dfrac{1}{8} - \dfrac{1}{8} + \dfrac{3}{64} - \dfrac{1}{160}\right] \\\\ & = 20 \left[\dfrac{15-2}{320}\right] \\\\ & = 20 \times \dfrac{13}{320} = \dfrac{13}{16} \end{aligned}$