Find k so that the function $f ( x ) = \left\{ \begin{array} { l l }
k x ( 1 - x ) & \text { if } 0

Question

Find k so that the function $f ( x ) = \left\{ \begin{array} { l l } 
k x ( 1 - x ) & \text { if } 0 < x < 1 \
0 & \text { otherwise }
\end{array} \right.$ can serve as the probability density function of a random variable X.
A) $\frac { 1 } { 3 }$ 
B) $\frac { 1 } { 6 }$ 
C) $\frac { 1 } { 2 }$ 
D)1
E)3
F)6
G)2
H)0

Quizplus · Accepted Answer

To be a probability density function, the integral of $f(x)$ from 0 to 1 must equal 1. Solving $\int_0^1 kx(1-x) \, dx = 1$ gives $k = 6$.

Find K So That the Function f(x)={kx(1−x) if 0<x<10 otherwise f ( x ) = \left\{ \begin{array} { l l } k x ( 1 - x ) & \text { if } 0 < x < 1 \\0 & \text { otherwise }\end{array} \right.f(x)={kx(1−x)0​ if 0<x<1 otherwise ​

Find K So That the Function $f ( x ) = \left\{ \begin{array} { l l } k x ( 1 - x ) & \text { if } 0 < x < 1 \\0 & \text { otherwise }\end{array} \right.$