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If the function $f$ defined by $f(x)=K\left(x-x^{2}\right)$ if $0 < x < 1$
$=0 \quad, \quad$ other wise
is the p.d.f. of a r. v. $X$, then the value of $P\left(X < \frac{1}{2}\right)$ is
Options:
$=0 \quad, \quad$ other wise
is the p.d.f. of a r. v. $X$, then the value of $P\left(X < \frac{1}{2}\right)$ is
Solution:
2373 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
Given $f$ to a the p.df of a r.v.x.
$\begin{aligned}
& \int f(x) d x=1 \Rightarrow \int_{0}^{1} K\left(x-x^{2}\right) d x=1 \\
& K\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=1 \Rightarrow K\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow \frac{K}{6}=1 \\
K &=6 \\
\therefore & P\left(x < \frac{1}{2}\right)=\int_{0}^{\frac{1}{2}} 6\left(x-x^{2}\right) d x=\left[\frac{6 x^{2}}{2}\right]_{0}^{1}-\left[\frac{6 x^{3}}{3}\right]_{0}^{1} \\
&=\left[3 x^{2}-2 x^{3}\right]_{0}^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}
\end{aligned}$
$\begin{aligned}
& \int f(x) d x=1 \Rightarrow \int_{0}^{1} K\left(x-x^{2}\right) d x=1 \\
& K\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=1 \Rightarrow K\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow \frac{K}{6}=1 \\
K &=6 \\
\therefore & P\left(x < \frac{1}{2}\right)=\int_{0}^{\frac{1}{2}} 6\left(x-x^{2}\right) d x=\left[\frac{6 x^{2}}{2}\right]_{0}^{1}-\left[\frac{6 x^{3}}{3}\right]_{0}^{1} \\
&=\left[3 x^{2}-2 x^{3}\right]_{0}^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}
\end{aligned}$
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