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If the function $\mathrm{f}$ defined by $\mathrm{f}(\mathrm{x})= \begin{cases}\mathrm{K}\left(\mathrm{x}-\mathrm{x}^2\right) & \text { if } 0 < \mathrm{x} < 1 \\ 0 & \text {, other wise }\end{cases}$ is the p.d.f. of a r.v.x, then the value of $\mathrm{P}\left(\mathrm{X} < \frac{1}{2}\right)$ is
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Given $\mathrm{f}$ is a the p.d.f. of a r.v.x.
$$
\begin{aligned}
& \therefore \int_0^1 \mathrm{f}(\mathrm{x}) \mathrm{dx}=1 \Rightarrow \int_0^1 \mathrm{~K}\left(\mathrm{x}-\mathrm{x}^2\right) \mathrm{dx}=1 \\
& \mathrm{~K}\left[\frac{\mathrm{x}^2}{2}-\frac{\mathrm{x}^3}{3}\right]_0^1=1 \Rightarrow \mathrm{K}\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow \frac{\mathrm{K}}{6}=1 \\
& \mathrm{~K}=6 \\
& \left.\therefore \mathrm{P}\left(\mathrm{X} < \frac{1}{2}\right)=\int_0^{\frac{1}{2}} 6\left(\mathrm{x}-\mathrm{x}^2\right) \mathrm{dx}=\left[\frac{6 \mathrm{x}^2}{2}\right]_0^1-\left[\frac{6 \mathrm{x}^3}{3}\right]_0^1\right] \\
& =\left[3 \mathrm{x}^2-2 \mathrm{x}^3\right]_0^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \int_0^1 \mathrm{f}(\mathrm{x}) \mathrm{dx}=1 \Rightarrow \int_0^1 \mathrm{~K}\left(\mathrm{x}-\mathrm{x}^2\right) \mathrm{dx}=1 \\
& \mathrm{~K}\left[\frac{\mathrm{x}^2}{2}-\frac{\mathrm{x}^3}{3}\right]_0^1=1 \Rightarrow \mathrm{K}\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow \frac{\mathrm{K}}{6}=1 \\
& \mathrm{~K}=6 \\
& \left.\therefore \mathrm{P}\left(\mathrm{X} < \frac{1}{2}\right)=\int_0^{\frac{1}{2}} 6\left(\mathrm{x}-\mathrm{x}^2\right) \mathrm{dx}=\left[\frac{6 \mathrm{x}^2}{2}\right]_0^1-\left[\frac{6 \mathrm{x}^3}{3}\right]_0^1\right] \\
& =\left[3 \mathrm{x}^2-2 \mathrm{x}^3\right]_0^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}
\end{aligned}
$$
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