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Question: A continuous random variable X has the P.D.F \(f\left( x \right)=3{{x}^{2}},0\le x\le 1\). The value...

A continuous random variable X has the P.D.F f(x)=3x2,0x1f\left( x \right)=3{{x}^{2}},0\le x\le 1. The value of a constant λ\lambda that satisfies the relation P\left\\{ X\le \lambda \right\\}=P\left\\{ X>\lambda \right\\} is
A. (13)12{{\left( \dfrac{1}{3} \right)}^{\dfrac{1}{2}}}
B. (12)13{{\left( \dfrac{1}{2} \right)}^{\dfrac{1}{3}}}
C. (23)12{{\left( \dfrac{2}{3} \right)}^{\dfrac{1}{2}}}
D. (23)13{{\left( \dfrac{2}{3} \right)}^{\dfrac{1}{3}}}

Explanation

Solution

We first check if the given P.D.F satisfies the condition of f(x)dx=1\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=1. We take the integration of 013x2dx\int\limits_{0}^{1}{3{{x}^{2}}dx}. We then use the given condition of P\left\\{ X\le \lambda \right\\}=P\left\\{ X>\lambda \right\\} to form the integration part. We solve the equation to find the final solution.

Complete step by step answer:
The given P.D.F is f(x)=3x2,0x1f\left( x \right)=3{{x}^{2}},0\le x\le 1. The condition for P.D.F is f(x)dx=1\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=1.
Here we take the integration for 0x10\le x\le 1.
We get 013x2dx=1\int\limits_{0}^{1}{3{{x}^{2}}dx}=1.

& \int\limits_{0}^{1}{3{{x}^{2}}dx}=1 \\\ & \Rightarrow \left[ {{x}^{3}} \right]_{0}^{1}=1-0=1 \\\ \end{aligned}$$ The given equation satisfies the condition. It is given that $P\left\\{ X\le \lambda \right\\}=P\left\\{ X>\lambda \right\\}$. We express them in the form of integration and get $$\begin{aligned} & P\left\\{ X\le \lambda \right\\}=P\left\\{ X>\lambda \right\\} \\\ & \Rightarrow \int\limits_{0}^{\lambda }{3{{x}^{2}}dx}=\int\limits_{\lambda }^{1}{3{{x}^{2}}dx} \\\ & \Rightarrow \left[ {{x}^{3}} \right]_{0}^{\lambda }=\left[ {{x}^{3}} \right]_{\lambda }^{1} \\\ \end{aligned}$$ We now solve the equation and get $$\begin{aligned} & \left[ {{x}^{3}} \right]_{0}^{\lambda }=\left[ {{x}^{3}} \right]_{\lambda }^{1} \\\ & \Rightarrow {{\lambda }^{3}}-0=1-{{\lambda }^{3}} \\\ & \Rightarrow 2{{\lambda }^{3}}=1 \\\ & \Rightarrow {{\lambda }^{3}}=\dfrac{1}{2} \\\ & \Rightarrow \lambda ={{\left( \dfrac{1}{2} \right)}^{\dfrac{1}{3}}} \\\ \end{aligned}$$ **Therefore, the correct option is B.** **Note:** It is a function whose value at any given sample in the sample space can be interpreted as providing a relative likelihood that the value of the random variable would equal that sample. For cases of P.M.F we have to use the distinct values of the masses.