Questions: The median value (y) of a continuous random variable is that value such that (F(y)=0.5). Find the median value of the random variable in [ f(y)=leftbeginarrayll frac118 y, 0 leq y leq 6 0, text elsewhere. endarrayright. ]

Solution Steps

To find the median value of the given continuous random variable, we need to determine the value \( y \) such that the cumulative distribution function (CDF) \( F(y) \) equals 0.5. We first integrate the probability density function (PDF) \( f(y) \) to find the CDF, and then solve for \( y \) when \( F(y) = 0.5 \).

The given probability density function (PDF) is: \[ f(y) = \begin{cases} \frac{1}{18} y, & 0 \leq y \leq 6 \\ 0, & \text{elsewhere} \end{cases} \]

To find the cumulative distribution function (CDF), we integrate the PDF from 0 to \( y \): \[ F(y) = \int_0^y \frac{1}{18} t \, dt = \frac{1}{18} \int_0^y t \, dt = \frac{1}{18} \left[ \frac{t^2}{2} \right]_0^y = \frac{1}{36} y^2 \]

The median value \( y \) is the value such that \( F(y) = 0.5 \): \[ F(y) = \frac{1}{36} y^2 = 0.5 \]

Solving for \( y \): \[ \frac{1}{36} y^2 = 0.5 \implies y^2 = 18 \implies y = \pm \sqrt{18} = \pm 3\sqrt{2} \]

Since \( y \) must be within the range \( 0 \leq y \leq 6 \), we discard the negative solution: \[ y = 3\sqrt{2} \]

Final Answer