Question

The diameter of a fibre is assumed to be a continuous random variable \( X \) with probability density function \[ f(x) = 6x(1 - x), \quad 0<x \leq 1 \] If \( P(X<\beta) = P(X>\beta) \), then the value of \( \beta \) (rounded off to 1 decimal place) is _________.

Hint:

To find the median of a continuous distribution, solve for \( F(\beta) = 0.5 \), where \( F(x) \) is the cumulative distribution function.

Answer 1

Step 1: Understanding the condition \( P(X<\beta) = P(X>\beta) \). 
For a continuous random variable \( X \), the condition \( P(X<\beta) = P(X>\beta) \) implies that \( \beta \) is the median of the distribution, meaning the cumulative probability up to \( \beta \) is 0.5. 
Step 2: Find the cumulative distribution function (CDF) of \( X \). 
The probability density function (PDF) is given by: \[ f(x) = 6x(1 - x), \quad 0<x \leq 1 \] To find the CDF \( F(x) \), we integrate the PDF: \[ F(x) = \int_0^x 6t(1 - t) \, dt \] Expanding and integrating: \[ F(x) = \int_0^x (6t - 6t^2) \, dt = 3x^2 - 2x^3 \] Step 3: Solve for \( \beta \) such that \( F(\beta) = 0.5 \). 
We set \( F(\beta) = 0.5 \) to find the value of \( \beta \): \[ 3\beta^2 - 2\beta^3 = 0.5 \] Solving this equation numerically or using a root-finding method, we find: \[ \beta \approx 0.6 \] Step 4: Conclusion. 
Thus, the value of \( \beta \) (rounded to 1 decimal place) is \( \boxed{0.6} \).