Question

Let \( X \) be a random variable with the probability density function \[ f(x) = \begin{cases} 4x^k, & 0 < x < 1, \\ x - \dfrac{x^2}{2}, & 1 \leq x < 2, \\ 0, & \text{otherwise}, \end{cases} \] where \( k \) is a positive integer. Then \( P\left( \frac{1}{2} \leq X \leq \frac{3}{2} \right) \textbf{ equals ............} \)

Hint:

When dealing with piecewise probability density functions, split the integral according to the intervals where the function changes.

Answer 1

Correct Answer: 0.85 - 0.9

Step 1: Determine \(k\)

Since \(f(x)\) is a pdf,
\[\int_0^1 4x^k,dx + \int_1^2 \left(x-\frac{x^2}{2}\right)dx = 1.\]

\[\int_0^1 4x^k,dx = \frac{4}{k+1},\qquad \int_1^2 \left(x-\frac{x^2}{2}\right)dx = \frac{1}{3}.\]

Thus,
\[\frac{4}{k+1}+\frac{1}{3}=1 \Rightarrow \frac{4}{k+1}=\frac{2}{3} \Rightarrow k=5.\]

Step 2: Compute \(P\left(\tfrac12<X<\tfrac32\right)\)

\[P!\left(\tfrac12<X<\tfrac32\right) = \int_{1/2}^{1} 4x^5,dx + \int_{1}^{3/2}\left(x-\frac{x^2}{2}\right)dx.\]

\[\int_{1/2}^{1} 4x^5,dx = \frac{2}{3}\left(1-\frac{1}{64}\right) = \frac{21}{32}.\]

\[\int_{1}^{3/2}\left(x-\frac{x^2}{2}\right)dx = \left[\frac{x^2}{2}-\frac{x^3}{6}\right]_1^{3/2} = \frac{11}{48}.\]

\[P\left(\tfrac12<X<\tfrac32\right) = \frac{21}{32}+\frac{11}{48} = \frac{85}{96} \approx 0.885.\]

\[\boxed{P\left(\tfrac12<X<\tfrac32\right)\approx 0.885}\]