Question

The p.d.f. of a continuous random variable \(X\) is \(f(x)\).
\[ f(x)= \begin{cases} \dfrac{x^2}{18}, & -3 \le x \le 3 \\ 0, & \text{otherwise} \end{cases} \] Then find \(P(|X|<2)\).

• A. \(1/27\)
• B. \(2/13\)
• C. \(8/27\)
• D. \(4/27\)

Hint:

For modulus probability: \[ |X|<a \Rightarrow -a<X<a \] Then use: \[ P(a<X<b)=\int_a^b f(x)\,dx \]

Answer 1

The Correct Option is C

Step 1: Understand the probability expression  

We need to find: \[ P(|X|<(2) \] We know: \[ |X|<2 \Rightarrow -2 \]
 

Step 2: Use the probability density function

For a continuous random variable, \[ P(a 
 

Step 3: Take constant outside the integral

\[ =\frac{1}{18}\int_{-2}^{2}x^2\,dx \]

 

Step 4: Integrate \(x^2\)

We know: \[ \int x^2\,dx=\frac{x^3}{3} \] So, \[ =\frac{1}{18}\left[\frac{x^3}{3}\right]_{-2}^{2} \]

 

Step 5: Apply limits

Substitute upper and lower limits: \[ =\frac{1}{18} \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right) \] \[ =\frac{1}{18} \left( \frac{8}{3} - \frac{-8}{3} \right) \] \[ =\frac{1}{18} \left( \frac{16}{3} \right) \]

 

Step 6: Simplify

\[ =\frac{16}{54} \] \[ =\frac{8}{27} \]

 

Final Answer:

Therefore, \[ P(|X|<2)=\boxed{\frac{8}{27}} \]