Question

The probability density function is given by \( f(x) = \begin{cases} C(x-1), & \text{for } 1<x<4 \\ 0, & \text{otherwise} \end{cases} \). Find \(P(2<X<3)\).

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{2}{9} \)
• D. \( \frac{2}{2} \) (This is 1, likely typo)

Hint:

For a continuous probability density function (PDF) \(f(x)\), \(\int_{-\infty}^{\infty} f(x)dx = 1\).
To find \(P(a<X<b)\), calculate \(\int_a^b f(x)dx\).
First, normalize the PDF by finding the constant C.

Answer 1

The Correct Option is A

First, we need to find the constant C using the property that the total probability is 1: \[ \int_{-\infty}^{\infty} f(x) dx = 1 \] \[ \int_{1}^{4} C(x-1) dx = 1 \] \[ C \left[ \frac{x^2}{2} - x \right]_{1}^{4} = 1 \] \[ C \left[ \left(\frac{4^2}{2} - 4\right) - \left(\frac{1^2}{2} - 1\right) \right] = 1 \] \[ C \left[ \left(\frac{16}{2} - 4\right) - \left(\frac{1}{2} - 1\right) \right] = 1 \] \[ C \left[ (8 - 4) - (-\frac{1}{2}) \right] = 1 \] \[ C \left[ 4 + \frac{1}{2} \right] = 1 \] \[ C \left[ \frac{8+1}{2} \right] = C \left[ \frac{9}{2} \right] = 1 \] \[ C = \frac{2}{9} \] So, the probability density function is \( f(x) = \begin{cases} \frac{2}{9}(x-1), & \text{for } 1<x<4
0, & \text{otherwise} \end{cases} \). Now, we need to find \(P(2<X<3)\): \[ P(2<X<3) = \int_{2}^{3} f(x) dx = \int_{2}^{3} \frac{2}{9}(x-1) dx \] \[ = \frac{2}{9} \left[ \frac{x^2}{2} - x \right]_{2}^{3} \] \[ = \frac{2}{9} \left[ \left(\frac{3^2}{2} - 3\right) - \left(\frac{2^2}{2} - 2\right) \right] \] \[ = \frac{2}{9} \left[ \left(\frac{9}{2} - 3\right) - (2 - 2) \right] \] \[ = \frac{2}{9} \left[ \left(\frac{9-6}{2}\right) - 0 \right] = \frac{2}{9} \left[ \frac{3}{2} \right] \] \[ = \frac{2 \times 3}{9 \times 2} = \frac{6}{18} = \frac{1}{3} \] Thus, \(P(2<X<3) = \frac{1}{3}\). This matches option (a). \[ \boxed{\frac{1}{3}} \]