3. Given that \[ f(x)=\left\{\begin{array}{cc}1 \sqrt{x}, & 0

Ask by Turner Pope. in the United States

Feb 20,2025

To solve the problem, we need to evaluate the constant \( k \) and then find the probability \( P(0.3 < X < 0.6) \) using the given density function. ### Part a: Evaluate \( k \) The function \( f(x) \) is defined as: \[ f(x) = \begin{cases} k \sqrt{x}, & 0 < x < 1 \\ 0, & \text{elsewhere} \end{cases} \] To find \( k \), we need to ensure that the total area under the probability density function (PDF) equals 1. This means we need to solve the following integral: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Since \( f(x) = 0 \) outside the interval \( (0, 1) \), we can simplify the integral to: \[ \int_0^1 k \sqrt{x} \, dx = 1 \] Now, we will calculate the integral: \[ \int_0^1 \sqrt{x} \, dx \] The integral of \( \sqrt{x} \) can be computed as follows: \[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \] Evaluating this from 0 to 1: \[ \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3} \] Now substituting back into the equation for \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] ### Part b: Evaluate \( P(0.3 < X < 0.6) \) Now that we have \( k = \frac{3}{2} \), we can find the probability \( P(0.3 < X < 0.6) \): \[ P(0.3 < X < 0.6) = \int_{0.3}^{0.6} f(x) \, dx = \int_{0.3}^{0.6} \frac{3}{2} \sqrt{x} \, dx \] Calculating the integral: \[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \] Now we evaluate this from 0.3 to 0.6: \[ \int_{0.3}^{0.6} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0.3}^{0.6} \] Calculating the values: 1. For \( x = 0.6 \): \[ \frac{2}{3} (0.6)^{3/2} = \frac{2}{3} \cdot 0.6^{1.5} = \frac{2}{3} \cdot 0.4641 \approx 0.3081 \] 2. For \( x = 0.3 \): \[ \frac{2}{3} (0.3)^{3/2} = \frac{2}{3} \cdot 0.3^{1.5} = \frac{2}{3} \cdot 0.1732 \approx 0.1155 \] Now, substituting these values back into the integral: \[ \int_{0.3}^{0.6} \sqrt{x} \, dx \approx 0.3081 - 0.1155 = 0.1926 \] Finally, multiplying by \( \frac{3}{2} \): \[ P(0.3 < X < 0.6) = \frac{3}{2} \cdot 0.1926 \approx 0.289 \] ### Summary of Results a) \( k = \frac{3}{2} \) b) \( P(0.3 < X < 0.6) \approx 0.289 \)