91Ó°ÊÓ

To find the value of c, we will integrate the probability density function f(x) over its domain [0, 2] and set the result equal to 1, since the integral of a pdf must be 1. So, we have the equation: \( \int_{0}^{2} c(4x-2x^2) dx = 1\) Now, let's evaluate this integral: \( c\int_{0}^{2} (4x-2x^2) dx = c \left[ 2x^2 - \frac{2}{3}x^3 \right]_0^2 = 1\) Now, plug in the limits and we get: \( c\left( 2(2)^2 - \frac{2}{3} (2)^3\right) = c\left(8 - \frac{16}{3}\right) = 1\) Now, solve for c: \( c\left(\frac{8}{3}\right) = 1 \) \(c = \frac{3}{8}\) The value of c is 3/8.

Now that we have the value of c, we can calculate the probability P(1/2 < X < 3/2) by evaluating the definite integral: \(P\left(\frac{1}{2}<X<\frac{3}{2}\right) = \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{3}{8}(4x-2x^2) dx\) Let's evaluate the integral by substituting \(c = \frac{3}{8}\): \( \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{3}{8}(4x-2x^2) dx = \frac{3}{8} \int_{\frac{1}{2}}^{\frac{3}{2}} (4x-2x^2) dx = \frac{3}{8} \left[ 2x^2 - \frac{2}{3}x^3 \right]_{\frac{1}{2}}^{\frac{3}{2}}\) Now, plug in the limits and we get: \( \frac{3}{8} \left[ \left(2(\frac{3}{2})^2 - \frac{2}{3} (\frac{3}{2})^3\right) - \left(2(\frac{1}{2})^2 - \frac{2}{3} (\frac{1}{2})^3\right) \right] = \frac{3}{8}\left(\frac{9}{2} - 9 - 1 + \frac{1}{3}\right)\) \(P\left(\frac{1}{2}<X<\frac{3}{2}\right) = \frac{3}{8}\left(-\frac{1}{2}\right) = - \frac{3}{16}\) However, a probability cannot be negative. This error was caused by a miscalculation in one of the steps: - Correct calculation: \( \frac{3}{8} \left[ \left(2(\frac{3}{2})^2 - \frac{2}{3} (\frac{3}{2})^3\right) - \left(2(\frac{1}{2})^2 - \frac{2}{3} (\frac{1}{2})^3\right) \right] = \frac{3}{8}\left(\frac{9}{2} - 9 + 1 - \frac{1}{3}\right)\) \(P\left(\frac{1}{2}<X<\frac{3}{2}\right) = \frac{3}{8}\left(\frac{1}{2}\right) = \frac{3}{16}\) Thus, the probability P(1/2 < X < 3/2) is 3/16.