91Ó°ÊÓ

The given probability density function \(f_i(x)\) is defined by a piecewise function: $$ f_{i}(x)=\left\{\begin{array}{ll} c\left(4 x-2 x^{2}\right), & 0<x<2 \\ 0, & \text { otherwise } \end{array}\right. $$ This means that the function has a non-zero value defined by the function \(c(4x-2x^2)\) for the range \(0 < x < 2\), and a value of zero elsewhere.

To find the value of \(c\), we need to use the property that the total probability under the curve of a probability density function must be equal to 1. This is mathematically expressed as: $$ \int_{-\infty}^{\infty} f_i(x) \, dx = 1 $$ Since the function is non-zero only in the interval \((0, 2)\), we can rewrite the integral as: $$ \int_{0}^{2} c(4x-2x^2) \, dx = 1 $$ Now, evaluate the integral and solve for \(c\): $$ c \int_{0}^{2} (4x-2x^2) \, dx = 1 $$ $$ c\left[\int_{0}^{2} 4x \, dx - \int_{0}^{2} 2x^2 \, dx \right] = 1 $$ $$ c\left[ (2x^2\Big|_{0}^{2}) - \frac{2}{3}x^3\Big|_{0}^{2} \right] = 1 $$ $$ c\left[ (2(2)^2) - \frac{2}{3}(2)^3 \right] = 1 $$ $$ c\left[8 - \frac{16}{3}\right] = 1 $$ $$ c \times \frac{8}{3} = 1 $$ By solving for \(c\), we get \(c = \frac{3}{8}\).

Now that we have found the value of \(c\), we can find the probability $$ P\left\{\frac{1}{2}<X<\frac{3}{2}\right\} $$ by integrating the probability density function \(f_i(x)\) over the interval \(\left(\frac{1}{2},\frac{3}{2}\right)\): $$ P\left\{\frac{1}{2}<X<\frac{3}{2}\right\} = \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{3}{8}(4x-2x^2) \, dx $$ Evaluate the integral: $$ \frac{3}{8}\left[\int_{\frac{1}{2}}^{\frac{3}{2}} 4x \, dx - \int_{\frac{1}{2}}^{\frac{3}{2}} 2x^2 \, dx \right] $$ $$ \frac{3}{8}\left[ (2x^2\Big|_{\frac{1}{2}}^{\frac{3}{2}}) - \frac{2}{3}x^3\Big|_{\frac{1}{2}}^{\frac{3}{2}} \right] $$ Calculate the definite integrals: $$ P\left\{\frac{1}{2}<X<\frac{3}{2}\right\} = \frac{3}{8}\left[\left(2\left(\frac{3}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2\right) - \frac{2}{3}\left(\left(\frac{3}{2}\right)^3 - \left(\frac{1}{2}\right)^3\right)\right] $$ Simplify the result: $$ P\left\{\frac{1}{2}<X<\frac{3}{2}\right\} = \frac{3}{8}\left[\left(\frac{9}{2} - \frac{1}{2}\right) - \frac{2}{3}\left(\frac{27}{8} - \frac{1}{8}\right)\right] = \frac{15}{16} $$ So, the probability \(P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}\) is \(\frac{15}{16}\).