Suppose you choose a real number X from the interval [2, 10] with a density function of the form where C is a constant\[f(x) = Cx\]
a. find \(C\): \[ \int_{2}^{10} Cx \, dx = 1 \] \[ C\int_{2}^{10} x \, dx = 1 \] \[ C\left[\frac{x^2}{2}\right]_{2}^{10} = 1 \] \[ C\left(\frac{10^2}{2} - \frac{2^2}{2}\right) = 1 \] \[ C\left(\frac{100}{2} - \frac{4}{2}\right) = 1 \] \[ C(50 - 2) = 1 \] \[ 48C = 1 \]
\[ C = \frac{1}{48} \]
b. Find \(P(E)\), Where \(E = [a, b]\) be a subinterval of \([2, 10]\). \[ P(E) = \int_{a}^{b} \frac{x}{48} \, dx \] \[ P(E) = \frac{1}{48} \int_{a}^{b} x \, dx \] \[ P(E) = \frac{1}{48} \left[\frac{x^2}{2}\right]_{a}^{b} \] \[ P(E) = \frac{1}{48} \left[\frac{b^2}{2} - \frac{a^2}{2}\right] = \frac{1}{48} \left[\frac{{b^2}- {a^2}}{2} \right] = \left[\frac{{b^2}- {a^2}}{96} \right] \]
c. To find \(P(X > 5)\), \(P(X < 7)\), and \(P(X^2 - 12X + 35 > 0)\), we integrate the density function over the appropriate intervals: \[ P(X > 5) = \int_{5}^{10} \frac{x}{48} \, dx \frac{1}{48} \left[\frac{10²}{2} - \frac{5²}{2}\right] = \frac{1}{48} \left[\frac{{100}- {25}}{2} \right] = \left[\frac{75}{96} \right] = = \left[\frac{25}{32} \right] \] \[ P(X < 7) = \int_{2}^{7} \frac{x}{48} \, dx =\frac{1}{48} \left[\frac{7²}{2} - \frac{2²}{2}\right] = \frac{1}{48} \left[\frac{{49}- {4}}{2} \right] = \left[\frac{45}{96} \right] = \left[\frac{15}{32} \right] \] \[ P(X^2 - 12X + 35 > 0) = \frac{1}{48}\int _0^{x^2-12x+35}xdx = \frac{1}{48}\left[\frac{x^2}{2}\right]_0^{x^2-12x+35} =\frac{1}{48}\left[\frac{\left(x^2-12x+35\right)^2}{2}\right] = \left[\frac{\left(x^2-12x+35\right)^2}{96}\right] \]
not sure about the answer for \[ P(X^2 - 12X + 35 > 0) \], but i would assume that this would be true for all values of x that makes this equation positive.