1. **Problem Statement:** We have a continuous random variable $X$ representing the total number of hours (in units of 100 hours) a family runs a vacuum cleaner in a year. The probability density function (pdf) is given by: $$ f(x) = \begin{cases} x, & 0 < x < 1 \\ 2 - x, & 1 \leq x < 2 \\ 0, & \text{elsewhere} \end{cases} $$ We want to find: (a) $P(X < 1.2)$ (less than 120 hours) (b) $P(0.5 < X < 1)$ (between 50 and 100 hours) 2. **Formula and Rules:** The probability for a continuous random variable over an interval $[a,b]$ is: $$ P(a \leq X \leq b) = \int_a^b f(x) \, dx $$ Since $f(x)$ is piecewise, we integrate over the relevant intervals. 3. **Calculate (a) $P(X < 1.2)$:** Split the integral at $x=1$ because the pdf changes: $$ P(X < 1.2) = \int_0^1 x \, dx + \int_1^{1.2} (2 - x) \, dx $$ Calculate each: $$ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} $$ $$ \int_1^{1.2} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_1^{1.2} = \left(2(1.2) - \frac{(1.2)^2}{2}\right) - \left(2(1) - \frac{1^2}{2}\right) = (2.4 - 0.72) - (2 - 0.5) = 1.68 - 1.5 = 0.18 $$ Sum: $$ P(X < 1.2) = 0.5 + 0.18 = 0.68 $$ 4. **Calculate (b) $P(0.5 < X < 1)$:** Since $0.5$ and $1$ are in the first interval where $f(x) = x$: $$ P(0.5 < X < 1) = \int_{0.5}^1 x \, dx = \left[ \frac{x^2}{2} \right]_{0.5}^1 = \frac{1}{2} - \frac{(0.5)^2}{2} = 0.5 - 0.125 = 0.375 $$ **Final answers:** - (a) $P(X < 1.2) = 0.68$ - (b) $P(0.5 < X < 1) = 0.375$