1. **Problem statement:** We have a random variable $T$ representing time, and a score $P = a - bT$ where $a,b > 0$. Given $E(P) = 100$, maximum score $= 150$, $E(T) = 4.723$, and $\mathrm{Var}(T) = 0.906$, find $a$, $b$ (nearest integers), and $\mathrm{Var}(P)$. 2. **Formulas and rules:** - Expectation linearity: $E(P) = E(a - bT) = a - bE(T)$. - Maximum score occurs when $T$ is minimum; since $P = a - bT$, max $P = a - b \times \min(T)$. - Variance scaling: $\mathrm{Var}(P) = \mathrm{Var}(a - bT) = b^2 \mathrm{Var}(T)$ because variance of a constant is zero. 3. **Find $a$ and $b$ using $E(P)$ and max score:** Given max score $= 150$, and since $P = a - bT$, max score occurs at minimum $T$. From the piecewise function, $T$ ranges from $2.25$ to $7.5$, so minimum $T = 2.25$. Set max score: $$150 = a - b \times 2.25$$ From expectation: $$100 = a - b \times 4.723$$ 4. **Solve the system:** Subtract second from first: $$150 - 100 = (a - 2.25b) - (a - 4.723b)$$ $$50 = -2.25b + 4.723b = 2.473b$$ So, $$b = \frac{50}{2.473} \approx 20.21$$ Plug $b$ back to find $a$: $$100 = a - 20.21 \times 4.723$$ $$a = 100 + 20.21 \times 4.723 = 100 + 95.44 = 195.44$$ Rounded to nearest integers: $$a = 195, \quad b = 20$$ 5. **Find $\mathrm{Var}(P)$:** $$\mathrm{Var}(P) = b^2 \mathrm{Var}(T) = 20^2 \times 0.906 = 400 \times 0.906 = 362.4$$ Rounded: $$\mathrm{Var}(P) = 362$$ **Final answers:** $$a = 195, \quad b = 20, \quad \mathrm{Var}(P) = 362$$