1. **Stating the problem:** A coffee machine fills cups with an amount of coffee uniformly distributed between 130 ml and 160 ml. 2. **Understanding the uniform distribution:** If $X$ is uniformly distributed on $[a,b]$, then: - The mean (expected value) is $E(X) = \frac{a+b}{2}$. - The variance is $Var(X) = \frac{(b-a)^2}{12}$. - The probability of any exact value is 0 because it is a continuous distribution. - The 95th percentile (quantile) $x_{0.95}$ satisfies $P(X \leq x_{0.95}) = 0.95$. 3. **Given:** $a=130$, $b=160$. 4. **a) Type of variable:** The amount of coffee is a continuous random variable because it can take any value in the interval $[130,160]$. 5. **b) Mean and standard deviation:** $$E(X) = \frac{130 + 160}{2} = \frac{290}{2} = 145$$ $$Var(X) = \frac{(160 - 130)^2}{12} = \frac{30^2}{12} = \frac{900}{12} = 75$$ $$\sigma = \sqrt{75} \approx 8.66$$ 6. **c) Probability of exactly 150 ml:** For a continuous uniform distribution, $P(X = 150) = 0$. 7. **d) Cup size for 95% capacity:** The 95th percentile is: $$x_{0.95} = a + 0.95(b - a) = 130 + 0.95 \times 30 = 130 + 28.5 = 158.5$$ **Final answers:** - a) Continuous random variable - b) Mean = 145 ml, Standard deviation = 8.66 ml - c) Probability = 0 - d) Cup size = 158.5 ml