1. **State the problem:** We need to find the sample standard deviation of the mileage (MPG) for the sample of convertibles: 25, 25, 24, 21, 21, 21. 2. **Recall the formula for sample standard deviation:** $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ where $n$ is the sample size, $x_i$ are the data points, and $\bar{x}$ is the sample mean. 3. **Calculate the sample mean $\bar{x}$:** $$\bar{x} = \frac{25 + 25 + 24 + 21 + 21 + 21}{6} = \frac{137}{6} \approx 22.8333$$ 4. **Calculate each squared deviation $(x_i - \bar{x})^2$:** - $(25 - 22.8333)^2 = (2.1667)^2 \approx 4.6944$ - $(25 - 22.8333)^2 = 4.6944$ - $(24 - 22.8333)^2 = (1.1667)^2 \approx 1.3611$ - $(21 - 22.8333)^2 = (-1.8333)^2 \approx 3.3611$ - $(21 - 22.8333)^2 = 3.3611$ - $(21 - 22.8333)^2 = 3.3611$ 5. **Sum the squared deviations:** $$4.6944 + 4.6944 + 1.3611 + 3.3611 + 3.3611 + 3.3611 = 20.8332$$ 6. **Divide by $n-1 = 5$:** $$\frac{20.8332}{5} = 4.16664$$ 7. **Take the square root to find $s$:** $$s = \sqrt{4.16664} \approx 2.0412$$ 8. **Round to at least one decimal place:** $$s \approx 2.0$$ **Final answer:** The sample standard deviation of the mileage for the convertibles is approximately **2.0**.