1. **State the problem:** We have 5 trials of a model car's distance traveled: 3, 6, 7, 5, 6 feet. 2. The students want to remove Trial 1 (value 3) as practice and analyze how the standard deviation changes. 3. **Recall the formula for standard deviation:** $$\sigma = \sqrt{\frac{1}{n}\sum_{i=1}^n (x_i - \mu)^2}$$ where $n$ is the number of data points and $\mu$ is the mean. 4. **Calculate the mean and standard deviation with all 5 trials:** - Mean: $$\mu = \frac{3 + 6 + 7 + 5 + 6}{5} = \frac{27}{5} = 5.4$$ - Variance: $$\frac{(3-5.4)^2 + (6-5.4)^2 + (7-5.4)^2 + (5-5.4)^2 + (6-5.4)^2}{5}$$ $$= \frac{(2.4)^2 + (0.6)^2 + (1.6)^2 + (0.4)^2 + (0.6)^2}{5}$$ $$= \frac{5.76 + 0.36 + 2.56 + 0.16 + 0.36}{5} = \frac{9.2}{5} = 1.84$$ - Standard deviation: $$\sigma = \sqrt{1.84} \approx 1.356$$ 5. **Calculate the mean and standard deviation after removing Trial 1 (value 3):** - New data: 6, 7, 5, 6 - New mean: $$\mu = \frac{6 + 7 + 5 + 6}{4} = \frac{24}{4} = 6$$ - Variance: $$\frac{(6-6)^2 + (7-6)^2 + (5-6)^2 + (6-6)^2}{4}$$ $$= \frac{0 + 1 + 1 + 0}{4} = \frac{2}{4} = 0.5$$ - Standard deviation: $$\sigma = \sqrt{0.5} \approx 0.707$$ 6. **Compare the standard deviations:** - Original standard deviation: approximately 1.356 - New standard deviation: approximately 0.707 7. **Conclusion:** The standard deviation decreases when the Trial 1 value (3) is removed because the value 3 was farther from the mean and increased the spread of the data. **Final answer:** The standard deviation of the data will **decrease** if the Trial 1 value is removed. This is because the removed value was farther from the mean, reducing the overall spread of the data.