1. The problem is to simplify and evaluate the expression: $$\sqrt{\frac{(1.16-1.154)^2+(1.14-1.154)^2+(1.15-1.154)^2+(1.16-1.154)^2+(1.15-1.154)^2+(1.15-1.154)^2+(1.16-1.154)^2+(1.15-1.154)^2+(1.17-1.154)^2+(1.16-1.154)^2}{10 \times (10-1)}}$$ 2. This expression resembles the formula for the sample standard deviation, which is: $$s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n(n-1)}}$$ where $x_i$ are the data points, $\bar{x}$ is the mean, and $n$ is the number of data points. 3. Calculate each difference: - $1.16 - 1.154 = 0.006$ - $1.14 - 1.154 = -0.014$ - $1.15 - 1.154 = -0.004$ - $1.16 - 1.154 = 0.006$ - $1.15 - 1.154 = -0.004$ - $1.15 - 1.154 = -0.004$ - $1.16 - 1.154 = 0.006$ - $1.15 - 1.154 = -0.004$ - $1.17 - 1.154 = 0.016$ - $1.16 - 1.154 = 0.006$ 4. Square each difference: - $(0.006)^2 = 0.000036$ - $(-0.014)^2 = 0.000196$ - $(-0.004)^2 = 0.000016$ - $(0.006)^2 = 0.000036$ - $(-0.004)^2 = 0.000016$ - $(-0.004)^2 = 0.000016$ - $(0.006)^2 = 0.000036$ - $(-0.004)^2 = 0.000016$ - $(0.016)^2 = 0.000256$ - $(0.006)^2 = 0.000036$ 5. Sum all squared differences: $$0.000036 + 0.000196 + 0.000016 + 0.000036 + 0.000016 + 0.000016 + 0.000036 + 0.000016 + 0.000256 + 0.000036 = 0.00066$$ 6. Calculate the denominator: $$10 \times (10 - 1) = 10 \times 9 = 90$$ 7. Divide the sum by the denominator: $$\frac{0.00066}{90} = 0.0000073333...$$ 8. Take the square root: $$\sqrt{0.0000073333} \approx 0.00271$$ 9. Therefore, the value of the expression is approximately $0.00271$. This represents the sample standard deviation of the given data points.