Set A: {7, 3, 4, 9, 2}
Finding the Mean of Set A: \(\bar{x} = \frac{7 + 3 + 4 + 9 + 2}{5}\)
\(\bar{x} = \frac{25}{5}\)
\(\bar{x} = 5\)
Finding the Standard of Set A: \(\sigma = \sqrt{\frac{(\bar{x} - x_{1})^{2} + (\bar{x} - x_{2})^{2} + (\bar{x} - x_{3})^{2} + (\bar{x} - x_{4})^{2} + (\bar{x} - x_{5})^{2}}{n}}\)
\(\sigma = \sqrt{\frac{(5 - 7)^{2} + (5 - 3)^{2} + (5 - 4)^{2} + (5 - 9)^{2} + (5 - 2)^{2}}{5}}\)
\(\sigma = \sqrt{\frac{(-2)^{2} + (2)^{2} + (1)^{2} + (-4)^{2} + (3)^{2}}{5}}\)
\(\sigma = \sqrt{\frac{4 + 4 + 1 + 16 + 9}{5}}\)
\(\sigma = \sqrt{\frac{34}{5}}\)
\(\sigma = \sqrt{6.8}\)
\(\sigma \approx 2.6\)
Finding the Mean of Set B: \(\bar{x} = \frac{5 + 8 + 7 + 6 + 4}{5}\)
\(\bar{x} = \frac{30}{5}\)
\(\bar{x} = 6\)
Finding the Standard Deviation of Set B: \(\sigma = \sqrt{\frac{(\bar{x} - x_{1})^{2} + (bar{x} - x_{2})^{2} + (\bar{x} - x_{3})^{2} + (\bar{x} - x_{4})^{2} + (\bar{x} - x_{5})}{n}}\)
\(\sigma = \sqrt{\frac{(6 - 5)^{2} + (6 - 8)^{2} + (6 - 7)^{2} + (6 - 6)^{2} + (6 - 4)^{2}}{5}}\)
\(\sigma = \sqrt{\frac{(1)^{2} + (-2)^{2} + (-1)^{2} + (0)^{2} + (2)^{2}}{5}}\)
\(\sigma = \sqrt{\frac{1 + 4 + 1 + 0 + 4}{5}}\)
\(\sigma = \sqrt{\frac{10}{2}}\)
\(\sigma = \sqrt{5}\)
\(\sigma \approx 2.236\)
The mean and standard deviation of Sets A and B are different.
