Two samples are given. Find each sample's a) standard deviation and b) coefficient of variation. Then decide sample has the higher dispersion, and d) which sample has the higher relative dispersion. A: $8,3,7,6,5$ B: $7,3,9,3,7,9,6$ a) What are the sample standard deviations? \[ \mathrm{s}_{\mathrm{A}}=\square \] (Do not round until the final answer. Then round to the nearest hundredth as needed.) \[ \mathrm{s}_{\mathrm{B}}=\square \] (Do not round until the final answer. Then round to the nearest hundredth as needed.) b) What are the sample coefficients of variation? \[ \mathrm{V}_{\mathrm{A}}=\square \] (Round the final answer to the nearest tenth as needed. Round all intermediate values to the nearest thousandth needed.) \[ \mathrm{V}_{\mathrm{B}}=\square \] (Round the final answer to the nearest tenth as needed. Round all intermediate values to the nearest thousandth needed.)

Step 1 ：Calculate the mean of each sample. For sample A: \(\mu_A = \frac{8+3+7+6+5}{5} = 5.8\). For sample B: \(\mu_B = \frac{7+3+9+3+7+9+6}{7} = 6.29\)

Step 2 ：Calculate the variance of each sample. For sample A: \(\sigma_A^2 = \frac{(8-5.8)^2 + (3-5.8)^2 + (7-5.8)^2 + (6-5.8)^2 + (5-5.8)^2}{5-1} = 3.7\). For sample B: \(\sigma_B^2 = \frac{(7-6.29)^2 + (3-6.29)^2 + (9-6.29)^2 + (3-6.29)^2 + (7-6.29)^2 + (9-6.29)^2 + (6-6.29)^2}{7-1} = 6.24\)

Step 3 ：Calculate the standard deviation of each sample. For sample A: \(s_A = \sqrt{3.7} = 1.92\). For sample B: \(s_B = \sqrt{6.24} = 2.50\)

Step 4 ：Calculate the coefficient of variation for each sample. For sample A: \(V_A = \frac{s_A}{\mu_A} \times 100 = \frac{1.92}{5.8} \times 100 = 33.1\%\). For sample B: \(V_B = \frac{s_B}{\mu_B} \times 100 = \frac{2.50}{6.29} \times 100 = 39.7\%\)

Step 5 ：The sample with the higher dispersion is sample B, since it has a higher standard deviation: \(\boxed{s_B = 2.50}\)

Step 6 ：The sample with the higher relative dispersion is also sample B, since it has a higher coefficient of variation: \(\boxed{V_B = 39.7\%}\)