The lengths of plate glass parts are measured to the nearest tenth of a millimeter. The lengths are uniformly distributed with values at every tenth of a millimeter starting at 590.1, and continuing through 590.8. Determine the mean and variance of the lengths. (a) mean (in tenths of millimeters) Round your answer to two decimal places (e.g. 98.76). (b) variance (in tenths of millimeters2) Round your answer to three decimal places (e.g. 98.765). (a)

Answer:A) μ = 590.45 mm; B) σ² = 0.053 mmStep-by-step explanation:To find the mean, we add together all of the values and divide by the number of data values:590.1+590.2+590.3+590.4+590.5+590.6+590.7+590.8 = 4723.6There are 8 data values; this gives us4723.6/8 = 590.45To find the variance, we subtract each of the data values from the mean.  We then square this difference, and find the mean of the squares:590.1-590.45 = -0.35; (-0.35)² = 0.1225590.2-590.45 = -0.25; (-0.25)² = 0.0625590.3-590.45 = -0.15; (-0.15)² = 0.0225590.4-590.45 = -0.05; (-0.05)² = 0.0025590.5-590.45 = 0.05; (0.05)² = 0.0025590.6-590.45 = 0.15; (0.15)² = 0.0225590.7-590.45 = 0.25; (0.25)² = 0.0625590.8-590.45 = 0.35; (0.35)² = 0.1225(0.1225+0.0625+0.0225+0.0025+0.0025+0.0225+0.0625+0.1225)/8= 0.42/8 = 0.0525 ≈ 0.053