Question

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 1

Answer:

A) μ = 590.45 mm; B) σ² = 0.053 mm

Step-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 = 4723.6

There are 8 data values; this gives us

4723.6/8 = 590.45

To 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 = 0.1225

590.2-590.45 = -0.25; (-0.25)² = 0.0625

590.3-590.45 = -0.15; (-0.15)² = 0.0225

590.4-590.45 = -0.05; (-0.05)² = 0.0025

590.5-590.45 = 0.05; (0.05)² = 0.0025

590.6-590.45 = 0.15; (0.15)² = 0.0225

590.7-590.45 = 0.25; (0.25)² = 0.0625

590.8-590.45 = 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