One operation of a mill is to cut pieces of steel into parts that will later be

Question

One operation of a mill is to cut pieces of steel into parts that will later be used as the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts to be within ±0.005 inch of the length specified by the automobile company. Data are collected from a sample of 50 steel parts and are shown in the following table. The measurement reported is the difference in inches between the actual length of the steel part, as measured by a laser measurement device, and the specified length of the steel part. For example, the first value, -0.003, represent a steel part that is 0.003 inch shorter than the specified length. Complete parts a through c. E Click the icon to view the data table a. Construct a frequency distribution. Difference in Length Frequency - 0.005 but less than -0.003 - 0.003 but less than -0.001 -0.001 but less than 0.0011 i Difference Between Actual and Specified Lengths 0.001 but less than 0.003 0.003 but less than 0.0051 Full data set - 0.003 0.0015 0.0015 0.002 0.003 0.001 0.004 -0.002 0 - 0.0015-0.004 0.001-0.0005 0 0.002 - 0.0020.002 0.0005 0.0035-0.0015 -0.0035 -0.0015 - 0.0025 -0.002 0.0025 Construct a percentage distribution. 0.001 0.001 0.001 0.002 -0.003 0.002 -0.0015-0.0015 0 -0.004 -0.002 0.002 -0.001 0 0.004 0.0025 0 Difference in Length -0.005 but less than -0.003 -0.003 but less than -0.001 Percentage 0.0005 0.002 0.0005 0.002 0.0035 0.001 0 1)% 1% Print Done -0.001 but less than 0.001 0.001 but less than 0.003 1% 0% 0.003 but less than 0.005

Explanation / Answer

please revet for any clariifcation required)

difference in length frequency percentage cumulative percentage -0.005 but less than -0.003 3 6.00 6.00 -0.003 but less than -0.001 14 28.00 34.00 -0.001 but less than 0.001 12 24.00 58.00 0.001 but less than 0.003 16 32.00 90.00 0.003 but less than -0.005 5 10.00 100.00

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