Question

2.2.19 Question Help 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 automoble company Data are colected 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.一0.003, represents a steel part that is 0003 inch shorter than the specified length. Complete parts a through c EEB Click the icon to view the data table. a. Construct a frequency distribution. Difference in Length -0.005 but lessthan -0.003 -0.003 but less than-0.001 -0.001 but less than 0.001 0.001 but less than 0.003 0.003 but less than 0.005 Frequency Difference Between Actual and Specified Lengths Full data set 0.002 0.003 0.0005 0.0015 00 0.002 0.001 0004 0.003 0 0.003-0.001 0.0005 0.0035-0.0015 -0.0015 -0.0015 -0.0015-0.001 0.0015 -0.0025 -0.002 0.0.0015 0 0.003 0.002 0.001 0.0020.003 -00030.0025 -0.0005 0 0.004 -0.002 0.0020.001 0 0.0015 0.002 0.0015 0.001 0.0025 0.001 0.003 0.0025 0 Print Done

Answer 1

Arrange the data in ascending data.

Count all the number that lies in the range, greater than -0.005 but less than -0.003 = 1

Similarly do it for the remaining bins.

The solution is shown below