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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 10.005 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, represents a steel part that is 0.003 inch shorter than the specified length. Complete parts a through c Click the icon to view the data table. a. Construct a frequency distribution Difference in Length -0.005 but less than -0.003 - -0.003but less than -0.001 -0.001but less than 0.001 0.001but less than 0.003 0.003but less than 0.005 Frequency Difference Between Actual and Specified Lengths 0.001 0.002 0.0015 0.0035 0 0.001 -0.0015 -0.003 0.001 0.002 -0.003 0.002 -0.001 0.0005 0.001 0.001 0.0015 -0.003 -0.0025 0.001 0.001 -0.004 -0.001 0.0015 0.001 0.004 -0.002 -0.003 -0.001 0.0015 -0.002 -0.0025 -0.0015 0 0.003 0 0.0025 0-0.0035 0.001 Print Done Full data set o 0 0.002 0.0035 -0.0025 0-0.0015 0-0.0015 -0.001 0.0035

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 10.005 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, represents a steel part that is 0.003 inch shorter than the specified length. Complete parts a through c Click the icon to view the data table. a. Construct a frequency distribution Difference in Length -0.005 but less than -0.003 - -0.003but less than -0.001 -0.001but less than 0.001 0.001but less than 0.003 0.003but less than 0.005 Frequency Difference Between Actual and Specified Lengths 0.001 0.002 0.0015 0.0035 0 0.001 -0.0015 -0.003 0.001 0.002 -0.003 0.002 -0.001 0.0005 0.001 0.001 0.0015 -0.003 -0.0025 0.001 0.001 -0.004 -0.001 0.0015 0.001 0.004 -0.002 -0.003 -0.001 0.0015 -0.002 -0.0025 -0.0015 0 0.003 0 0.0025 0-0.0035 0.001 Print Done Full data set o 0 0.002 0.0035 -0.0025 0-0.0015 0-0.0015 -0.001 0.0035

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter1: Expressions And Functions
Section1.5: Descriptive Modeling And Accuracy
Problem 4CYU
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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 10.005 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, represents a steel part that is 0.003 inch shorter than the specified length. Complete parts a through c Click the icon to view the data table. a. Construct a frequency distribution Difference in Length -0.005 but less than -0.003: -0.003but less than -0.001 -0.001but less than 0.001 0.001but less than 0.003 0.003but less than 0.005 Frequency Difference Between Actual and Specified Lengths 0.002 0 -0.003 -0.0025 0.001 0.0015 0.0035 -0.003 0.001 -0.0015 0.001 0.001 0.001 0.002 -0.003 0 0.002 -0.001 -0.004 -0.001 0.0015 0.001 0.0005 0.001 0.0015 Print 0.001 0.004 -0.002 -0.003 -0.001 0.0015 -0.002 -0.0025 -0.0015 0.003 0 0.0025 0-0.0035 0.001 Done Full data set o 0 0.002 0.0035 -0.0025 0-0.0015 0 -0.0015 -0.001 0.0035
Transcribed Image Text: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 10.005 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, represents a steel part that is 0.003 inch shorter than the specified length. Complete parts a through c Click the icon to view the data table. a. Construct a frequency distribution Difference in Length -0.005 but less than -0.003: -0.003but less than -0.001 -0.001but less than 0.001 0.001but less than 0.003 0.003but less than 0.005 Frequency Difference Between Actual and Specified Lengths 0.002 0 -0.003 -0.0025 0.001 0.0015 0.0035 -0.003 0.001 -0.0015 0.001 0.001 0.001 0.002 -0.003 0 0.002 -0.001 -0.004 -0.001 0.0015 0.001 0.0005 0.001 0.0015 Print 0.001 0.004 -0.002 -0.003 -0.001 0.0015 -0.002 -0.0025 -0.0015 0.003 0 0.0025 0-0.0035 0.001 Done Full data set o 0 0.002 0.0035 -0.0025 0-0.0015 0 -0.0015 -0.001 0.0035
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