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 us the automobile company. Data are collected from a sample of 50 steel parts measurement device, and the specified length of the steel part. For example Click the icon to view the data table. Construct a percentage distribution. Difference in Length Percentage - 0.005 but less than - 0.003 6 % - 0.003 but less than - 0.001 26 % - 0.001 but less than 0.001 24 % 0.001 but less than 0.003 38 % 0.003 but less than 0.005 6 % Construct a cumulative percentage distribution. Difference In Length Cumulative Percentage -0.005 % -0 003 % -0 001 % -0.001 % 0.003 % 0.005 %

Explanation / Answer

(a) since total frequency is n= 50(sample size) , so to get the frequency we find the corresponding by the

formula=n*percent

for class -0.005 but less than -0.003, the freqncy=50*6%=50*6/100=3

for the class -0.003 but less than -0.001, freqnency=50*26%=50*26/100=13

and so on

  The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the total for all observations, since allfrequencies will already have been added to the previous total.

Difference in length Percentage frequency -0.005 but less than -0.003 6 3 -0.003 but less than -0.001 26 13 -0.001 but less than 0.001 24 12 0.001 but less than 0.003 38 19 0.003 but less than 0.005 6 3 total 100 50 (b) Difference in length Percentage cumulative perecnetate frequency cumulative frequency -0.005 6 6 3 3 -0.003 26 32 13 16 -0.001 24 56 12 28 0.001 38 94 19 47 0.003 6 100 3 50 0.005 100 50

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