Allowable mechanical properties for structural designof metallic aerospace vehicles requires an approvedmethod for statistically analyzing empirical test data.The article “Establishing Mechanical PropertyAllowables for Metals” (J. of Testing and Evaluation,1998: 293–299) used the accompanying data on tensileultimate strength (ksi) as a basis for addressingthe difficulties in developing such a method.122.2 124.2 124.3 125.6 126.3 126.5 126.5 127.2 127.3127.5 127.9 128.6 128.8 129.0 129.2 129.4 129.6 130.2130.4 130.8 131.3 131.4 131.4 131.5 131.6 131.6 131.8131.8 132.3 132.4 132.4 132.5 132.5 132.5 132.5 132.6132.7 132.9 133.0 133.1 133.1 133.1 133.1 133.2 133.2133.2 133.3 133.3 133.5 133.5 133.5 133.8 133.9 134.0134.0 134.0 134.0 134.1 134.2 134.3 134.4 134.4 134.6134.7 134.7 134.7 134.8 134.8 134.8 134.9 134.9 135.2135.2 135.2 135.3 135.3 135.4 135.5 135.5 135.6 135.6135.7 135.8 135.8 135.8 135.8 135.8 135.9 135.9 135.9135.9 136.0 136.0 136.1 136.2 136.2 136.3 136.4 136.4136.6 136.8 136.9 136.9 137.0 137.1 137.2 137.6 137.6137.8 137.8 137.8 137.9 137.9 138.2 138.2 138.3 138.3138.4 138.4 138.4 138.5 138.5 138.6 138.7 138.7 139.0139.1 139.5 139.6 139.8 139.8 140.0 140.0 140.7 140.7140.9 140.9 141.2 141.4 141.5 141.6 142.9 143.4 143.5143.6 143.8 143.8 143.9 144.1 144.5 144.5 147.7 147.7 a. Construct a stem-and-leaf display of the data byfirst deleting (truncating) the tenths digit and thenrepeating each stem value five times (once forleaves 1 and 2, a second time for leaves 3 and 4,etc.). Why is it relatively easy to identify a representativestrength value?b. Construct a histogram using equal-width classeswith the first class having a lower limit of 122 and anupper limit of 124. Then comment on any interestingfeatures of the histogram.
Allowable mechanical properties for structural design
of metallic aerospace vehicles requires an approved
method for statistically analyzing empirical test data.
The article “Establishing Mechanical Property
Allowables for Metals” (J. of Testing and Evaluation,
1998: 293–299) used the accompanying data on tensile
ultimate strength (ksi) as a basis for addressing
the difficulties in developing such a method.
122.2 124.2 124.3 125.6 126.3 126.5 126.5 127.2 127.3
127.5 127.9 128.6 128.8 129.0 129.2 129.4 129.6 130.2
130.4 130.8 131.3 131.4 131.4 131.5 131.6 131.6 131.8
131.8 132.3 132.4 132.4 132.5 132.5 132.5 132.5 132.6
132.7 132.9 133.0 133.1 133.1 133.1 133.1 133.2 133.2
133.2 133.3 133.3 133.5 133.5 133.5 133.8 133.9 134.0
134.0 134.0 134.0 134.1 134.2 134.3 134.4 134.4 134.6
134.7 134.7 134.7 134.8 134.8 134.8 134.9 134.9 135.2
135.2 135.2 135.3 135.3 135.4 135.5 135.5 135.6 135.6
135.7 135.8 135.8 135.8 135.8 135.8 135.9 135.9 135.9
135.9 136.0 136.0 136.1 136.2 136.2 136.3 136.4 136.4
136.6 136.8 136.9 136.9 137.0 137.1 137.2 137.6 137.6
137.8 137.8 137.8 137.9 137.9 138.2 138.2 138.3 138.3
138.4 138.4 138.4 138.5 138.5 138.6 138.7 138.7 139.0
139.1 139.5 139.6 139.8 139.8 140.0 140.0 140.7 140.7
140.9 140.9 141.2 141.4 141.5 141.6 142.9 143.4 143.5
143.6 143.8 143.8 143.9 144.1 144.5 144.5 147.7 147.7
a. Construct a stem-and-leaf display of the data by
first deleting (truncating) the tenths digit and then
repeating each stem value five times (once for
leaves 1 and 2, a second time for leaves 3 and 4,
etc.). Why is it relatively easy to identify a representative
strength value?
b. Construct a histogram using equal-width classes
with the first class having a lower limit of 122 and an
upper limit of 124. Then comment on any interesting
features of the histogram.
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
