Construct a Weibull probability plot. Observation 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.0 0.2 0.4 0.6 0.8 1.0 In(-In(1-p:)) Observation 1.1 p 1.0 0.9 0.8 0.7 0.6 0.5 0.4 -4 -3 In(-In(1-p.)) Observation 1.1. 1.0 0.9 0.8 0.7 0.6 0.5 0.4 -4-3 In(-In(1-P)) Observation. 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.0 0.2 0.4 Comment on the plot. The plot shows no relationship. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens. The plot shows a strong curved relationship throughout. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens. The plot shows a strong curved relationship throughout. This indicates a Weibull distribution might be a good fit to the population distribution of fracture toughness in concrete specimens. The plot is quite linear. This indicates a Weibull distribution might be a good fit to the population distribution of fracture toughness in concrete specimens. The plot is quite linear. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens. 0.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 1GP
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An article gives arguments for why fracture toughness in concrete specimens should have a Weibull distribution and presents several histograms of data that appear well fit by superimposed Weibull curves. Consider the following sample of size n = 18 observations on
(i - 0.5)
toughness for high-strength concrete (consistent with one of the histograms); values of p; =
are also given.
18
Observation
O
P₁
Observation
Pi
Observation
0.47 0.60 0.65 0.69 0.72
0.74
0.0278 0.0833 0.1389 0.1944 0.25 0.3056
0.78 0.79 0.80 0.81 0.82
0.3611 0.4167 0.4722 0.5278 0.5833
0.86
0.6944
Pi
Construct a Weibull probability plot.
Observation
1.1 p
1.0
0.9
0.89 0.91 0.95 1.01
0.75 0.8056 0.8611 0.9167
0.8
0.7
0.6
0.5
0.4
0.0 0.2 0.4 0.6 0.8 1.0
0.84
0.5833 0.6389
In(-In(1-pi))
Observation
1.1
1.0
0.9
0.8
0.7
0.6
1.05
0.9722
●
·
******
0.5
0.4
-4 -3 -2 -1 0 1 2
In(-In(1 p.))
N
In(-In(1-P))
-3 -2 -1 012
Observation
1.1 p
1.0
0.9
0.8
0.7
0.6
0.5
0.4
-4
......
Observation
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.0 0.2 0.4
Comment on the plot.
The plot shows no relationship. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens.
● The plot shows a strong curved relationship throughout. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens.
The plot shows a strong curved relationship throughout. This indicates a Weibull distribution might be a good fit to the population distribution of fracture toughness in concrete specimens.
The plot is quite linear. This indicates a Weibull distribution might be a good fit to the population distribution of fracture toughness in concrete specimens.
The plot is quite linear. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens.
0.6
0.8
In(-In(1-p:))
1.0
Transcribed Image Text:An article gives arguments for why fracture toughness in concrete specimens should have a Weibull distribution and presents several histograms of data that appear well fit by superimposed Weibull curves. Consider the following sample of size n = 18 observations on (i - 0.5) toughness for high-strength concrete (consistent with one of the histograms); values of p; = are also given. 18 Observation O P₁ Observation Pi Observation 0.47 0.60 0.65 0.69 0.72 0.74 0.0278 0.0833 0.1389 0.1944 0.25 0.3056 0.78 0.79 0.80 0.81 0.82 0.3611 0.4167 0.4722 0.5278 0.5833 0.86 0.6944 Pi Construct a Weibull probability plot. Observation 1.1 p 1.0 0.9 0.89 0.91 0.95 1.01 0.75 0.8056 0.8611 0.9167 0.8 0.7 0.6 0.5 0.4 0.0 0.2 0.4 0.6 0.8 1.0 0.84 0.5833 0.6389 In(-In(1-pi)) Observation 1.1 1.0 0.9 0.8 0.7 0.6 1.05 0.9722 ● · ****** 0.5 0.4 -4 -3 -2 -1 0 1 2 In(-In(1 p.)) N In(-In(1-P)) -3 -2 -1 012 Observation 1.1 p 1.0 0.9 0.8 0.7 0.6 0.5 0.4 -4 ...... Observation 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.0 0.2 0.4 Comment on the plot. The plot shows no relationship. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens. ● The plot shows a strong curved relationship throughout. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens. The plot shows a strong curved relationship throughout. This indicates a Weibull distribution might be a good fit to the population distribution of fracture toughness in concrete specimens. The plot is quite linear. This indicates a Weibull distribution might be a good fit to the population distribution of fracture toughness in concrete specimens. The plot is quite linear. This indicates a Weibull distribution might not be a good fit to the population distribution of fracture toughness in concrete specimens. 0.6 0.8 In(-In(1-p:)) 1.0
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