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Statistics / Statistics for Engineers and Scientists 4 / Chapter 2.6 / Problem 30E

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Solved: The oxygen equivalence number of a weld is a

Chapter 2, Problem 30E

(choose chapter or problem)

QUESTION:

The oxygen equivalence number of a weld is a number that can be used to predict properties such as hardness, strength, and ductility. The article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds" (D. Harwig, W. Ittiwattana, and H. Castner, The Welding Journal, $2001:126\ s-136\ s$ ) presents several equations for computing the oxygen equivalence number of a weld. An equation designed to predict the strength of a weld is $X = 1.12C + 2.69N + O − 0.21\ Fe$, where $X$ is the oxygen equivalence, and $C,N,O$ , and $Fe$ are the amounts of carbon, nitrogen, oxygen, and iron, respectively, in weight percent, in the weld. Suppose that for welds of a certain type, $\mu_C=0.0247$, $\mu_N=0.0255$, $\mu_O=0.1668$, $\mu_{Fe}=0.0597$, $\sigma_C=0.01351), \(\sigma_N=0.0194), \(\sigma_O=0.0340), \(\sigma_{Fe}=0.0413) . Furthermore assume that correlations are given by \(\rho_{C,N}=-0.44$, $\rho_{C,O}=-0.58$, $\rho_{C,Fe}=-0.39$, $\rho_{N,O}=-0.32$, $\rho_{N,Fe}=-0.09$, and $\rho_{O,Fe}=-0.35$,.
a. Find $\mu_X$.

b. Find $\text {Cov}(C,N)$, $\text {Cov}(C,O)$, $\text {Cov}(C,Fe)$, $\text {Cov}(N,O)$, $\text {Cov}(N,Fe)$, and $\text {Cov}(O,Fe)$

c. Find $\sigma_X$

Equation Transcription:

$X=1.12C+2.69N+O-0.21\ Fe$

$C,N,O$

$Fe$

${\mu{}}_{C}=0.0247$

${\mu{}}_{N}=0.0255$

${\mu{}}_{O}=0.1668$

${\mu{}}_{Fe}=0.0597$

${\sigma{}}_{C}=0.0131$

${\sigma{}}_{N}=0.0194$

${\sigma{}}_{O}=0.0340$

${\sigma{}}_{Fe}=0.0413$

${\rho{}}_{C,N}=-0.44$

${\rho{}}_{C,O}=0.58$

${\rho{}}_{C,Fe}=0.39$

${\rho{}}_{N,O}=-0.32$

${\rho{}}_{N,Fe}=0.09$

${\rho{}}_{O,Fe}=-0.35$

${\mu{}}_{X}$

$Cov(C,N)$

$Cov(C,O)$

$Cov(C,Fe)$

$Cov(N,O)$

$Cov(N,Fe)$

$Cov(O,Fe)$

${\sigma{}}_{X}$

Text Transcription:

X=1.12C+2.69N+O-0.21 Fe

C,N,O

mu_C=0.0247

mu_N=0.0255

mu_O=0.1668

mu_Fe=0.0597

sigma_C=0.0131

sigma_N=0.0194

sigma_O=0.0340

sigma_Fe=0.0413

rho_C,N=-0.44

rho_C,O=0.58

rho_C,Fe=0.39

rho_N,O=-0.32

rho_N,Fe=0.09

rho_O,Fe=-0.35

mu_X

Cov(C,N)

Cov(C,O)

Cov(C,Fe)

Cov(N,O)

Cov(N,Fe)

Cov(O,Fe)

sigma_X

Questions & Answers

QUESTION:

b. Find $\text {Cov}(C,N)$, $\text {Cov}(C,O)$, $\text {Cov}(C,Fe)$, $\text {Cov}(N,O)$, $\text {Cov}(N,Fe)$, and $\text {Cov}(O,Fe)$

c. Find $\sigma_X$

Equation Transcription:

$X=1.12C+2.69N+O-0.21\ Fe$

$C,N,O$

$Fe$

${\mu{}}_{C}=0.0247$

${\mu{}}_{N}=0.0255$

${\mu{}}_{O}=0.1668$

${\mu{}}_{Fe}=0.0597$

${\sigma{}}_{C}=0.0131$

${\sigma{}}_{N}=0.0194$

${\sigma{}}_{O}=0.0340$

${\sigma{}}_{Fe}=0.0413$

${\rho{}}_{C,N}=-0.44$

${\rho{}}_{C,O}=0.58$

${\rho{}}_{C,Fe}=0.39$

${\rho{}}_{N,O}=-0.32$

${\rho{}}_{N,Fe}=0.09$

${\rho{}}_{O,Fe}=-0.35$

${\mu{}}_{X}$

$Cov(C,N)$

$Cov(C,O)$

$Cov(C,Fe)$

$Cov(N,O)$

$Cov(N,Fe)$

$Cov(O,Fe)$

${\sigma{}}_{X}$

Text Transcription:

X=1.12C+2.69N+O-0.21 Fe

C,N,O

mu_C=0.0247

mu_N=0.0255

mu_O=0.1668

mu_Fe=0.0597

sigma_C=0.0131

sigma_N=0.0194

sigma_O=0.0340

sigma_Fe=0.0413

rho_C,N=-0.44

rho_C,O=0.58

rho_C,Fe=0.39

rho_N,O=-0.32

rho_N,Fe=0.09

rho_O,Fe=-0.35

mu_X

Cov(C,N)

Cov(C,O)

Cov(C,Fe)

Cov(N,O)

Cov(N,Fe)

Cov(O,Fe)

sigma_X

ANSWER:

Step 1 of 4:

Here an equation designed to measure the strength of a weld is given.

The equation is X=1.12C+2.69N+O-0.21Fe, where X is the oxygen equivalence,C stands for amount of carbon,N stands for amount of nitrogen, O stands for amount of oxygen,Fe stands for amount of iron.

Also,the means of the amounts of C,N,O and Fe are given.

${\mu{}}_{C}=0.0247$

${\mu{}}_{N}$ =0.0255