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# Refer to Exercise 30. An equation to predict the ductility ISBN: 9780073401331 38

## Solution for problem 31E Chapter 2.6

Statistics for Engineers and Scientists | 4th Edition

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Problem 31E

Refer to Exercise 30. An equation to predict the ductility of a titanium weld is Y = 7.84C + 11.44N + O - 1.58Fe, where Y is the oxygen equivalence used to predict ductility, and C, N, O. and Fe are the amounts of carbon, nitrogen, oxygen, and iron, respectively, in weight percent, in the weld. Using the means, standard deviations, and correlations presented in Exercise 30, find μY and σY.

Problem 30E

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, 2001:126s-136s) 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, μC = 0.0247, μN = 0.0255, μ0 = 0.1668,  μFe = 0.0597, σC = 0.0131, σN = 0.0194, σO = 0.0340, and σFe = 0.0413.  Furthermore assume that correlations are given by ρC,N = -0.44, ρC,O = 0.58, ρC,Fe = 0.39, ρN,O = -0.32, ρN,Fe = 0.09, and ρO,Fe = -0.35.

a. Find μX.

b. Find Cov(C, N), Cov(C, O), Cov(C, Fe), Cov(N, O), Cov(N, Fe), and Cov(O, Fe).

c. Find σX.

Step-by-Step Solution:

Step 1 of 3:

Here an equation Y=7.84C+11.44N+O-1.58Fe is given which is used to predict the ductility of a titanium weld.

Here C represents the amount of carbon,N denotes the amount of nitrogen, O denotes the amount of oxygen and Fe denotes the amount of iron.

The mean of the amounts of C,N,O,Fe are also given.The values are =0.0247 =0.0255 =0.1668 =0.0597

The standard deviations of the amounts of C,N,O,Fe are =0.0132 =0.0194 =0.0340 =0.0413

Also,correlation coefficients are also given as, =-0.44 =0.58 =0.39 =-0.32 =0.09 =-0.35

Using these given values we have to find the required values.

Step 2 of 3:

Here we have to find We have Y=7.84C+11.44N+O-1.58Fe.

Now we have to find We know that =a +b Thus, becomes, =7.84( )+11.44( )+ -1.58( )

=7.84(0.0247)+11.44(0.0255)+(0.1668)-1.58(0.0597)

=0.1936+0.2917+0.1668-0.0943

=0.6521-0.0943

=0.5578

Thus, =0.5578.

Step 3 of 3

##### ISBN: 9780073401331

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