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Chapter 4

Q. 4.23

In the article “Advances in Oxygen Equivalent Equations for Predicting the Properties of Titanium Welds” (D. Harwig, W. lttiwattana, and H. Castner, The Welding Journal, 2001: 126s-136s), the authors propose an oxygen equivalence equation to predict the strength, ductility, and hardness of welds made from nearly pure titanium. The equation is E = 2C + 3.5N + O, where E is the oxygen equivalence, and C, N, and O are the proportions by weight, in parts per million, of carbon, nitrogen, and oxygen, respectively (a constant term involving iron content has been omitted) . Assume that for a particular grade of commercially pure titanium, the quantities C, N, and O are approximately independent and normally distributed with means μ_{C} = 150, μ_{N} = 200, μ_{O} = 1500, and standard deviations σ_{C} = 30, σ_{N} = 60, σ_{O} = 100. Find the distribution of E. Find P(E > 3000).

Step-by-Step

Verified Solution

Since E is a linear combination of independent normal random variables, its distribution is normal. We must now find the mean and variance of E. Using Equation ( 4.12),

{c_{1}X_{1}+c_{2}X_{2}+…+c_{n}X_{n}} ∼ N(c_{1}\mu_{1}+c_{2}\mu_{2}+…+c_{n}\mu _{n}, c^{2}_{1}\sigma^{2}_{1}+c^{2}_{2}\sigma^{2}_{2}+…+c^{2}_{n}\sigma^{2}_{n})        (4,12)

we compute

μ_{E}  =  2μ_{C}  +  3.5μ_{N}  +  1μ_{O}

= 2(150) + 3.5(200) + 1(1500)
= 2500

\sigma^{2}_{E}  =  2²\sigma^{2}_{C}  +  3.5²\sigma^{2}_{N}  +  1²\sigma^{2}_{O}

= 2²(30²) + 3.5²(60²) + 1²(100²)
= 57,700

We conclude that E ~ N(2500, 57,700).

To find P(E > 3000), we compute the z-score: z = (3000- 2500)/\sqrt{57,700} = 2.08. The area to the right of z = 2.08 under the standard normal curve is 0.0188. So P(E > 3000) = 0.0188.