The article &ldquo;Stochastic Modeling for Pavement Warranty Cost Estimation&rdquo; (J. of Constr. Engr. and Mgmnt., 2009: 352&ndash;359) proposes the following model for the distribution of Y = time to pavement failure. Let (displaystyle{X}_{{{1}}}) be the time to failure due to rutting, and (displaystyle{X}_{{{2}}}) be the time to failure due to transverse cracking, these two rvs are assumed independent. Then (displaystyle{Y}=min{left({X}_{{{1}}},{X}_{{{2}}}right)}). The probability of failure due to either one of these distress modes is assumed to be an increasing function of time t. After making certain distributional assumptions, the following form of the cdf for each mode is obtained: (displaystylePhi{left[frac{{{a}+{b}{t}}}{{left({c}+{left.{d}{t}right.}+{e}{t}^{{{2}}}right)}^{{frac{{1}}{{2}}}}}right]}) where (Uparrow Phi) is the standard normal cdf. Values of the five parameters a, b, c, d, and e are -25.49, 1.15, 4.45, -1.78, and .171 for cracking and -21.27, .0325, .972, -.00028, and .00022 for rutting. Determine the probability of pavement failure within (displaystyle{t}={5}) years and also (displaystyle{t}={10}) years.

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The article &amp;ldquo;Stochastic Modeling for Pavement Warranty Cost Estimation&amp;rdquo; (J. of Constr. Engr. and Mgmnt., 2009: 352&amp;ndash;359) proposes the following model for the distribution of Y = time to pavement failure. Let (displaystyle{X}_{{{1}}}) be the time to failure due to rutting, and (displaystyle{X}_{{{2}}}) be the time to failure due to transverse cracking, these two rvs are assumed independent. Then (displaystyle{Y}=min{left({X}_{{{1}}},{X}_{{{2}}}right)}). The probability of failure due to either one of these distress modes is assumed to be an increasing function of time t. After making certain distributional assumptions, the following form of the cdf for each mode is obtained: (displaystylePhi{left[frac{{{a}+{b}{t}}}{{left({c}+{left.{d}{t}right.}+{e}{t}^{{{2}}}right)}^{{frac{{1}}{{2}}}}}right]}) where (Uparrow Phi) is the standard normal cdf. Values of the five parameters a, b, c, d, and e are -25.49, 1.15, 4.45, -1.78, and .171 for cracking and -21.27, .0325, .972, -.00028, and .00022 for rutting. Determine the probability of pavement failure within (displaystyle{t}={5}) years and also (displaystyle{t}={10}) years.

Macsen Nixon · Accepted Answer

Step 1 Cracking Φ((a+bt)(c+df+et2))12 a=−25.49 b=1.15 c=4.45 d=−1.78 e=0.171 Determine the corresponding probability using the normal probability table. t=5Φ(a+bt(c+dt+et2)12) =Φ(−25.49+1.15(5)(4.45+(−1.78)(5)+0.171(5)2)12) =Φ(−19.74(−0.175)12) =Underfined t=10Φ(a+bt(c+dt+et2)12) =Φ(−25.49+1.15(10)(4.45+(−1.78)(10)+0.171(10)2)12) =Φ(−7.22) ≈0 Note that the probability is undefined when t=5, because the expression under the square root (1/2) is negative and the square root of a negative number doesn&#039;t exist. Step 2 Cracking Φ(a+bt(c+dt+et2)12) a=−21.27 b=0.0325 c=0.972 d=−0.00028 e=0.00022 Determine the corresponding probability using the normal probability y table. Φ(x) is approximately when x is a value smaller than all z-scores in the table t=5 Φ(a+bt(c+dt+et2)12) =Φ(−21.27+0.0325(5)(0.972+(−0.00028)(5)+0.00022(5)2)12} =Φ(−21.36) ≈0

The article “Stochastic Modeling for Pavement Warranty Cost Estimation” (J. of Constr. Engr. and Mgmnt., 2009: 352–359) proposes the

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