document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1169639>Question 1169639</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A construction project is at present 30 days away from the scheduled completion date. Depending on the weather condition next month ,the time required for the remaining construction are as follows :\r<BR>\n" );
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document.write( "If weather is good, the time required distribution would be: normal, mean=25           ,Standard deviation = 4\r<BR>\n" );
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document.write( "If weather is Bad, the time required distribution would be: Log-Normal      mean=30,Standard deviation =6 \r<BR>\n" );
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document.write( "The weather next month would be equally likely to be either good or bad. What is the probability that there will be a delay in the completion of the project<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851307 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851307\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down this problem step by step.\r<BR>\n" );
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document.write( "**Understanding the Problem**\r<BR>\n" );
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document.write( "We need to calculate the probability that the construction project will be delayed beyond the scheduled completion date, given two possible weather scenarios and their associated time requirements.\r<BR>\n" );
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document.write( "**Given Information**\r<BR>\n" );
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document.write( "* Current time until scheduled completion: 30 days<BR>\n" );
document.write( "* Weather scenarios:<BR>\n" );
document.write( "    * Good weather: Normal distribution, mean (μ) = 25 days, standard deviation (σ) = 4 days<BR>\n" );
document.write( "    * Bad weather: Log-normal distribution, mean (μ) = 30 days, standard deviation (σ) = 6 days<BR>\n" );
document.write( "* Weather likelihood: Equally likely (50% good, 50% bad)\r<BR>\n" );
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document.write( "**Solution**\r<BR>\n" );
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document.write( "1.  **Good Weather Scenario:**\r<BR>\n" );
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document.write( "    * We want to find the probability that the time required is greater than 30 days (delay).<BR>\n" );
document.write( "    * Since the distribution is normal, we need to calculate the z-score:<BR>\n" );
document.write( "        * z = (x - μ) / σ = (30 - 25) / 4 = 5 / 4 = 1.25<BR>\n" );
document.write( "    * We need to find P(Z > 1.25) using a standard normal distribution table or calculator.<BR>\n" );
document.write( "        * P(Z > 1.25) = 1 - P(Z ≤ 1.25) ≈ 1 - 0.8944 = 0.1056\r<BR>\n" );
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document.write( "2.  **Bad Weather Scenario:**\r<BR>\n" );
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document.write( "    * We want to find the probability that the time required is greater than 30 days.<BR>\n" );
document.write( "    * Since the distribution is log-normal, we need to convert the mean and standard deviation to the parameters of the underlying normal distribution.<BR>\n" );
document.write( "    * Let Y = ln(X), where X follows a log-normal distribution.<BR>\n" );
document.write( "    * E(X) = 30, SD(X) = 6.<BR>\n" );
document.write( "    * Var(X) = 36<BR>\n" );
document.write( "    * We use the following formulas to find the mean (μy) and standard deviation (σy) of Y:<BR>\n" );
document.write( "        * σy^2 = ln(1 + (Var(X) / E(X)^2)) = ln(1 + 36 / 900) = ln(1 + 0.04) = ln(1.04) ≈ 0.03922<BR>\n" );
document.write( "        * σy = √0.03922 ≈ 0.198<BR>\n" );
document.write( "        * μy = ln(E(X)) - 0.5 * σy^2 = ln(30) - 0.5 * 0.03922 ≈ 3.4012 - 0.01961 ≈ 3.3816<BR>\n" );
document.write( "    * Now, we need to find P(X > 30), which is equivalent to P(ln(X) > ln(30)) = P(Y > ln(30)).<BR>\n" );
document.write( "    * ln(30) ≈ 3.4012<BR>\n" );
document.write( "    * We calculate the z-score:<BR>\n" );
document.write( "        * z = (ln(30) - μy) / σy = (3.4012 - 3.3816) / 0.198 ≈ 0.0196 / 0.198 ≈ 0.09899<BR>\n" );
document.write( "    * We need to find P(Z > 0.099) using a standard normal distribution table or calculator.<BR>\n" );
document.write( "        * P(Z > 0.099) = 1 - P(Z ≤ 0.099) ≈ 1 - 0.5393 = 0.4607\r<BR>\n" );
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document.write( "3.  **Combined Probability:**\r<BR>\n" );
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document.write( "    * Since the weather is equally likely, we average the probabilities of delay for each scenario:<BR>\n" );
document.write( "        * P(delay) = (0.5 * 0.1056) + (0.5 * 0.4607) = 0.0528 + 0.23035 = 0.28315\r<BR>\n" );
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document.write( "**Final Answer**\r<BR>\n" );
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document.write( "The probability that there will be a delay in the completion of the project is approximately 0.2832.<BR>\n" );
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