document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1177278>Question 1177278</A>: <I><SPAN STYLE=\"word-break: break-all;\"> At a construction site, the daily requirement of gneiss<BR>\n" );
document.write( "(in metric tons) is a random variable having a gamma distribution with α = 2 and β = 5. If their supplier’s<BR>\n" );
document.write( "daily supply capacity is 25 metric tons, what is the<BR>\n" );
document.write( "probability that this capacity will be inadequate on<BR>\n" );
document.write( "any given day? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850485 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=850485\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's solve this problem step-by-step.\r<BR>\n" );
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document.write( "**Understanding Gamma Distribution**\r<BR>\n" );
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document.write( "* A random variable X has a gamma distribution with parameters α and β if its probability density function (pdf) is:<BR>\n" );
document.write( "    * f(x) = (1 / (β^α * Γ(α))) * x^(α-1) * e^(-x/β) for x > 0<BR>\n" );
document.write( "* In this case, α = 2 and β = 5.<BR>\n" );
document.write( "* The expected value is E(X) = αβ = 2 * 5 = 10.<BR>\n" );
document.write( "* The variance is Var(X) = αβ² = 2 * 5² = 50.\r<BR>\n" );
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document.write( "**The Problem**\r<BR>\n" );
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document.write( "We need to find the probability that the daily requirement (X) exceeds the supplier's capacity (25 metric tons). In other words, we need to find P(X > 25).\r<BR>\n" );
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document.write( "**Calculating P(X > 25)**\r<BR>\n" );
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document.write( "* P(X > 25) = ∫[25, ∞] f(x) dx<BR>\n" );
document.write( "* P(X > 25) = ∫[25, ∞] (1 / (5² * Γ(2))) * x^(2-1) * e^(-x/5) dx<BR>\n" );
document.write( "* P(X > 25) = ∫[25, ∞] (1 / 25) * x * e^(-x/5) dx\r<BR>\n" );
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document.write( "Since Γ(2) = 1! = 1.\r<BR>\n" );
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document.write( "Now, we need to evaluate the integral:\r<BR>\n" );
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document.write( "* P(X > 25) = (1/25) * ∫[25, ∞] x * e^(-x/5) dx\r<BR>\n" );
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document.write( "We'll use integration by parts:\r<BR>\n" );
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document.write( "* Let u = x, dv = e^(-x/5) dx<BR>\n" );
document.write( "* Then du = dx, v = -5e^(-x/5)\r<BR>\n" );
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document.write( "* ∫ x * e^(-x/5) dx = -5xe^(-x/5) - ∫ -5e^(-x/5) dx<BR>\n" );
document.write( "* ∫ x * e^(-x/5) dx = -5xe^(-x/5) + 5 ∫ e^(-x/5) dx<BR>\n" );
document.write( "* ∫ x * e^(-x/5) dx = -5xe^(-x/5) - 25e^(-x/5)\r<BR>\n" );
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document.write( "Now, we evaluate the definite integral:\r<BR>\n" );
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document.write( "* ∫[25, ∞] x * e^(-x/5) dx = [-5xe^(-x/5) - 25e^(-x/5)] from 25 to ∞\r<BR>\n" );
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document.write( "As x approaches infinity, e^(-x/5) approaches 0, so the upper limit is 0.\r<BR>\n" );
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document.write( "* = 0 - [-5(25)e^(-25/5) - 25e^(-25/5)]<BR>\n" );
document.write( "* = 125e^(-5) + 25e^(-5)<BR>\n" );
document.write( "* = 150e^(-5)\r<BR>\n" );
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document.write( "Now, plug this back into the probability equation:\r<BR>\n" );
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document.write( "* P(X > 25) = (1/25) * 150e^(-5)<BR>\n" );
document.write( "* P(X > 25) = 6e^(-5)<BR>\n" );
document.write( "* P(X > 25) ≈ 6 * 0.006737947<BR>\n" );
document.write( "* P(X > 25) ≈ 0.040427682\r<BR>\n" );
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document.write( "**Therefore, the probability that the capacity will be inadequate on any given day is approximately 0.0404.**<BR>\n" );
document.write( " </SPAN>\n" );
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