document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1178063>Question 1178063</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that 2500 customers subscribe to a telephone exchange. There are 80 trunk lines available. Any one customer has the probability of 0.03 of needing a trunk line on a given call. Consider the situation as 2500 trials with probability of “success” p = 0.03. What is the approximate probability that the 2500 customers will “tie up” the 80 trunk lines at any given time?\r<BR>\n" );
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document.write( "Thank you :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850366 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</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=850366\">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 the Problem**\r<BR>\n" );
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document.write( "* **Number of Customers (n):** 2500<BR>\n" );
document.write( "* **Number of Trunk Lines:** 80<BR>\n" );
document.write( "* **Probability of Needing a Trunk Line (p):** 0.03<BR>\n" );
document.write( "* **Goal:** Find the probability that at least 81 customers need a trunk line (since there are only 80 available).\r<BR>\n" );
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document.write( "**Approximation Using Normal Distribution**\r<BR>\n" );
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document.write( "Since n is large and p is relatively small, we can approximate the binomial distribution with a normal distribution.\r<BR>\n" );
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document.write( "1.  **Calculate the Mean (μ) and Variance (σ²) of the Binomial Distribution:**<BR>\n" );
document.write( "    * μ = np = 2500 * 0.03 = 75<BR>\n" );
document.write( "    * σ² = np(1 - p) = 2500 * 0.03 * 0.97 = 72.75<BR>\n" );
document.write( "    * σ = √72.75 ≈ 8.529\r<BR>\n" );
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document.write( "2.  **Continuity Correction:**<BR>\n" );
document.write( "    * We want to find P(X ≥ 81), where X is the number of customers needing a trunk line.<BR>\n" );
document.write( "    * Since we are approximating a discrete distribution with a continuous one, we need to apply a continuity correction.<BR>\n" );
document.write( "    * We will use P(X > 80.5) to approximate P(X ≥ 81).\r<BR>\n" );
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document.write( "3.  **Calculate the Z-score:**<BR>\n" );
document.write( "    * Z = (X - μ) / σ<BR>\n" );
document.write( "    * Z = (80.5 - 75) / 8.529 = 5.5 / 8.529 ≈ 0.6448\r<BR>\n" );
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document.write( "4.  **Find the Probability Using the Standard Normal Table:**<BR>\n" );
document.write( "    * We want to find P(Z > 0.6448).<BR>\n" );
document.write( "    * From the standard normal table, P(Z ≤ 0.6448) ≈ 0.7397.<BR>\n" );
document.write( "    * Therefore, P(Z > 0.6448) = 1 - P(Z ≤ 0.6448) = 1 - 0.7397 ≈ 0.2603.\r<BR>\n" );
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document.write( "**Therefore, the approximate probability that the 2500 customers will \"tie up\" the 80 trunk lines at any given time is approximately 0.2603.**<BR>\n" );
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