document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1172799>Question 1172799</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Delta International delivers approximately one mil-<BR>\n" );
document.write( "lion packages a day between East Asia and the United<BR>\n" );
document.write( "States. A random sample of the daily number of pack-<BR>\n" );
document.write( "age delivery failures over the past six months pro-<BR>\n" );
document.write( "vided the following results: 15, 10, 8, 16, 12, 11, 9, 8,<BR>\n" );
document.write( "12, 9, 10, 8, 7, 16, 14, 12, 10, 9, 8, 11.<BR>\n" );
document.write( "unusual about the operations during these days and,<BR>\n" );
document.write( "thus, the results can be considered typical. Using these<BR>\n" );
document.write( "data and your understanding of the delivery process<BR>\n" );
document.write( "answer the following:<BR>\n" );
document.write( "a. What probability model should be used and why?<BR>\n" );
document.write( "b. What is the probability of 10 or more failed deliv-<BR>\n" );
document.write( "eries on a typical future day?<BR>\n" );
document.write( "c. What is the probability of less than 6 failed deliveries?<BR>\n" );
document.write( "d. Find the number of failures such that the probabil-<BR>\n" );
document.write( "ity of exceeding this number is 10% or less. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850801 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=850801\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's analyze this problem step-by-step.\r<BR>\n" );
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document.write( "**a. Probability Model**\r<BR>\n" );
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document.write( "* **Poisson Distribution:**<BR>\n" );
document.write( "    * The Poisson distribution is the appropriate probability model here.<BR>\n" );
document.write( "    * **Why?**<BR>\n" );
document.write( "        * We are dealing with the number of occurrences (delivery failures) within a fixed interval (a day).<BR>\n" );
document.write( "        * The events (failures) are assumed to be independent.<BR>\n" );
document.write( "        * The average rate of failures is relatively small compared to the total number of deliveries (one million packages).<BR>\n" );
document.write( "        * The events occur randomly.\r<BR>\n" );
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document.write( "**b. Probability of 10 or More Failed Deliveries**\r<BR>\n" );
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document.write( "1.  **Calculate the Mean (λ):**<BR>\n" );
document.write( "    * Sum the data: 15 + 10 + 8 + 16 + 12 + 11 + 9 + 8 + 12 + 9 + 10 + 8 + 7 + 16 + 14 + 12 + 10 + 9 + 8 + 11 = 215<BR>\n" );
document.write( "    * Divide by the number of data points (20): 215 / 20 = 10.75<BR>\n" );
document.write( "    * λ = 10.75\r<BR>\n" );
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document.write( "2.  **Poisson Probability Formula:**<BR>\n" );
document.write( "    * P(X = k) = (e^(-λ) * λ^k) / k!<BR>\n" );
document.write( "    * Where:<BR>\n" );
document.write( "        * X = number of failures<BR>\n" );
document.write( "        * k = specific number of failures<BR>\n" );
document.write( "        * λ = mean (10.75)<BR>\n" );
document.write( "        * e ≈ 2.71828\r<BR>\n" );
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document.write( "3.  **Calculate P(X ≥ 10):**<BR>\n" );
document.write( "    * P(X ≥ 10) = 1 - P(X < 10)<BR>\n" );
document.write( "    * P(X < 10) = P(X = 0) + P(X = 1) + ... + P(X = 9)<BR>\n" );
document.write( "    * It is easier to use a calculator or statistical software to calculate this.<BR>\n" );
document.write( "    * Using a calculator or software, we find that P(X<10) = 0.4633.<BR>\n" );
document.write( "    * P(X ≥ 10) = 1-0.4633 = 0.5367.<BR>\n" );
document.write( "    * Therefore, the probability of 10 or more failed deliveries is approximately 0.5367.\r<BR>\n" );
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document.write( "**c. Probability of Less Than 6 Failed Deliveries**\r<BR>\n" );
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document.write( "1.  **Calculate P(X < 6):**<BR>\n" );
document.write( "    * P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)<BR>\n" );
document.write( "    * Again, using a calculator or statistical software:<BR>\n" );
document.write( "    * P(X < 6) = 0.0579.<BR>\n" );
document.write( "    * Therefore, the probability of less than 6 failed deliveries is approximately 0.0579.\r<BR>\n" );
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document.write( "**d. Number of Failures with 10% or Less Probability of Exceeding**\r<BR>\n" );
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document.write( "1.  **Find k such that P(X > k) ≤ 0.10:**<BR>\n" );
document.write( "    * This is equivalent to finding k such that P(X ≤ k) ≥ 0.90.<BR>\n" );
document.write( "    * We need to use a Poisson cumulative distribution table or a calculator/software to find the value of k.<BR>\n" );
document.write( "    * Using statistical software, we find that when k=15, P(X<=15) = 0.9079. When k=14, P(X<=14)=0.8494.<BR>\n" );
document.write( "    * Therefore the number of failures is 15.\r<BR>\n" );
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document.write( "**Summary:**\r<BR>\n" );
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document.write( "* **a.** Poisson distribution<BR>\n" );
document.write( "* **b.** Approximately 0.5367<BR>\n" );
document.write( "* **c.** Approximately 0.0579<BR>\n" );
document.write( "* **d.** 15 failures<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );