document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1176232>Question 1176232</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help please.<BR>\n" );
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document.write( "A machine fills a 3 pound coffee can with a measured amount of coffee. The weight of coffee in each can has a known normal distribution and a standard deviation. By law, the contents of any product must be contained in 90% of the products sold. in the case, 90% of the 3 pound coffee cans must contain at least 3 pounds of coffee. Assuming that the set point of the filling machine can be varied without changing the standard deviation of 2.2 ounces, and that the coffee costs $0.75 per pound at filling time, what is the annual cost of the coffee needed to satisfy the 90% rule? Assume the filling machine operates 24 hours per day for 250 days per year and that filling a can requires 2 seconds.<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 #803125 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=803125\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i understand this problem as follows:\r<BR>\n" );
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document.write( "since the distribution of the pounds of coffee is normal, then you can use the z-score formula to help solve this problem.\r<BR>\n" );
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document.write( "the z-score formula is:\r<BR>\n" );
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document.write( "z = (x - m) / s\r<BR>\n" );
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document.write( "z is the z-score<BR>\n" );
document.write( "x is the raw score<BR>\n" );
document.write( "m is the raw mean<BR>\n" );
document.write( "s is the standard deviation.\r<BR>\n" );
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document.write( "the critical x-score for 10% of thde normal distribution to be less than it, is found by looking up an area of .10 in the z-score table, or using a normal distribution calculator to find the same z-score.\r<BR>\n" );
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document.write( "i used the z-score calculator from the ti-84 plus scientific calculator.\r<BR>\n" );
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document.write( "i looked for the inverse z-score for 10% of the area under the normal distribution curve to be equal to 10%.<BR>\n" );
document.write( "the calculator told me that the z-score that has 10% of the area under the normal distribution curve to the left of it was equal to -1.281551567.\r<BR>\n" );
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document.write( "that was my critical z-score.\r<BR>\n" );
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document.write( "the raw score associated with that z-score was 3 pounds.\r<BR>\n" );
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document.write( "the standard deviation was given as 2.2 ounces which is equal to 2.2 / 16 = .175 pounds.\r<BR>\n" );
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document.write( "i used the z-score formula to find the mean.\r<BR>\n" );
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document.write( "the formula of z = (x - m) / s became:\r<BR>\n" );
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document.write( "-1.281551567 = (3 - m) / .1375.\r<BR>\n" );
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document.write( "i solved for m to get:\r<BR>\n" );
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document.write( "m = 3 - .1375 * -1.281551567 = 3 + .1762133404 = 3.1762133404.\r<BR>\n" );
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document.write( "that is the average pounds of coffee required per can so that 90% of the cans will have at least 3 pounds in them.\r<BR>\n" );
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document.write( "the mean is the set point.\r<BR>\n" );
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document.write( "the annual cost of the coffee is calculated as follows.\r<BR>\n" );
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document.write( "the filling machine operates continuously for 24 hours a day for 250 days each year.\r<BR>\n" );
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document.write( "this means that it is operating continuously for 24 * 250 = 6,000 hours each year.\r<BR>\n" );
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document.write( "since there are 60 * 60 = 3600 seconds each hour, this means that it is operating continuously for 6000 * 3600 = 21,600,000 seconds each year.\r<BR>\n" );
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document.write( "since it takes 2 seconds to fill a can, then it is filling 21,600,000 / 2 = 10,800,000 cans each year.\r<BR>\n" );
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document.write( "since each can requires an average of  3.17621334 pounds of coffee, then it is requiring 10,800,000 * 3.17621334  = 34,303,104.08 pounds of coffee each year.\r<BR>\n" );
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document.write( "since the coffee costs .75 per pound, then the total cost of the coffee each year is equal to .75 * 34,303,104.08 = 25,727,328.06 each year.\r<BR>\n" );
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document.write( "an annual cost of 25,727,318.06 each year is your solution.\r<BR>\n" );
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document.write( "graphically, the probability looks loke this.\r<BR>\n" );
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