document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1099250>Question 1099250</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can someone help me? Thank you.<BR>\n" );
document.write( "A production line produces bags of sugar that follow a normal distribution with a mean weight of 1.01kg and a standard deviation of 0.02kg. <BR>\n" );
document.write( "If P(X&#8805;b)=0.9330, then what is the value of b? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #713836 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=713836\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean is 1.01 kg.<BR>\n" );
document.write( "standard deviation is .02 kg.\r<BR>\n" );
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document.write( "you want to find b in the formula p(x >= b) = .9330\r<BR>\n" );
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document.write( "you can use a z-score calculator find p(x >= b) = .9330.\r<BR>\n" );
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document.write( "that will tell you the z-score associated with an area under the normal distribution curve that is to the right of that z-score.\r<BR>\n" );
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document.write( "the area to the right of any z-score is equal to 1 minus the area to the left of that z-score.\r<BR>\n" );
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document.write( "therefore you are looking for an area to the left of the z-score equal to 1 - .9330 = .0670.\r<BR>\n" );
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document.write( "without getting into the details of how i found that just yet, the z-score would be equal to -1.499\r<BR>\n" );
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document.write( "that would be the z-score that had an area to the right of it of .9330, or equivalently, an area to the left of it of .0670.\r<BR>\n" );
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document.write( "the following pictures from an online z-score calculator show that to be true.<BR>\n" );
document.write( "the first picture looks for an area to the right of .9330.<BR>\n" );
document.write( "the second picture looks for an area to the left of .0670\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2017/102901.jpg\" alt=\"$$$\" >\r<BR>\n" );
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document.write( "i used my TI-84 PLUS scientific calculator to find the z-score and i got a z-score of -1.498513067 which can be rounded to -1.499.\r<BR>\n" );
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document.write( "the online z-score distribution calculator can be found at <a  rel=nofollow HREF= \"http://davidmlane.com/hyperstat/z_table.html\" target = \"_blank\">http://davidmlane.com/hyperstat/z_table.html</a>\r<BR>\n" );
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document.write( "the TI-84 PLUS can be found on EBAY at a reasonable cost (used).\r<BR>\n" );
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document.write( "i paid around $40 for mine.\r<BR>\n" );
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document.write( "you can also use z-score tables to find the result manually.\r<BR>\n" );
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document.write( "one such table can be found at <a  rel=nofollow HREF= \"http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf\" target = \"_blank\">http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf</a>\r<BR>\n" );
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document.write( "using this table, i looks for an area under the normal distribution curve to the left of the indicated z-score of .0670.\r<BR>\n" );
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document.write( "the translation was necessary since the table only shows the area to the left of the indicated z-score.\r<BR>\n" );
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document.write( "i found that an area to the left of .0681 gave me a z-score of -1.49 and an area to the left of .0668 gave me a z-score of -1.5.\r<BR>\n" );
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document.write( "a manual interpolation to find the z-score for an area to the left of that z-score of .0670 yielded -1.498 which is very close to the mechanically generated z-score of -1.499.\r<BR>\n" );
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document.write( "the difference has to do with rounding and interpolation techniques.<BR>\n" );
document.write( "the mechanically generated results are more accurate.\r<BR>\n" );
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document.write( "once the z-score is found, then you need to find the raw score correlated to that.\r<BR>\n" );
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document.write( "the formula for z-score is z = (x-m)/s\r<BR>\n" );
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document.write( "x is the raw score.<BR>\n" );
document.write( "z is the z-score.<BR>\n" );
document.write( "m is the mean.<BR>\n" );
document.write( "s is the stnadard deviation.\r<BR>\n" );
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document.write( "you are given the mean is 1.01 kg.<BR>\n" );
document.write( "you are given the standard deviation is .02 kg.<BR>\n" );
document.write( "you have found that the z-score is -1.499\r<BR>\n" );
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document.write( "the formula becomes -1.499 = (x-1.01) / .02\r<BR>\n" );
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document.write( "solve for x to get x = .02 * -1.499 + 1.01.\r<BR>\n" );
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document.write( "this gets you x = .98002 kg which can be rounded to .98 kg\r<BR>\n" );
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document.write( "the online normal distribution calculator shows this to be true.\r<BR>\n" );
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document.write( "using the calculator this way, you simply replace 0 with the actual mean and 1 with the actual standard deviation.\r<BR>\n" );
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document.write( "here's what the online normal distribution calculator shows when you do that.\r<BR>\n" );
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document.write( "any questions about how to do any of this, just send me an email with your questions and i'll explain as best i can.\r<BR>\n" );
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document.write( "your questions asked you the following:\r<BR>\n" );
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document.write( "A production line produces bags of sugar that follow a normal distribution with a mean weight of 1.01kg and a standard deviation of 0.02kg. <BR>\n" );
document.write( "If P(X&#8805;b)=0.9330, then what is the value of b?\r<BR>\n" );
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document.write( "if b is the z-score, then the answer would be b = -1.499.\r<BR>\n" );
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document.write( "if b is the raw score, then the answer would be b = .98 kg.\r<BR>\n" );
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document.write( "i believe they're looking for the raw score, so b = .98 is probably that.\r<BR>\n" );
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