document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1147305>Question 1147305</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have a homework question that I am having trouble with understanding:\r<BR>\n" );
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document.write( "Using the unit normal table, find the proportion under the standard normal curve that lies between the following values.\r<BR>\n" );
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document.write( "(a) the mean and z= 0<BR>\n" );
document.write( "(b) the mean and z= 1.96<BR>\n" );
document.write( "(c) z= -1.50 and z= 1.50<BR>\n" );
document.write( "(d) z= -0.08 and z= -0.50<BR>\n" );
document.write( "(e) z= 1.00 and z= 2.00 \r<BR>\n" );
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document.write( ".........<BR>\n" );
document.write( "If someone could please explain in a step by step way to solve this problem, I have already read the chapters and I am lost. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #768637 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=768637\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If you can, use a table of positive and negative z-values which lists the associated probabilities for a Normal Distribution, if you have a table with only positive z-values, you can subtract the corresponding probability from 1 and get the negative z-score probability<BR>\n" );
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document.write( "For a unit normal table, z = 1 corresponds to a proportion of 0.3413 from the mean, z = 2 corresponds to a proportion of 0.4772 from the mean and z = 0.5 corresponds to a proportion of 0.19.15 from the mean<BR>\n" );
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document.write( "(a) mean is 50% = 0.50 proportion(probability), a z=0 corresponds to 0.50 and 0.50 - 0.50 = 0<BR>\n" );
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document.write( "(b) z = 1.96, then probability(P) = 0.9750, 0.9750 - 0.50 = 0.4750<BR>\n" );
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document.write( "(c) z = -1.50 then P = 0.0668, z = 1.50 then P = 0.9332 and 0.9332 - 0.0668 = 0.8664<BR>\n" );
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document.write( "(d) z = -0.08 then P = 0.4681, z = -0.50 then P = 0.3085 and 0.4681 - 0.3085 = 0.1596<BR>\n" );
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document.write( "(e) z = 1.00 then P = 0.8413, z = 2 then P = 0.9772 and 0.9772 - 0.8413 = 0.1359<BR>\n" );
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document.write( "Note mean of a Normal Distribution means 50% of the values are above the mean and 50% of the values are below the mean<BR>\n" );
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document.write( "Note z-values correspond to standard deviations above(positive) and below(negative) from the mean<BR>\n" );
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document.write( "Note z-value = (X - mean)/standard deviation, where X is the test value, for example, suppose you want to compare someone’s weight(150 pounds) to the “average” of a population, the z-score can tell you where that person’s weight is compared to the average population’s mean weight.<BR>\n" );
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