document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161766>Question 1161766</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can you help me solve these.  The instructor says I should be able to put it in my TI-89 calculator but I cannot get it to come out right.\r<BR>\n" );
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document.write( "1.) A manufacturer knows that their items have a normally distributed length, with a mean of 15.9 inches, and standard deviation of 1.2 inches.\r<BR>\n" );
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document.write( "If 17 items are chosen at random, what is the probability that their mean length is less than 15.8 inches? \r<BR>\n" );
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document.write( "2.) A manufacturer knows that their items have a normally distributed length, with a mean of 17.2 inches, and standard deviation of 2.6 inches.\r<BR>\n" );
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document.write( "If 4 items are chosen at random, what is the probability that their mean length is less than 17.2 inches?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #785843 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=785843\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean - 15.9 inches.<BR>\n" );
document.write( "standard deviation is 1.2 inches.<BR>\n" );
document.write( "sample is 17 items.<BR>\n" );
document.write( "standard deviation of distribution of sample means (standard error) is equal to standard deviation / sqrt (sample size) = 1.2 / sqrt(17) = .29104275.<BR>\n" );
document.write( "i have a ti-84 which is similar to ti-89<BR>\n" );
document.write( "you can find it by z-score or you can find it by raw score.<BR>\n" );
document.write( "to find it by z-score, then calculate z-score.<BR>\n" );
document.write( "formula is z = (x-m)/s = (15.8-15.9)/.29104275 = -.3435921355<BR>\n" );
document.write( "find area to the left of that.<BR>\n" );
document.write( "command is normalcdf(-E9,-.3435921355) = .365576582.<BR>\n" );
document.write( "that's your answer.<BR>\n" );
document.write( "alternatively, you can enter the raw score and the mean and the standard error directly as normalcdf(-E9,15.8,15.9,.29104275) = .365576582.<BR>\n" );
document.write( "first method used z-score.<BR>\n" );
document.write( "second method used raw score.<BR>\n" );
document.write( "-E9 means -10^9 which is the area to the left of the z-score (first method), or area to the left of the raw score (second method).<BR>\n" );
document.write( "if you did everything else correctly, i suspect you didn't get the standard error correctly, and possible used the standard deviation instead, but that's only conjecture.\r<BR>\n" );
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document.write( "for your second problem:<BR>\n" );
document.write( "mean is 17.2 and standard deviation is 2.6.<BR>\n" );
document.write( "4 items are chosen at random, therefore sample size is 4.<BR>\n" );
document.write( "probability of mean length should be .5 since,in a normal distribution, the mean is at the midpoint of the distribution.<BR>\n" );
document.write( "still, ........<BR>\n" );
document.write( "standard error = standard deviation divided by sqrt of sample size = 2.6 / sqrt(4) = 1.3<BR>\n" );
document.write( "use the second method to make entries as normalcdf(-E9,17.2,17.2,1.3) = .5<BR>\n" );
document.write( "in this particular case, since you were at the mean, the standard error didn't matter because the numerator in the z-score equation came out as 0.<BR>\n" );
document.write( "for example:<BR>\n" );
document.write( "z = (17.2 - 17.2) / se = 0 / se = 0.<BR>\n" );
document.write( "se stands for standard error which is also called the standard deviation of the distribution of samplemeans.<BR>\n" );
document.write( "the z-score of 0 is the same as the mean.<BR>\n" );
document.write( "the mean in a normal distribution is always at the midpoint of the distribution curve with half of the area to the left and half of the area to the right.<BR>\n" );
document.write( "i also used an online calculator to confirm the results.<BR>\n" );
document.write( "here are the displays from that calculator.\r<BR>\n" );
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document.write( "4 displays.<BR>\n" );
document.write( "2 from first problem and 2 from second problem.<BR>\n" );
document.write( "first of each set of 2 uses z-score.<BR>\n" );
document.write( "second of each set of 2 uses raw score.<BR>\n" );
document.write( "in both cases, the standard error is calculated as standard deviation / square root of sample size.\r<BR>\n" );
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document.write( "these answers should be correct if i understood the problem correctly.<BR>\n" );
document.write( "let me know is this satisfies your inquiry.<BR>\n" );
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