document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204331>Question 1204331</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A particular fruit's weights are normally distributed, with a mean of 375 grams and a standard deviation of 21 grams.<BR>\n" );
document.write( "If you pick 13 fruits at random, then 6% of the time, their mean weight will be greater than how many grams?<BR>\n" );
document.write( "Give your answer to the nearest gram. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #840519 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=840519\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the mean is 375 grams.<BR>\n" );
document.write( "the standard devitiion is 21 grams.<BR>\n" );
document.write( "your sample size is 13.<BR>\n" );
document.write( "since you are looking for the mean of a distribution of sample means of size 13, then use the standard error rther than the standard deviation.\r<BR>\n" );
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document.write( "standard error = standard devation / sqrt(sample size) = 21 / sqrt(13) = 5.824352.\r<BR>\n" );
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document.write( "z-score = (x - m) / s = (x -375) / 5.824352.\r<BR>\n" );
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document.write( "you don't know the z-score and you don't know the raw score, but you do know tha the probability of the z-score being greater than a certain value is .06.\r<BR>\n" );
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document.write( "yoou would either look into a table or use a calculator to find the z-score that have .06 probability that other z-scores will be greater than it.\r<BR>\n" );
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document.write( "i used the ti-84 plus to find that the z-score that has .06 of the area under the normal distribution curve greater than it is equal to 1.554774.\r<BR>\n" );
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document.write( "i used the z-score formula to find the raw score.<BR>\n" );
document.write( "z = (x - m) / s becomes 1.554774 = (x - 375) / 5.824352.<BR>\n" );
document.write( "i solved for x to get:<BR>\n" );
document.write( "x = 1.554774 * 5.824352 + 375 = 384.055551.<BR>\n" );
document.write( "that's your solution.\r<BR>\n" );
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document.write( "If you pick 13 fruits at random, then 6% of the time, their mean weight will be greater than 384.055551 grams.<BR>\n" );
document.write( "round to the nearest gram to get 384 grams.\r<BR>\n" );
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document.write( "here's a visual representation of what that looks like on a normal distribution graph.\r<BR>\n" );
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document.write( "when you are looking for a value from an area .....\r<BR>\n" );
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document.write( "when you are looking for an area from a value .....\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2023/101812.jpg\">\r<BR>\n" );
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document.write( "if you are looking for the vlue (x in the formula of z = (x - m) / s, thn you need to put in the standard devation or standard error, whichever is approapriate (in this case the standard error because you are looking for the mean of a sample).\r<BR>\n" );
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document.write( "if you are looking for the z-score (z in the formula of z = (x - m) / s, then you need to put in a 0 form the mean and a 1 for the standard deviation.\r<BR>\n" );
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document.write( "the z-score will be the same whether you are looking for an element of a distribution of the mean of a distribuion of sample elements.\r<BR>\n" );
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document.write( "what makes the raw scores different is whether you used the standard deviation (looking for a single element) or the standard error (looking for the mean of a sample of specified size).\r<BR>\n" );
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document.write( "in this particular problem, you were looking for the mean of a sample of a specified size.<BR>\n" );
document.write( "that's why you are using the standard error, rather than the standard deviation.\r<BR>\n" );
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