document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197769>Question 1197769</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.62 and a standard deviation of 0.4 Using the empirical rule, what percentage of the students have grade point averages that are at least 1.82?  Please do not round your answer. </SPAN>\n" );
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document.write( "mu = 2.62 = mean<BR>\n" );
document.write( "sigma = 0.4 = standard deviation\r<BR>\n" );
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document.write( "Let's find the z score when x = 1.82<BR>\n" );
document.write( "z = (x - mu)/sigma<BR>\n" );
document.write( "z = (1.82 - 2.62)/0.4<BR>\n" );
document.write( "z = -2<BR>\n" );
document.write( "This score is exactly 2 standard deviations below the mean.\r<BR>\n" );
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document.write( "The task of computing P(X > 1.82) is equivalent to finding P(Z > -2).\r<BR>\n" );
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document.write( "The empirical rule says roughly 95% of the normal distribution is within 2 standard deviations of the mean.\r<BR>\n" );
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document.write( "That means there is roughly 2.5% of the distribution in the left tail since (100%-95%)/2 = 2.5%<BR>\n" );
document.write( "The remaining portion is then 100% - 2.5% = 97.5%\r<BR>\n" );
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document.write( "Or basically there's 95% in the middle and 2.5% in the right tail, so 95% + 2.5% = 97.5%<BR>\n" );
document.write( "and P(Z > -2) = 0.975\r<BR>\n" );
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document.write( "Check out this diagram below<BR>\n" );
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document.write( "Answer: <font color=red>Approximately 97.5%</font><BR>\n" );
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