document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168184>Question 1168184</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For a statistics assignment, a student observed the cars entering the school staff parking lot during a 1-hour period. The results are shown:<BR>\n" );
document.write( "40 cars had only a driver<BR>\n" );
document.write( "28 cars had the driver and one passenger<BR>\n" );
document.write( "10 cars had the driver and two passengers<BR>\n" );
document.write( "2 cars had the driver and three passengers\r<BR>\n" );
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document.write( "a) Compute the mean, median, and mode number of people per car.<BR>\n" );
document.write( "b) The student presented her statistics to the staff. Write a sentence that is appropriate to use for each measure of central tendency. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #792812 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=792812\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1-40<BR>\n" );
document.write( "2-28<BR>\n" );
document.write( "3-10<BR>\n" );
document.write( "4-2<BR>\n" );
document.write( "the mean is (1)(40)+2(28)+3(10)+2(4)=134 in 80 cars for a mean of 1.675 people per car<BR>\n" );
document.write( "the variance is 0.675^2*40+0.325^2*28 +1.325^2*10+2.325^2*2 all divided by 80=18.225+2.9575+17.55625+10.81125/80=0.619 people^2 per car<BR>\n" );
document.write( "SD is sqrt(V)=0.787 people per car\r<BR>\n" );
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document.write( "median is 1.5 per car. There are 80, so the median is half way between the 40th and 41st ordered observation value. 40th is 1 and 41st is 2, so the median is In between.\r<BR>\n" );
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document.write( "The mode is 1 person per car\r<BR>\n" );
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document.write( "This is a right skewed distribution. The median is often the best measure of central tendency to use in these instances, although the mean is not too much different. The mean is more sensitive to extreme values, but there are not too many extreme ones here. The mode is a descriptor and here gives a sense of where a significant portion of the data were.\r<BR>\n" );
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document.write( "If normally distributed, mean and median.<BR>\n" );
document.write( "If significantly skewed, median.\r<BR>\n" );
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document.write( "There is no reason not to use both when describing data. If they are close, there is a decent estimation of central tendency.  If they are not, the difference will show that the distribution is not normal and likely skewed. Remember that the mean can be a very unlikely occurrence with bimodal distributions, so knowing the sd is also useful. </SPAN>\n" );
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