document.write( "Question 979122: The number of sick days due to cold and flu last year was recorded by a sample of 15 adults. The data are
\n" ); document.write( " 5; 7; 0; 3; 15; 6; 5; 9; 3; 8; 10; 5; 2; 0; 12:
\n" ); document.write( "a) Compute the mean, median and mode.
\n" ); document.write( "b) Describe what you have learned from the three statistics calculated in part (a).
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Algebra.Com's Answer #600410 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
00\r
\n" ); document.write( "\n" ); document.write( "2
\n" ); document.write( "33\r
\n" ); document.write( "\n" ); document.write( "555
\n" ); document.write( "6
\n" ); document.write( "7
\n" ); document.write( "8
\n" ); document.write( "9
\n" ); document.write( "10\r
\n" ); document.write( "\n" ); document.write( "12
\n" ); document.write( "15\r
\n" ); document.write( "\n" ); document.write( "Mean is sum divided by 15. It is 90/15=6
\n" ); document.write( "Median is middle number which is 8th from bottom if numbers are ordered. It is 5
\n" ); document.write( "Mode is the most common number. It is 5.\r
\n" ); document.write( "\n" ); document.write( "Mode is generally the least useful, except as a descriptor.
\n" ); document.write( "The fact that the mean is greater than the median means that the distribution has a right skew, and that a larger number is affecting it. Those numbers are 12 and 15.\r
\n" ); document.write( "\n" ); document.write( "The central portion of this distribution is 5 (median) and 6 (mean).
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