document.write( "Question 472559: The average value of a distribution is 150. If the standard deviation is 25, what percentage of the distribution lies within 100 and 200?
\n" ); document.write( "
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #324056 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
The distribution is not necessarily normal, so we use Chebyshev's inequality, \r
\n" ); document.write( "\n" ); document.write( " $\"P%28abs%28X+-+mu%29+%3C=+k%2Asigma%29+%3E=+1-1%2Fk%5E2\"$ . Since in this case, k = 2, we get\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"P%28abs%28X+-+150%29+%3C=+50%29+%3E=+1-1%2F4+=+3%2F4\"$ .\r
\n" ); document.write( "\n" ); document.write( "Hence at least 75% of the distribution lies with 100 and 200.\r
\n" ); document.write( "\n" ); document.write( "Note that if the distribution is normal, then by the empirical rule, around 95% would lie within 2 sd's of the mean, or between 100 and 200.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );