document.write( "Question 1041007: A sample of the hourly wages of employees who work in a restaurant in Siem Reap has the mean of $5.02 and a standard deviation of $0.09. Using the Chebyshev's Theorem, find the range in which at least 75% of the date values fall. \n" ); document.write( "

Algebra.Com's Answer #655928 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
Chebyshev's Theorem: $\"P%28abs%28X+-+mu%29+%3C=+k%2Asigma%29+%3E=+1-1%2Fk%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"P%28abs%28X+-+5.02%29+%3C=+0.09k%29+%3E=+1-1%2Fk%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "Letting k = 2,\r
\n" ); document.write( "\n" ); document.write( "==> $\"P%28abs%28X+-+5.02%29+%3C=+0.18%29+%3E=+1-1%2F2%5E2+=+3%2F4\"$
\n" ); document.write( "==> $\"P%28-0.18+%3C=+X+-+5.02+%3C=+0.18%29+%3E=+3%2F4\"$
\n" ); document.write( "==> $\"P%284.84+%3C=+X+%3C=+5.20%29+%3E=+3%2F4+=+0.75\"$ \r
\n" ); document.write( "\n" ); document.write( "==> the range in which at least 75% of the values fall is the interval
\n" ); document.write( "[4.84, 5.20]. \n" ); document.write( "

\n" );