document.write( "Question 1183149: \"A study of the nicotine contents of a certain brand of cigarette shows that on the average, one cigarette contains 1.52 milligrams of nicotine with a standard deviation of 0.07 milligrams\". Between what values must the nicotine content be for: (Hint: Use Chebyshev’s Theorem)
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\n" ); document.write( "a. At least 24/25 of all cigarettes of this brand?
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\n" ); document.write( "b. At least 48/49 of all cigarettes of this brand?
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Algebra.Com's Answer #813363 by robertb(5830)\"\" \"About 
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Let \"X\" represent the random variable for the nicotine content of a cigarette.\r
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\n" ); document.write( "\n" ); document.write( "Then Chebyshev's theorem states that\r
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\n" ); document.write( "\n" ); document.write( "\"P%28abs%28X+-+mu%29+%3C=+k%2Asigma%29+=+P%28abs%28X+-+1.52%29+%3C=+0.07k%29%3E=+1-1%2Fk%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "a. Let \"1-1%2Fk%5E2+=+24%2F25\" ==> \"k%5E2+=+25\" ===> \"k=5\".\r
\n" ); document.write( "\n" ); document.write( "===> \"abs%28X+-+1.52%29+%3C=+0.07%2A5+=+0.35\"\r
\n" ); document.write( "\n" ); document.write( "==> \"-0.35+%3C=+X+-+1.52+%3C=+0.35\" <==> \"1.17+%3C=+X+%3C=+1.87\".\r
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\n" ); document.write( "\n" ); document.write( "b. Apply the same procedure as in part (a), but now letting \"1-1%2Fk%5E2+=+48%2F49\", which gives \"k+=+7\".
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