SOLUTION: "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 mi

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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)


a. At least 24/25 of all cigarettes of this brand?


b. At least 48/49 of all cigarettes of this brand?


Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let X represent the random variable for the nicotine content of a cigarette.

Then Chebyshev's theorem states that

P%28abs%28X+-+mu%29+%3C=+k%2Asigma%29+=+P%28abs%28X+-+1.52%29+%3C=+0.07k%29%3E=+1-1%2Fk%5E2.

a. Let 1-1%2Fk%5E2+=+24%2F25 ==> k%5E2+=+25 ===> k=5.
===> abs%28X+-+1.52%29+%3C=+0.07%2A5+=+0.35
==> -0.35+%3C=+X+-+1.52+%3C=+0.35 <==> 1.17+%3C=+X+%3C=+1.87.

b. Apply the same procedure as in part (a), but now letting 1-1%2Fk%5E2+=+48%2F49, which gives k+=+7.