Question 964391: A symmetric die is thrown 720 times. Use Chebyshev’s inequality to find
the lower bound for the probability of getting 100 to 140 sixes.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Note that X is binomial with n=720,p=0.17 (approx 1/6) so
E[X] = np = 720*0.17 = 122.4,
σ^2 = Var(X) = np(1−p) = 122.4 * 0.83 = 101.59 , giving
σ = sqrt(101.59) = 10.079285689 approx 10.08
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Chebychev's Inequality is P(|X − u| ≥ kσ ) ≤ 1/k^2 for k > 0, therefore
We want a lower bound on Pr( 100 < X < 140). The complementary event is |X − 120| ≥ 20. We first find an upper bound for Pr(|X − 120| ≥ 20)
from P(|X − u| ≥ kσ ) we have kσ = 20 and k = 20 / σ and
1 / k^2 = σ^2 / 20^2 = 101.59 / 400 = 0.253975 approx 0.25
then Pr(| X − 120| ≥ 20) is ≤ 0.25
Pr( 100 < X < 140) ≥ 1 - 0.25 = 0.75
That is the lower bound given by the Chebyshev Inequality.
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