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OK, so here's the question (not my homework - which is line segments -_-)

According to Chebyshev's theorem, how many standard deviations from the mean would make up the central 60% of scores for this class? [What are the corresponding grades? Answer the same questions for central 80%. Do these values capture more than the desired amount? Does this agree with Chebyshev's theorem?]

(The stuff in brackets doesn't need to be answered. I included it there just in case.)

Thanks in advance.

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So $P(|X-\mu| \geq k \sigma) \leq \frac{1}{k^2}$. The central $60 \%$ is $1-P(|X-\mu| \leq k \sigma) = 0.4$.

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This is the one that says the probability of being outside $k$ standard deviations is less than $\frac{1}{k^2}$. So $1-0.6=\frac{1}{k^2}$. Solve for $k$. Then change .6 to .8.

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Starting with $0.6=1-1/k^2,$ then $1/k^s=1-0.6,$ then $1/k^2=0.4,$ then $1=0.4k^2,$ then $1/0.4=k^2,$ then $2.5=k^2,$ then you square root each side and you will find $k$.

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    Welcome to MSE! It helps readability to $f$ormat questions/answers using MathJax (See FAQ). Regards2013-04-23