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Math Help - Guidance r.e. Chebyshev's Inequlity.

  1. #16
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Hmm, I got a different result for the two one tailed inequalities... what did you calculate?
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  2. #17
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Which inequalities can you write down now about Pr(1 ≤ X ≤ 19)?
    (When you have the right value for the one-tailed version.)
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  3. #18
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    Hmm, I got a different result for the two one tailed inequalities... what did you calculate?
    I got k=10/3 for 19 and k= 8/3 for 1

    Using P(X-E[X]>=k[std dev])<=1/(1+k^2)

    P(X-9>=10/3[std dev])<=1/1+((10/3)^2) = 0.05

    P(X-9>=8/3[std dev])<=1/1+((10/3)^2)=0.07

    I then added them together and took them from 1, although I don't know if that is the correct procedure.
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  4. #19
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    I think you meant 1/(1+(10/3)^2).
    (Careful with the parentheses!)

    My calculator say 0.0826...


    Correctly written, you would have (with an application of the sum rule for disjoint events):

    P(1 ≤ X ≤ 19) = 1 - P(X < 1 ∨ X > 19) = 1 - (P(X < 1) + P(X > 19))

    with:

    P(X > 19) = P(X - 9 > (10/3)σ ) < 1/(1+(10/3)^2) = ?
    P(X < 1) = P(X - 9 < -(8/3)σ) < 1/(1+(8/3)^2) = ?
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  5. #20
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    I think you meant 1/(1+(10/3)^2).
    (Careful with the parentheses!)

    My calculator say 0.0826...


    Correctly written, you would have (with an application of the sum rule for disjoint events):

    P(1 ≤ X ≤ 19) = 1 - P(X < 1 ∨ X > 19) = 1 - (P(X < 1) + P(X > 19))

    with:

    P(X > 19) = P(X - 9 > (10/3)σ ) < 1/(1+(10/3)^2) = ?
    P(X < 1) = P(X - 9 < -(8/3)σ) < 1/(1+(8/3)^2) = ?
    Ah yes I now get 0.08 as well, I think I did (1+10/3)^2 the first time which is obviously incorrect, silly mistake.

    P(X > 19) = P(X - 9 > (10/3)σ ) < 1/(1+(10/3)^2) = 0.08

    P(X < 1) = P(X - 9 < -(8/3)σ) < 1/(1+(8/3)^2) = 0.12

    P(1 ≤ X ≤ 19) = 1 - P(X < 1 ∨ X > 19) = 1 - (P(X < 1) + P(X > 19))

    1-(0.12+0.08)

    1-0.2

    =0.8

    Perhaps?

    Can I confirm you made 8/3 a negative to account for the change in the inequality sign?
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  6. #21
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    That looks much better!

    So which inequalities can you write down now for Pr(1 ≤ X ≤ 19)?
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  7. #22
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    That looks much better!

    So which inequalities can you write down now for Pr(1 ≤ X ≤ 19)?
    I am not sure what you mean sorry, all of the ones above?
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  8. #23
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Yes, just write down all the inequalities you have found.
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  9. #24
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    Yes, just write down all the inequalities you have found.
    So:

    Pr(|X-9| >= 3.3[std dev]) <= 1/(3.3^2) = 0.91

    P(|X-9| >= 8/3[std dev])<=1/(8/3)^2 = 0.86

    P(X > 19) = P(X - 9 > (10/3)σ ) < 1/(1+(10/3)^2) = 0.08

    P(X < 1) = P(X - 9 < -(8/3)σ) < 1/(1+(8/3)^2) = 0.12
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  10. #25
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Can you rewrite this as inequalities of Pr(-1 ≤ X ≤ 19), Pr(1 ≤ X ≤ 17), and Pr(1 ≤ X ≤ 19)?
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  11. #26
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    Can you rewrite this as inequalities of Pr(-1 ≤ X ≤ 19), Pr(1 ≤ X ≤ 17), and Pr(1 ≤ X ≤ 19)?
    Pr(|X-9| >= 3.3[std dev]) <= 1/(3.3^2) = 0.91 < Pr(-1 ≤ X ≤ 19)

    P(|X-9| >= 8/3[std dev])<=1/(8/3)^2 = 0.86 < Pr(1 ≤ X ≤ 17)

    P(X > 19) = P(X - 9 > (10/3)σ ) < 1/(1+(10/3)^2) = 0.08 < P(X > 19)

    P(X < 1) = P(X - 9 < -(8/3)σ) < 1/(1+(8/3)^2) = 0.12 < P(X < 1)

    ? Sorry if I am being stupid hehe.
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  12. #27
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Okay, so we have:
    Pr(-1 ≤ X ≤ 19) > 0.91
    Pr(1 ≤ X ≤ 17) > 0.86
    Pr(1 ≤ X ≤ 19) > 0.80

    And we want to find the highest minimum for Pr(1 ≤ X ≤ 19).
    Do you see what the highest minimum is that we can claim with certainty?
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  13. #28
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    Okay, so we have:
    Pr(-1 ≤ X ≤ 19) > 0.91
    Pr(1 ≤ X ≤ 17) > 0.86
    Pr(1 ≤ X ≤ 19) > 0.80

    And we want to find the highest minimum for Pr(1 ≤ X ≤ 19).
    Do you see what the highest minimum is that we can claim with certainty?
    0.80?
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  14. #29
    Super Member ILikeSerena's Avatar
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    Re: Guidance r.e. Chebyshev's Inequlity.

    What is the inequality between Pr(1 ≤ X ≤ 17) and Pr(1 ≤ X ≤ 19)?
    Which one is biggest?
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  15. #30
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    Re: Guidance r.e. Chebyshev's Inequlity.

    Quote Originally Posted by ILikeSerena View Post
    What is the inequality between Pr(1 ≤ X ≤ 17) and Pr(1 ≤ X ≤ 19)?
    Which one is biggest?
    I must look so stupid right now... Pr(1 ≤ X ≤ 18)? haha.

    out of the two you state its Pr(1 ≤ X ≤ 17)=0.86
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