# Guidance r.e. Chebyshev's Inequlity.

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• Jan 3rd 2012, 12:44 PM
ILikeSerena
Re: Guidance r.e. Chebyshev's Inequlity.
Hmm, I got a different result for the two one tailed inequalities... what did you calculate?
• Jan 3rd 2012, 12:47 PM
ILikeSerena
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.)
• Jan 3rd 2012, 12:49 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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.
• Jan 3rd 2012, 12:57 PM
ILikeSerena
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) = ?
• Jan 3rd 2012, 01:07 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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?
• Jan 3rd 2012, 01:09 PM
ILikeSerena
Re: Guidance r.e. Chebyshev's Inequlity.
That looks much better!

So which inequalities can you write down now for Pr(1 ≤ X ≤ 19)?
• Jan 3rd 2012, 01:15 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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?
• Jan 3rd 2012, 01:17 PM
ILikeSerena
Re: Guidance r.e. Chebyshev's Inequlity.
Yes, just write down all the inequalities you have found.
• Jan 3rd 2012, 01:40 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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
• Jan 3rd 2012, 01:42 PM
ILikeSerena
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)?
• Jan 3rd 2012, 01:47 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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.
• Jan 3rd 2012, 01:49 PM
ILikeSerena
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?
• Jan 3rd 2012, 01:54 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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?
• Jan 3rd 2012, 01:56 PM
ILikeSerena
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?
• Jan 3rd 2012, 02:05 PM
Wevans2303
Re: Guidance r.e. Chebyshev's Inequlity.
Quote:

Originally Posted by ILikeSerena
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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