# Thread: Guidance r.e. Chebyshev's Inequlity.

1. ## Re: Guidance r.e. Chebyshev's Inequlity.

Hmm, I got a different result for the two one tailed inequalities... what did you calculate?

2. ## 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.)

3. ## Re: Guidance r.e. Chebyshev's Inequlity.

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.

4. ## 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) = ?

5. ## Re: Guidance r.e. Chebyshev's Inequlity.

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?

6. ## Re: Guidance r.e. Chebyshev's Inequlity.

That looks much better!

So which inequalities can you write down now for Pr(1 ≤ X ≤ 19)?

7. ## Re: Guidance r.e. Chebyshev's Inequlity.

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?

8. ## Re: Guidance r.e. Chebyshev's Inequlity.

Yes, just write down all the inequalities you have found.

9. ## Re: Guidance r.e. Chebyshev's Inequlity.

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

10. ## 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)?

11. ## Re: Guidance r.e. Chebyshev's Inequlity.

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.

12. ## 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?

13. ## Re: Guidance r.e. Chebyshev's Inequlity.

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?

14. ## 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?

15. ## Re: Guidance r.e. Chebyshev's Inequlity.

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

Page 2 of 3 First 123 Last