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

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- January 3rd 2012, 12:44 PMILikeSerenaRe: Guidance r.e. Chebyshev's Inequlity.
Hmm, I got a different result for the two one tailed inequalities... what did you calculate?

- January 3rd 2012, 12:47 PMILikeSerenaRe: 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.) - January 3rd 2012, 12:49 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.
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. - January 3rd 2012, 12:57 PMILikeSerenaRe: 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) = ? - January 3rd 2012, 01:07 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.
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? - January 3rd 2012, 01:09 PMILikeSerenaRe: Guidance r.e. Chebyshev's Inequlity.
That looks much better!

So which inequalities can you write down now for Pr(1 ≤ X ≤ 19)? - January 3rd 2012, 01:15 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.
- January 3rd 2012, 01:17 PMILikeSerenaRe: Guidance r.e. Chebyshev's Inequlity.
Yes, just write down all the inequalities you have found.

- January 3rd 2012, 01:40 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.
- January 3rd 2012, 01:42 PMILikeSerenaRe: 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)?

- January 3rd 2012, 01:47 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.
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. - January 3rd 2012, 01:49 PMILikeSerenaRe: 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? - January 3rd 2012, 01:54 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.
- January 3rd 2012, 01:56 PMILikeSerenaRe: Guidance r.e. Chebyshev's Inequlity.
What is the inequality between Pr(1 ≤ X ≤ 17) and Pr(1 ≤ X ≤ 19)?

Which one is biggest? - January 3rd 2012, 02:05 PMWevans2303Re: Guidance r.e. Chebyshev's Inequlity.