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.
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?
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.
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?