Im am given a list of integers and asked to give the interval that contains at least 90 % of my values.
Values : $62,56,72,83,66,77,62,71,50,58, 74,81,76,67,70,70,69,67,80,81, 74,53,73,55,66,88,73,61,63,70, 72,63,75,68,78,75,61,69,80,82, 87,57,74,74,85,68,75,63,81,73$
At First, I found the following values :
$ avg = 70.56 $
$ variance = 80.55 $
$ standard deviation = 8.97 $
Now the part I am not sure. Chebychev inequality is $1 - \frac{1}{k^2}$
k * std var. being how far from the mean the numbers are.
I figured that with a k of 4, i get $15/16$ which is about $93 %$
Is that ok or do I have to figure out some fraction of k to get exactly 90 % that is asked ? How do I figure out precisely that fraction ?
Using a k of 4, I multiplied my std var by 4, and that gives me a minimum of 34.68 and maximum of 106.44
Looking at my numbers, the min and max seems a bit off...
So for now, my answer is : the interval is $[34.68,106.44]$ covers at least 93.75 %
Hint: Chebyshev's Inequality says $$ \mathrm{P}\left(|X-\mu|\ge k\sigma\right)\le\frac1{k^2} $$
If we set $\frac1{k^2}=0.10$, that is, $k\doteq3.162$, then at most $10\%$ of the samples would be outside of $3.162$ standard deviations from the mean.