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

Can someone point me in the right direction with this question? My textbook doesn't give an example of this type of Q.

Use Chebyshev's Inequality to calculate the minimum possible value of:

Pr(1<=X<=19)

Where the mean is 9 and the variance is 9.

Hi Wevans2303!

Since your interval is not symmetric around the mean, it appears that you need to calculate the one-side versions of Chebyshev's inequality and combine the results.

Did you try to calculate it?
How far did you get?

Originally Posted by ILikeSerena

Hi Wevans2303!

Since your interval is not symmetric around the mean, it appears that you need to calculate the one-side versions of Chebyshev's inequality and combine the results.

Did you try to calculate it?
How far did you get?

I will be totally honest and say that I wouldn't know where to begin. We never really covered this in class and my text book only gives examples where you are given a probability and solving for value of k.

I guess in this example you use the value for the mean and standard deviation and put it into the inequality, but that's where I would get stuck!

Originally Posted by Wevans2303

I will be totally honest and say that I wouldn't know where to begin. We never really covered this in class and my text book only gives examples where you are given a probability and solving for value of k.

I guess in this example you use the value for the mean and standard deviation and put it into the inequality, but that's where I would get stuck!

Perhaps you could start by writing the relevant inequality down?

Originally Posted by ILikeSerena

Perhaps you could start by writing the relevant inequality down?

Never seen the one sided version of the inequality before, is it as follows?

P(X-E[X]>=k[std dev]) <= 1/(1+k^2)

So I use mean = 9 and std dev = 3, substitute them in and work out k, then work out the probabilities and combine them?

I will try to write it down properly tomorrow it's a little too confusing on the computer and I don't know latex.

Originally Posted by Wevans2303

Never seen the one sided version of the inequality before, is it as follows?

P(X-E[X]>=k[std dev]) <= 1/(1+k^2)

So I use mean = 9 and std dev = 3, substitute them in and work out k, then work out the probabilities and combine them?

I will try to write it down properly tomorrow it's a little too confusing on the computer and I don't know latex.

That would work yes, and it will give you a minimum.

You should also use the two sided version of the inequality, but calculate Pr(-1<=X<=19) instead of Pr(1<=X<=19).
Since it has a bigger chance, that means it will give you another minimum, which may be sharper.

Originally Posted by ILikeSerena

That would work yes, and it will give you a minimum.

You should also use the two sided version of the inequality, but calculate Pr(-1<=X<=19) instead of Pr(1<=X<=19).
Since it has a bigger chance, that means it will give you another minimum, which may be sharper.

Okay I will do both in the morning and post my answers. It doesn't look too bad now.

Thank you!

Right okay here is what I got for Pr(-1<=x<=19)

Pr(|X-9|>=3.3[std dev]) <= 1/(3.3^2)

Need to do 1- 1/(3.3^2)

=0.91??

Where k = 3 and one third? as -1 and 19 are 10 away from the mean which is 3.3333 standard devations?

I will try doing the one tailed versions now.

And what about:

Originally Posted by ILikeSerena

And what about:

Again not too sure but I make k=8/3 and putting that into inequality I get:

P(|X-9|>=8/3[std dev])<=1/(8/3)^2

=0.14

so 1-0.14 = 0.86?

Originally Posted by Wevans2303

Right okay here is what I got for Pr(-1<=x<=19)

Pr(|X-9|>=3.3[std dev]) <= 1/(3.3^2)

Need to do 1- 1/(3.3^2)

=0.91??

Where k = 3 and one third? as -1 and 19 are 10 away from the mean which is 3.3333 standard devations?

I will try doing the one tailed versions now.

Originally Posted by Wevans2303

Again not too sure but I make k=8/3 and putting that into inequality I get:

P(|X-9|>=8/3[std dev])<=1/(8/3)^2

=0.14

so 1-0.14 = 0.86?

Yep.

Originally Posted by ILikeSerena

Yep.

So how do I go about getting a final answer to the question? For the one tailed version it's the same process?

So what do you have for the combination of the 2 one-tailed versions?

Originally Posted by ILikeSerena

So what do you have for the combination of the 2 one-tailed versions?

I will do now.

Originally Posted by Wevans2303

I will do now.

Right okay I get two values of 0.05 and 0.07 for the two one tailed inequalities.

0.07+0.05 = 0.12.

1-0.12=0.88.

? :S