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?
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?
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!
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.
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.