Hello Group, I am having trouble applying Chebyshev's inequality to a
certain problem.
Here is the problem:
Let X be the mean of a random sample of size n=15 from a distribution
with mean u=80 and variance v^2=60. (v is lower case sigma). Use
Chebyshev's inequality to find a lower bound for P(75<X<85).
I am okay with the fact that one of the properties of the inequality
are that no more than 1/k^2 of the values are more than k standard
deviations from the mean.
Can I assume for this problem that the 15 samples will have a mean and
variance equal to the original sample space?
If so, here is what I did:
I let the mean, u=80
I let the variance v^2=60
Thus, my standard deviation is sqrt(60)~7.746
From Chebyshev, P( |X=u| < k*v) >= 1-(1/k^2)
I note than 5*(0.6455)~7.746, (my standard deviation)
So:
P( |X-80|<5 )
= P (|X-80|< 0.6455 * v) >= 1 - (1/(0.6455)^2) = -1.4
But my probability can't be negative... so what I am I doing wrong?
Please help!
Thanks very much,
Phil
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