<pmalvoisin@[EMAIL PROTECTED]
> wrote in message
news:9ca6e97d-7753-412c-a7b9-badfe8435b4f@[EMAIL PROTECTED]
> 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).
> Can I assume for this problem that the 15 samples will have a mean and
> variance equal to the original sample space?
You can assume that the 15 samples have the mean of the distribution, but
NOT the variance of the distribution.
IF (Y_1, Y_2,...Y_n) represent 'n' samples from a distribution with
parameters:
mean=mu; and
variance = v^2
THEN a sample space (of size 'n') has parameters:
mean = mu
variance =(v^2)/n
> 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
Here it should be: standard deviation (of the sample) = sqrt(60/15)=2.
(Much prettier result, don't you think?)
> From Chebyshev, P( |X=u| < k*v) >= 1-(1/k^2)
Or in this case:
P( |X=u| < k*v/sqrt(15)) >= 1-(1/k^2)
More im****tantly is that you have chosen the correct form of Tchebysheff's
Thm. to obtain a lower bound.
Now all that is needed at this point is to calculate k.
I trust you can take it from here?
> Thanks very much,
> Phil
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