How to find sample variance given population mean and variance (no specific data)

#1
Hi, I am working on the following problem:confused::

Population Mean = 1000
Population Std. Dev. = 400

Sample N=25
How many, from the sample of 25, are greater than 1000 if the sample mean is 1100.

Can I assume the sample std. dev. is the same as the population?

Does anyone know how I would go about solving this?

Thanks a ton in advance.
 

BGM

TS Contributor
#4
Do you have more information? E.g. any distributional assumption? It may be better for you to type the whole question/whole piece of information you have.

Anyway it seems to ask for the following quantity:

\( E\left[\sum_{i=1}^N \mathbf{1}\{X_i > \mu\}\Bigg|\bar{X} = x\right]
= \sum_{i=1}^N \Pr\{X_i > \mu|\bar{X} = x\} \)

With the limited information given, it seems that at most you can obtain a bound for this like the Chebyshev Inequality.