**7. Correct. The answer is true.
**The mean of the sampling
distribution of means would be identical to the population mean, i.e. 1% for
both cases. The standard deviation for the case with replacement would be:

σ /√n = 0.02/√50 = 0.0028

and without replacement

(σ /√n)*( √((N-n)/9N-1)) = (0.02/√50)(( √100-50)/(100-1))
= 0.002