Answer:
I am not sure if you're accustomed to using a z-score and table of probabilities or technology, so I will answer using both!
a) Since you're choosing one person from the population, you will use the mean and standard deviation from the population. z= (210.9-197.5)/8.6 = 1.56 Using a table of normal probabilities,
P(x>210.9)= 1-P(z<1.56)= 1- 0.9406=0.0594
Using technology: normalcdf(210.9, 1E99, 197.5, 8.6)=0.0596
b) Since you're calculating the population for a sample to occur, you need to get a mean of the sampling distribution and standard deviation of the sampling distribution.
Meanpop=Meansampling dist.=197.5
St. Devsampling dist.= Stdpopulation/√n = 8.6/√15 = 2.22
Using technology, P(X>196.20)= normalcdf( 196.20, 1E99, 197.5, 2.22) = 0.7209
c) You only have to worry about the sample size in a sampling distribution if you either do not know the shape of the distribution of the population or if the distribution of the population is not normally distributed. Since this population is normally distributed (as stated in the given information), then your sample size does not have to be greater than 30.
Explanation: