Question

Question help the average income in a certain region in 2013 was ​$62 comma 00062,000 per person per year. suppose the standard deviation is ​$27 comma 00027,000 and the distribution is​ right-skewed. suppose we take a random sample of 100100 residents of the region.
a. is the sample size large enough to use the central limit theorem for​ means? explain.
b. what are the mean and standard error of the sampling​ distribution?
c. what is the probability that the sample mean will be more than ​$2 comma 7002,700 away from the population​ mean?

Answer 1

Part a
The sample size is large enough to use the central limit theorem because it is greater than 30.

Part b.
The mean of the sampling distribution is 62000.
The standard error is computed as
     $SE=\frac{s}{\sqrt{n}}=\frac{27000}{\sqrt{100}}=2700$

Part c.
Compute for the corresponding z-score of more than 2700 from the population mean. 
     $z=\frac{2700}{2700}=1$

Therefore,
     $P\left(z\ \textgreater \ 1\right)=0.1587$