Question

The average income in a certain region in 2013 was $68000 per person per year. Suppose the standard deviation is $33000 and the distribution is​ right-skewed. Suppose we take a random sample of 100 residents of the region.

What is the probability that the sample mean will be more than $3300 away from the population​ mean?

Answer 1

for normal distribution z score =(X-μ)/σ
here mean=       μ=	68000
std deviation   =σ=	33000
sample size       =n=	100
std error=σx̅=σ/√n=	3300.0000

  probability that the sample mean will be more than $3300 away from the population​ mean:

probability =P(X>71300)=P(Z>(71300-68000)/3300)=P(Z>1)=1-P(Z<1)=1-0.8413=0.1587