Question

The mean per capita income is 15,65415,654 dollars per annum with a standard deviation of 570570 dollars per annum.

What is the probability that the sample mean would differ from the true mean by less than  6363  dollars if a sample of 315315  persons is randomly selected? Round your answer to four decimal places.

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	15654
std deviation   =σ=	570.000

sample size       =n=	315
std error=σx̅=σ/√n=	32.1159

probability that the sample mean would differ from the true mean by less than  63 :

probability =P(15591<X<15717)=P((15591-15654)/32.116)<Z<(15717-15654)/32.116)=P(-1.96<Z<1.96)=0.975-0.025=0.9500