How many close friends do you have? Suppose that the number of close friends adults claim to have varies from person to person with mean = 9 and...

Here the random variable X is the number of friends a person has. The populatiom mean is M = 9 persons . The standard deviation of the population is Sigma = 2.5 persons. The ample size n = 1100. Let us call the sample mean x bar. Also we know that a sample mean x bar of a large sample  follows the nomal distrinbution, with mean M and the and standard deviation sigma/sqrt( n). Therefore, z = (x bar - M)/(sigma/(sqrt n) follows the standard  normal distribution with mean 0 and standard deviation 1.

So,  the sample mean, x bar follows normal distribution with mean 9 and standard deviation  2.5/sqrt 1100 , or x bar is N(9 , 2.5/sqrt(1100) ). Or Z = (x bar -9)/(9/sqrt100)  is a standard normal variate with mean 0 and standard deviation 1 or Z = (x bar -9)/(9/sqrt100) is a N(0,1)

Therefore, the probablity  P that the sample mean, x bar lies between 8 an 10 is given in usual notation,

P( 8 <= x bar < = 10) =  P {( 8-9)/(2.5/sqrt 100 <= (x bar - 9)/(2.5/sqrt1100) < (10-9)/(2.5/sqrt100) }

=P{-13.2665 <= Z < = 13.2665 }  = 1. It is a very high range for sample mean (with  sample standard deviation  of 2.5/sqrt 1100 =0.75377836) to be  between 8 and 10. It is almost very very sure that the statistic, the sample mean has the range, (8 < = x bar, the sample mean <= 10).