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).
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