# 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 standard deviation σ = 2.5. An opinion poll asks...

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

standard deviation *σ* = 2.5. An opinion poll asks this question of an SRS of 1100 adults. We will see in the next chapter that in this situation the sample mean response has approximately the Normal distribution with mean 9 and standard deviation 0.075. What is *P*(8 ≤ "line above x" ≤ 10), the probability that the statistic estimates the parameter to within ± 1?

"line above x"= mean

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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).