A sample of a population was determined to have a true mean

A sample of a population was determined to have a true mean of 15 and a standard deviation of 3. Using the 68-95-99.7 rule, what is the probability that a sample of size of 9 from the population will have a mean greater than 18?

Solution

We are given that A sample of a population was determined to have a

true mean (mean) = 15

and standard deviation (sigma) = 3

sample size (n) = 9

What is the probability that a sample of size of 9 from the population will have a mean greater than 18?

That is here we have to find the P(Xbar > 18)

And we know that the distribution of Xbar is also Normal with mean = 15

and sd = sigma / sqrt(n)

sd = 3 / sqrt(9) = 1

Now convert Xbar = 18 into z-sore.

z-score = (Xbar - mean) / sd

z = (18 - 15) / 1 = 3

Now we have to find P(Z > 3) = 1 - P(Z <=3)

Because EXCEL always gives us probability of left tail.

syntax :

NORMSDIST(z)

where z is the test statistic value.

P(Z<=3) = 0.9987

P(Z >3) = 1 - 0.9987

P(Z > 3) = 0.0012