A population of values has a normal distribution with mean 18.6 and standard deviation 57, sample size is \(\displaystyle{n}={25}\). The value of x is 17.5. The z-score is,

\(\displaystyle{z}={\frac{{{x}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\)

\(\displaystyle={\frac{{{17.5}-{18.6}}}{{{\frac{{{57}}}{{\sqrt{{{25}}}}}}}}}\)

\(\displaystyle={\frac{{-{1.1}}}{{{11.4}}}}=-{0.0965}\)

The area to the right of \(\displaystyle{z}=−{0.0965}\) under the standard normal curve is \(\displaystyle{P}{\left({z}{>}−{0.0965}\right)}={1}−{P}{\left({z}{<}−{0.0965}\right)}\).

The probability of z less than –0.0965 can be obtained using the excel formula “=NORM.S.DIST(–0.0965,TRUE)”. The probability value is 0.4616.

The probability that a sample of size \(\displaystyle{n}={25}\) is randomly selected with a mean greater than 17.5 is,

\(\displaystyle{P}{\left({z}\succ{0.0965}\right)}={1}-{P}{\left({z}{<}-{0.0965}\right)}\)

\(\displaystyle={1}-{0.4616}={0.5384}\)

Thus, the probability that a sample of size \(\displaystyle{n}={25}\) is randomly selected with a mean greater than 17.5 is 0.5384.