Question # A population of values has a normal distribution with mean 18.6 and standard deviation 57. If a random sample of size n=25 is selected, Find the probability that a sample of size n=25 is randomly selected with a mean greater than 17.5 round your answer to four decimal places. P=?

Random variables
ANSWERED A population of values has a normal distribution with mean 18.6 and standard deviation 57. If a random sample of size $$\displaystyle{n}={25}$$ is selected,
Find the probability that a sample of size $$\displaystyle{n}={25}$$ is randomly selected with a mean greater than 17.5 round your answer to four decimal places.
P=? 2020-11-09

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.