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=?

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asked 2020-11-08
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=?

Answers (1)

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.

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